# ActuaryProof: full free solution corpus Verified actuarial exam solutions. Every answer earns its proof. This file contains the complete independently written free solution layer. Official question wording is withheld. Read each question in the linked official source before using its reference solution. Canonical index: https://actuaryproof.inshihub.com/llms.txt Verification method: https://actuaryproof.inshihub.com/methodology/ ## SOA Exam P Sample Question #577 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-577/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Sums of Independent Random Variables; Coefficient of Variation The independent components have combined mean 250 and combined standard deviation sqrt(40^2+30^2)=50. Dividing 50 by 250 gives coefficient of variation 0.20, so choice A is correct. ### Independent solution ### Setup Add the two component means to obtain the expected combined loss. LaTeX: E[S]=100+150=250 ### Model Independence makes the component variances additive. LaTeX: \operatorname{Var}(S)=40^2+30^2=2500 LaTeX: \operatorname{SD}(S)=50 ### Compute Form the dimensionless ratio of standard deviation to mean. LaTeX: \operatorname{CV}(S)=\frac{\operatorname{SD}(S)}{E[S]} LaTeX: \operatorname{CV}(S)=\frac{50}{250}=0.20 ### Answer The combined coefficient of variation is 0.20. LaTeX: \boxed{0.20\quad\text{(A)}} ### Distractor analysis - B: This adds the component standard deviations to get 70 and divides by 250, instead of adding variances. - C: This averages the two individual coefficients of variation, 0.40 and 0.20, without weighting the independent aggregate correctly. - D: This reports the first component's coefficient of variation, 40/100, rather than that of the sum. - E: This adds the two component coefficients of variation, even though coefficients of variation are not additive. ### Verification record - Main value: 0.2 - Main method: Add the independent component means and variances, then divide the aggregate standard deviation by the aggregate mean. - Independent check: Independence gives E[S^2]=250^2+2500=65000; therefore CV^2=E[S^2]/E[S]^2-1=0.04 and CV=0.20. - Independent value: 0.2 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: unknown component spread Two independent laboratory costs A and B have means 40 and 60. The standard deviation of A is 6, while the coefficient of variation of A+B is 0.10. Calculate the standard deviation of B. - A: 4 - B: 6 - C: 8 - D: 10 - E: 16 The combined mean is 100, so CV 0.10 implies combined standard deviation 10 and variance 100. Independence gives Var(B)=100-36=64, hence SD(B)=8 and choice C is correct. #### Variant solution ### Setup Use the combined mean and coefficient of variation to recover the total standard deviation. LaTeX: E[A+B]=40+60=100 LaTeX: \operatorname{SD}(A+B)=0.10(100)=10 ### Model Independent component variances add to the total variance. LaTeX: 10^2=6^2+\operatorname{Var}(B) ### Compute Isolate the unknown variance and take its square root. LaTeX: \operatorname{Var}(B)=100-36=64 LaTeX: \operatorname{SD}(B)=8 ### Answer The unknown component standard deviation is 8. LaTeX: \boxed{8\quad\text{(C)}} ### Verification record - Main value: 8 - Main method: Convert the aggregate coefficient of variation to aggregate variance, subtract the known independent variance, and take the square root. - Independent check: Substituting SD(B)=8 gives aggregate standard deviation sqrt(6^2+8^2)=10 and aggregate CV 10/(40+60)=0.10. - Independent value: 8 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #578 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-578/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Normal Distribution; Sums of Independent Random Variables The independent profit components sum to a normal variable with mean 600 and standard deviation sqrt(400^2+300^2)=500. Standardizing zero gives -1.20, so the requested probability is P(Z>-1.20) and choice B is correct. ### Independent solution ### Setup Add the component means to obtain the center of overall profit. LaTeX: E[S]=200+400=600 ### Model The sum remains normal, and independence makes the variances additive. LaTeX: \operatorname{Var}(S)=400^2+300^2=250000 LaTeX: S\sim N(600,500^2) ### Compute Standardize the zero-profit boundary. LaTeX: \Pr(S>0)=\Pr\!\left(Z>\frac{0-600}{500}\right) LaTeX: \Pr(S>0)=\Pr(Z>-1.20)=0.8849303298\ldots ### Answer The standardized expression matches choice B. LaTeX: \boxed{\Pr(Z>-1.20)\quad\text{(B)}} ### Distractor analysis - A: This uses an aggregate spread that is too small, producing an excessively large standardized mean magnitude. - C: This adds the component standard deviations to get 700, yielding 600/700 about 0.86 rather than using the root-sum-of-squares spread. - D: The displayed z-score 0.83 implies an aggregate standard deviation near 600/0.83=723, which is even larger than the unsupported arithmetic sum 400+300=700. - E: This standardizes the first profit component alone, using 200/400=0.50 and omitting the second component. ### Verification record - Main value: 0.8849303297782918 - Main method: Add the independent normal means and variances, standardize zero, and evaluate the standard-normal upper tail. - Independent check: The sum is N(600,250000); direct evaluation of its CDF at zero gives 0.1150696702, whose complement is 0.8849303298. - Independent value: 0.8849303297782918 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: inferred reading correlation Two jointly normal calibration readings A and B have means 10 and 6 and standard deviations 3 and 4, respectively. The probability that A-B is positive is 0.8413447461. Calculate the correlation between A and B. - A: -0.375 - B: 0.000 - C: 0.375 - D: 0.750 - E: 1.000 The difference has mean 4, and the stated probability is Phi(1), so its standard deviation must be 4. Equating Var(A-B)=25-24rho to 16 gives rho=0.375, so choice C is correct. #### Variant solution ### Setup Let D be the difference and express its mean and variance in terms of the unknown correlation. LaTeX: D=A-B LaTeX: E[D]=10-6=4 LaTeX: \operatorname{Var}(D)=3^2+4^2-2\rho(3)(4)=25-24\rho ### Model The stated positive probability corresponds to a standard-normal score of one. LaTeX: \Pr(D>0)=\Phi\!\left(\frac{4}{\operatorname{SD}(D)}\right)=\Phi(1) LaTeX: \operatorname{SD}(D)=4 ### Compute Set the variance expression equal to 16 and solve for the correlation. LaTeX: 25-24\rho=16 LaTeX: \rho=\frac9{24}=0.375 ### Answer The readings have correlation 0.375. LaTeX: \boxed{\rho=0.375\quad\text{(C)}} ### Verification record - Main value: 0.375 - Main method: Infer the difference variance from its positive probability, then solve the correlated difference-variance formula for rho. - Independent check: At rho=0.375, Var(A-B)=9+16-2(0.375)(12)=16, so A-B is N(4,16) and P(A-B>0)=Phi(1)=0.8413447461. - Independent value: 0.375 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #579 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-579/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Conditional Probability; Survival Functions Integrating the density gives survival S(t)=e^(-t)(1+t). Conditional on lasting beyond one year, the probability of ending by year two is 1-S(2)/S(1)=1-3/(2e)=0.448181, so choice C is correct. ### Independent solution ### Setup Integrate the density tail to obtain the survival function. LaTeX: S(t)=\int_t^\infty ue^{-u}\,du=e^{-t}(1+t) ### Model Express the conditional interval probability using survival at the two endpoints. LaTeX: \Pr(T\le2\mid T\ge1)=\frac{S(1)-S(2)}{S(1)} ### Compute Evaluate and simplify the survival ratio. LaTeX: S(1)=2e^{-1},\qquad S(2)=3e^{-2} LaTeX: \Pr(T\le2\mid T\ge1)=1-\frac{3e^{-2}}{2e^{-1}} LaTeX: =1-\frac{3}{2e}=0.4481808382\ldots ### Answer The conditional probability rounds to 0.448. LaTeX: \boxed{0.448\quad\text{(C)}} ### Distractor analysis - A: This comes from an incorrect antiderivative or normalization of the density and is not a valid ratio of the two relevant probabilities. - B: This is the unconditional probability of lasting between one and two years, S(1)-S(2), without division by P(T>=1). - D: This is P(T>=1)=2/e, the probability of the conditioning event rather than the conditional interval probability. - E: This uses F(2)/S(1), putting the unconditional event T<=2 in the numerator without intersecting it with the conditioning event T>=1. ### Verification record - Main value: 0.4481808382428365 - Main method: Derive the survival function of the gamma-shaped lifetime and divide the probability mass between one and two by survival through one. - Independent check: Direct numerical integration gives integral_1^2 t exp(-t) dt=0.3297530326 and integral_1^infinity t exp(-t) dt=0.7357588823; their ratio is 0.4481808382. - Independent value: 0.4481808382428365 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: remaining filter service A two-stage filter has total service time T with a gamma distribution of shape 2 and rate 0.5 per hour. The filter is still operating after 4 hours. Calculate its expected additional operating time. - A: 2.000 hours - B: 2.667 hours - C: 4.000 hours - D: 6.667 hours - E: 8.000 hours For a shape-2 gamma lifetime with rate 0.5, survival is e^(-0.5t)(1+0.5t). Integrating the conditional survival beyond hour 4 gives mean residual life 8/3=2.6667 hours, so choice B is correct. #### Variant solution ### Setup Write the survival function for the shape-two gamma lifetime. LaTeX: S(t)=e^{-0.5t}(1+0.5t) LaTeX: S(4)=3e^{-2} ### Model The expected remaining life after age four is the future survival area divided by survival to age four. LaTeX: E[T-4\mid T>4]=\frac{\int_4^\infty S(t)\,dt}{S(4)} ### Compute Evaluate the survival integral and simplify the ratio. LaTeX: \int_4^\infty e^{-0.5t}(1+0.5t)\,dt=8e^{-2} LaTeX: E[T-4\mid T>4]=\frac{8e^{-2}}{3e^{-2}}=\frac83 ### Answer The expected additional operating time is approximately 2.667 hours. LaTeX: \boxed{2.667\text{ hours}\quad\text{(B)}} ### Verification record - Main value: 2.6666666666666665 - Main method: Integrate the gamma survival function beyond age four and normalize by the probability of surviving to age four. - Independent check: Directly integrating (t-4)(0.25t exp(-0.5t)) from 4 to infinity gives numerator 8exp(-2); division by P(T>4)=3exp(-2) yields 8/3. - Independent value: 2.6666666666666665 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #580 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-580/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Linear Combinations of Independent Random Variables; Variance Each observation has variance 9. Independence makes the variance of the linear combination equal to 3^2(9)+9+9+9=108, so choice E is correct. ### Independent solution ### Setup Convert the common standard deviation into variance. LaTeX: \operatorname{Var}(X_i)=3^2=9 ### Model For independent observations, square every coefficient and add the resulting variance contributions. LaTeX: \operatorname{Var}(3X_1-X_2-X_3-X_4) LaTeX: =3^2\operatorname{Var}(X_1)+\sum_{i=2}^{4}(-1)^2\operatorname{Var}(X_i) ### Compute Substitute the common variance. LaTeX: \operatorname{Var}(3X_1-X_2-X_3-X_4)=9(9)+9+9+9=108 ### Answer The variance is 108. LaTeX: \boxed{108\quad\text{(E)}} ### Distractor analysis - A: This multiplies the first variance by the coefficient 3 rather than its square and also omits the remaining observations. - B: This treats all four coefficients as having magnitude one, giving four times the common variance. - C: This multiplies variance by the sum of coefficient magnitudes, 3+1+1+1=6, instead of by the sum of squared coefficients. - D: This includes only the contribution 3^2Var(X1)=81 and omits the other three independent terms. ### Verification record - Main value: 108 - Main method: Square the coefficients of the independent linear combination and multiply their sum by the common variance. - Independent check: After centering each observation, the four independent zero-mean terms have variances 81, 9, 9, and 9; the second moment of their sum is therefore 108. - Independent value: 108 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: calibration coefficient Independent calibration errors A and B have variances 4 and 9. A combined error is Q=kA-B, where k is positive. The standard deviation of Q is 5. Calculate k. - A: 1 - B: 2 - C: 4 - D: 5 - E: 8 The target standard deviation gives variance 25. Independence yields Var(Q)=4k^2+9, so 4k^2+9=25 and the positive coefficient is k=2, which is choice B. #### Variant solution ### Setup Convert the given standard deviation of the combined error to variance. LaTeX: \operatorname{Var}(Q)=5^2=25 ### Model Use independence and square the linear-combination coefficients. LaTeX: \operatorname{Var}(kA-B)=k^2(4)+(-1)^2(9) ### Compute Solve the variance equation and retain the specified positive coefficient. LaTeX: 4k^2+9=25 LaTeX: k^2=4 LaTeX: k=2 ### Answer The positive calibration coefficient is 2. LaTeX: \boxed{k=2\quad\text{(B)}} ### Verification record - Main value: 2 - Main method: Set the independent linear-combination variance equal to 25 and solve the resulting quadratic for the positive coefficient. - Independent check: At k=2, the independent variance contributions are 2^2(4)=16 and 9, totaling 25 and reproducing standard deviation 5. - Independent value: 2 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #581 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-581/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Joint Distributions; Covariance and Correlation The joint table gives E[X]=4/3, E[Y]=1/3, and E[XY]=1/3. Therefore Cov(X,Y)=1/3-(4/3)(1/3)=-1/9, so choice A is correct. ### Independent solution ### Setup Compute the two marginal first moments from the nonzero joint masses. LaTeX: E[X]=0\left(\frac16\right)+2\left(\frac12\right)+1\left(\frac13\right)=\frac43 LaTeX: E[Y]=1\left(\frac13\right)=\frac13 ### Model Calculate the cross moment. Only the joint mass with both coordinates nonzero contributes. LaTeX: E[XY]=(1)(1)\left(\frac13\right)=\frac13 ### Compute Subtract the product of the marginal means from the cross moment. LaTeX: \operatorname{Cov}(X,Y)=E[XY]-E[X]E[Y] LaTeX: \operatorname{Cov}(X,Y)=\frac13-\frac43\left(\frac13\right)=-\frac19 ### Answer The covariance is negative one-ninth. LaTeX: \boxed{-\frac19\quad\text{(A)}} ### Distractor analysis - B: This assumes zero covariance without checking that the joint table differs from the product of its marginals. - C: This misses the factor 2 in the X=2 contribution, using E[X]=5/6 and obtaining 1/3-(5/6)(1/3)=1/18. - D: This reports the magnitude of the covariance but loses its negative sign. - E: This assigns the Y=1 mass to X=2 when calculating E[XY], giving 2/3-4/9=2/9. ### Verification record - Main value: -0.1111111111111111 - Main method: Compute both marginal means and the cross moment from the joint probability table, then apply the covariance identity. - Independent check: For binary Y with p=1/3, E[X|Y=1]=1 and E[X|Y=0]=3/2; p(1-p)(1-3/2)=-1/9 independently confirms the covariance. - Independent value: -0.1111111111111111 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: transformed calibration errors Independent calibration errors A and B have variances 4 and 9. Define U=A+B and V=2A-B. Calculate Cov(U,V). - A: -17 - B: -1 - C: 0 - D: 1 - E: 17 Covariance bilinearity and independence give Cov(A+B,2A-B)=2Var(A)-Var(B)=8-9=-1, so choice B is correct. #### Variant solution ### Setup Write the covariance of the two linear transformations. LaTeX: \operatorname{Cov}(U,V)=\operatorname{Cov}(A+B,2A-B) ### Model Expand bilinearly and use independence to remove both cross-covariance terms. LaTeX: \operatorname{Cov}(U,V)=2\operatorname{Var}(A)-\operatorname{Var}(B) LaTeX: \operatorname{Cov}(A,B)=0 ### Compute Substitute the two component variances. LaTeX: \operatorname{Cov}(U,V)=2(4)-9=-1 ### Answer The transformed errors have covariance -1. LaTeX: \boxed{-1\quad\text{(B)}} ### Verification record - Main value: -1 - Main method: Expand the covariance of the two linear transformations and use independence to retain only component-variance terms. - Independent check: Since U+V=3A, Var(U)=13, Var(V)=25, and Var(U+V)=36; the identity Cov(U,V)=[36-13-25]/2 gives -1. - Independent value: -1 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #582 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-582/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Variance and Standard Deviation; Sums of Independent Random Variables Independence makes the variance of the annual total equal to 52 times the common weekly variance. Equating this to the annual variance 9 gives weekly variance 9/52 and weekly standard deviation 3 divided by the square root of 52, so choice D is correct. ### Independent solution ### Setup Let W_i denote the amount in week i and let sigma be the common weekly standard deviation. LaTeX: S=\sum_{i=1}^{52}W_i LaTeX: \operatorname{SD}(W_i)=\sigma ### Model Because the weekly amounts are independent, their variances add in the annual total. LaTeX: \operatorname{Var}(S)=\sum_{i=1}^{52}\operatorname{Var}(W_i)=52\sigma^2 ### Compute Square the stated annual standard deviation, solve for the weekly variance, and then take its positive square root. LaTeX: 3^2=52\sigma^2 LaTeX: \sigma^2=\frac{9}{52} LaTeX: \sigma=\frac{3}{\sqrt{52}}=0.4160251472\ldots ### Answer Each weekly amount has standard deviation 3 divided by the square root of 52. LaTeX: \boxed{\frac{3}{\sqrt{52}}\quad\text{(D)}} ### Distractor analysis - A: This divides a standard deviation by 52 squared. Variances add across independent weeks, so only one factor of 52 belongs in the variance equation. - B: This divides the annual standard deviation directly by 52, as though standard deviations rather than variances added for independent amounts. - C: This treats the annual standard deviation 3 as though it were the annual variance, producing the square root of 3/52 instead of 9/52. - E: This scales the annual standard deviation upward by the square root of 52; the weekly component must be smaller than the annual total. ### Verification record - Main value: 0.41602514716892186 - Main method: Set the annual variance 9 equal to the sum of 52 identical independent weekly variances and take the positive square root. - Independent check: Substitution gives an annual standard deviation of sqrt(52) times 3/sqrt(52), which reproduces the stated value 3. - Independent value: 0.41602514716892186 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: weighted navigation errors Three independent navigation errors A, B, and C have standard deviations 2, 3, and 6 meters, respectively. A corrected position error is defined as T = 2A - B + 0.5C. Calculate the standard deviation of T. - A: 5.000 - B: 5.831 - C: 5.916 - D: 7.000 - E: 10.000 The three transformed errors have standard deviations 4, 3, and 3. Independence makes their variances add, so Var(T)=16+9+9=34 and SD(T)=sqrt(34)=5.831, which is choice B. #### Variant solution ### Setup Record the standard deviations and the coefficients in the corrected error. LaTeX: \operatorname{SD}(A)=2,\quad \operatorname{SD}(B)=3,\quad \operatorname{SD}(C)=6 LaTeX: T=2A-B+0.5C ### Model For independent inputs, each variance is multiplied by the square of its coefficient before the contributions are added. LaTeX: \operatorname{Var}(T)=2^2\operatorname{Var}(A)+(-1)^2\operatorname{Var}(B)+(0.5)^2\operatorname{Var}(C) ### Compute Insert the three component variances and take the square root of the resulting total variance. LaTeX: \operatorname{Var}(T)=4(2^2)+1(3^2)+0.25(6^2)=34 LaTeX: \operatorname{SD}(T)=\sqrt{34}=5.830951895\ldots ### Answer The corrected position error has standard deviation approximately 5.831 meters. LaTeX: \boxed{5.831\text{ meters}\quad\text{(B)}} ### Verification record - Main value: 5.830951894845301 - Main method: Apply the independent linear-combination variance formula to the heterogeneous weighted errors and take the square root. - Independent check: The transformed components have standard deviations 4, 3, and 3; their root-sum-square is sqrt(4^2+3^2+3^2)=sqrt(34). - Independent value: 5.830951894845301 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #583 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-583/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Uniform Distribution; Conditional Probability The two probability statements identify the uniform support as [4,20]. After conditioning on values above 6, the remaining interval has length 14, and the favorable part from 6 to 10 has length 4. Their ratio is 4/14=2/7, so choice C is correct. ### Independent solution ### Setup Let the unknown uniform support be [a,b] and translate each supplied probability into an interval-length ratio. LaTeX: \frac{12-a}{b-a}=0.50 LaTeX: \frac{6-a}{b-a}=1-0.875=0.125 ### Model Divide the first relation by the second to remove the unknown support length and solve for the lower endpoint. LaTeX: \frac{12-a}{6-a}=\frac{0.50}{0.125}=4 LaTeX: 12-a=24-4a LaTeX: a=4 ### Compute Use either probability relation to recover the upper endpoint, then form the conditional interval ratio. LaTeX: \frac{12-4}{b-4}=0.50\quad\Longrightarrow\quad b=20 LaTeX: \Pr(X<10\mid X>6)=\frac{\Pr(66)} LaTeX: \frac{(10-6)/16}{(20-6)/16}=\frac{2}{7} ### Answer The requested conditional probability is 2/7. LaTeX: \boxed{\frac{2}{7}\quad\text{(C)}} ### Distractor analysis - A: This uses a denominator of 18 instead of the conditioned interval length 14. Once X>6 is known, only the interval from 6 to 20 remains possible. - B: This is the unconditional probability of the intersection, 4/16, and omits division by the probability of the conditioning event. - D: This is the unconditional probability P(X<10)=6/16 and ignores both the intersection with X>6 and the conditional normalization. - E: This divides P(X<10)=6/16 by P(X>6)=14/16 without first intersecting the numerator event with the condition. ### Verification record - Main value: 0.2857142857142857 - Main method: Recover the uniform endpoints from the two cumulative probabilities and divide the favorable conditioned interval length by the full conditioned length. - Independent check: The value 6 is the 12.5th percentile and 12 is the median; the six-unit quantile gap is 37.5% of the support, giving width 16 and support [4,20], after which direct interval geometry gives 4/14. - Independent value: 0.2857142857142857 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: calibration acceptance cutoff A calibration score U is uniformly distributed from 0 to 30. An engineer knows that U exceeds 6 and wants a cutoff c for which the conditional probability P(U6) equals 0.625. Determine c. - A: 9.75 - B: 15.00 - C: 18.75 - D: 21.00 - E: 24.75 Conditioning on U>6 leaves a uniform interval of length 24. The cutoff must place 62.5% of that interval below c, so c-6=0.625(24)=15 and c=21, which is choice D. #### Variant solution ### Setup Restrict the uniform sample space to the interval that remains after the observed lower cutoff. LaTeX: U\mid(U>6)\sim\operatorname{Uniform}(6,30) ### Model For a cutoff between 6 and 30, the conditional probability is the favorable subinterval length divided by 24. LaTeX: \Pr(U6)=\frac{c-6}{30-6} ### Compute Set the conditional interval ratio equal to the specified probability and solve for the cutoff. LaTeX: \frac{c-6}{24}=0.625 LaTeX: c-6=15 LaTeX: c=21 ### Answer The required calibration cutoff is 21. LaTeX: \boxed{c=21\quad\text{(D)}} ### Verification record - Main value: 21 - Main method: Invert the conditional uniform CDF on the truncated interval [6,30]. - Independent check: At c=21, the favorable interval has length 15 and the conditioned interval has length 24, so their ratio is 15/24=0.625. - Independent value: 21 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #584 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-584/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Sums of Independent Random Variables A one-year zero-count probability of 0.90 implies a Poisson mean of -ln(0.90). Adding 15 independent yearly counts gives a Poisson total whose variance is 15[-ln(0.90)]=1.5804, so choice E is correct. ### Independent solution ### Setup Let lambda be the mean count for one year and convert the positive-count probability to a zero-count probability. LaTeX: \Pr(N=0)=1-0.10=0.90 ### Model For a Poisson count, the zero probability determines its parameter. Independent Poisson counts add by summing their parameters. LaTeX: e^{-\lambda}=0.90 LaTeX: \lambda=-\ln(0.90) LaTeX: S=\sum_{j=1}^{15}N_j\sim\operatorname{Poisson}(15\lambda) ### Compute Use equality of the Poisson mean and variance for the aggregated count. LaTeX: \operatorname{Var}(S)=15\lambda LaTeX: \operatorname{Var}(S)=-15\ln(0.90)=1.5804077349\ldots ### Answer The variance of the 15-year total rounds to 1.58. LaTeX: \boxed{1.58\quad\text{(E)}} ### Distractor analysis - A: This is approximately the one-year Poisson mean -ln(0.90), so it omits aggregation over all 15 years. - B: This scales the one-year parameter by the square root of 15, mixing a standard-deviation scaling rule with the Poisson variance calculation. - C: This treats each year as a Bernoulli indicator with variance 0.10(0.90) and adds 15 such variances, discarding the possibility of multiple counts. - D: This replaces the Poisson mean -ln(0.90) by the positive-count probability 0.10 and reports 15(0.10). ### Verification record - Main value: 1.5804077348673942 - Main method: Infer the annual Poisson parameter from its zero probability, sum 15 independent parameters, and use equality of Poisson variance and mean. - Independent check: The probability of zero events in all 15 independent years is 0.90^15; a Poisson total with that zero probability has parameter -ln(0.90^15)=-15ln(0.90). - Independent value: 1.5804077348673942 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: quiet monitoring window Independent daily alert counts for a monitoring system follow Poisson distributions with a common mean. The variance of the total alert count over eight days is 12. Calculate the probability of receiving no alerts during a separate three-day period. - A: 0.000006 - B: 0.011109 - C: 0.125000 - D: 0.223130 - E: 0.988891 The eight-day Poisson total has mean and variance 12, so the daily mean is 1.5. A three-day total therefore has mean 4.5, and its zero-count probability is exp(-4.5)=0.011109, which is choice B. #### Variant solution ### Setup Let lambda denote the common mean of one daily Poisson count. LaTeX: N_d\sim\operatorname{Poisson}(\lambda) ### Model The eight-day total is Poisson with parameter 8lambda, and its variance is therefore 8lambda. LaTeX: \operatorname{Var}\!\left(\sum_{d=1}^{8}N_d\right)=8\lambda=12 ### Compute Recover the daily parameter, form the three-day parameter, and evaluate the Poisson zero probability. LaTeX: \lambda=1.5 LaTeX: \sum_{d=1}^{3}N_d\sim\operatorname{Poisson}(4.5) LaTeX: \Pr\!\left(\sum_{d=1}^{3}N_d=0\right)=e^{-4.5}=0.0111089965\ldots ### Answer The probability that the three-day period is quiet is approximately 0.011109. LaTeX: \boxed{0.011109\quad\text{(B)}} ### Verification record - Main value: 0.011108996538242306 - Main method: Recover the daily Poisson parameter from the eight-day variance, scale it to three days, and evaluate the zero-count mass. - Independent check: One day is quiet with probability exp(-1.5); independence makes three quiet days have probability [exp(-1.5)]^3=exp(-4.5). - Independent value: 0.011108996538242306 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #585 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-585/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Continuous Distributions; Conditional Probability Intersecting the two interval events leaves 25)=\Pr\!\left(0.8(L-0.5)>5\right) ### Model Solve the payment inequality for the underlying loss threshold. LaTeX: L-0.5>\frac{5}{0.8}=6.25 LaTeX: L>6.75 ### Compute Use the exponential survival function with scale 6 at the transformed threshold. LaTeX: \Pr(L>\ell)=e^{-\ell/6} LaTeX: \Pr(Y>5)=e^{-6.75/6}=e^{-1.125}=0.3246524674\ldots ### Answer The payment exceeds the specified amount with probability approximately 0.325. LaTeX: \boxed{0.325\quad\text{(C)}} ### Distractor analysis - A: This reflects a severe unit or threshold conversion error; the payment condition should produce a loss threshold of 6.75 in the model's thousand-unit scale. - B: This models the payment as 0.8L-0.5, applying coinsurance before subtracting the deductible, rather than 0.8(L-0.5). - D: This is the complement of the correct exponential tail and therefore represents the probability that the payment does not exceed the threshold. - E: This is the complement of the value generated by the incorrect transformation 0.8L-0.5, combining both a payment-model error and a reversed tail. ### Verification record - Main value: 0.32465246735834974 - Main method: Invert the deductible-and-coinsurance payment inequality and evaluate the exponential survival function at the resulting loss threshold. - Independent check: By memorylessness, the probability factors as P(L>0.5) times P(L-0.5>6.25 | L>0.5)=exp(-0.5/6)exp(-6.25/6)=exp(-6.75/6). - Independent value: 0.32465246735834974 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: conditional repair-payment percentile A repair cost X follows an exponential distribution with mean 4 thousand dollars. A service plan pays 60% of the amount above a 1-thousand-dollar deductible. Given that the plan makes a positive payment, calculate the 90th percentile of the payment, in thousands. - A: 2.400 - B: 4.926 - C: 5.526 - D: 6.126 - E: 9.210 Given a positive payment, exponential memorylessness makes the excess over the deductible exponential with mean 4. Multiplying by 0.60 gives a conditional payment with mean 2.4. Its 90th percentile is 2.4ln(10)=5.526, choice C. #### Variant solution ### Setup Let R be the service-plan payment and condition on the event that the repair cost clears the deductible. LaTeX: R=0.6(X-1)_+ LaTeX: R>0\quad\Longleftrightarrow\quad X>1 ### Model Memorylessness makes the deductible excess exponential with the original mean; multiplying by 0.6 rescales that mean. LaTeX: X-1\mid(X>1)\sim\operatorname{Exponential}(\text{mean }4) LaTeX: R\mid(R>0)\sim\operatorname{Exponential}(\text{mean }2.4) ### Compute At the 90th percentile q, the conditional survival probability is 0.10. LaTeX: e^{-q/2.4}=0.10 LaTeX: q=2.4\ln(10)=5.526204223\ldots ### Answer The conditional 90th percentile of the payment is approximately 5.526 thousand dollars. LaTeX: \boxed{5.526\text{ thousand dollars}\quad\text{(C)}} ### Verification record - Main value: 5.52620422318571 - Main method: Use exponential memorylessness after conditioning on a positive payment, rescale the excess by coinsurance, and invert its survival function at 0.10. - Independent check: Substitution gives P(R>5.526204223 | R>0)=exp[-5.526204223/2.4]=0.10, so the conditional CDF equals 0.90. - Independent value: 5.52620422318571 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #587 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-587/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Joint Distributions; Marginal Distributions; Variance This is a marginal-distribution variance calculation. The stated mean fixes p=0.12, normalization gives q=0.16, and the afternoon-distance second moment is 48, so the variance is 48-6^2=12 and choice B is correct. ### Independent solution ### Setup Collapse the joint table over the morning categories. The afternoon distance A has masses q, 4p, and 3p at 0, 5, and 10, respectively. LaTeX: \Pr(A=0)=q,\qquad \Pr(A=5)=4p,\qquad \Pr(A=10)=3p ### Model Use the supplied mean to determine p, then use total probability to determine q. LaTeX: E[A]=5(4p)+10(3p)=50p=6 LaTeX: 7p+q=1 ### Compute After solving for the two table parameters, calculate the second moment and subtract the square of the mean. LaTeX: p=0.12,\qquad q=0.16 LaTeX: E[A^2]=25(4p)+100(3p)=400p=48 LaTeX: \operatorname{Var}(A)=48-6^2=12 ### Answer The variance of the afternoon distance is 12 square miles. LaTeX: \boxed{12.0\quad\text{(B)}} ### Distractor analysis - A: This results if the second moment is understated by one unit before subtracting the squared mean; the marginal probabilities should be recomputed from all nonzero table cells. - C: This can result from failing to combine both joint-table cells that contribute to the five-mile afternoon outcome. - D: This treats the symbols in the table as final probabilities without first imposing both the mean and normalization conditions. - E: This is E[A^2]-E[A]=48-6=42, which subtracts the mean rather than the square of the mean. ### Verification record - Main value: 12 - Main method: Marginalize the joint table, solve p from the afternoon mean, normalize to obtain q, and apply Var(A)=E[A^2]-E[A]^2. - Independent check: Using p=0.12 and q=0.16, sum squared deviations directly: 0.16(0-6)^2+0.48(5-6)^2+0.36(10-6)^2=12. - Independent value: 12 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: correlated sensor adjustment Two calibrated sensor errors X and Y have standard deviations 2 and 3, with correlation 0.5. A correction score is defined as Z=2X-Y. Calculate the variance of Z. - A: 1 - B: 9 - C: 13 - D: 25 - E: 37 The covariance is 0.5(2)(3)=3. Applying the variance formula for 2X-Y gives 4(4)+9-4(3)=13, so choice C is correct. #### Variant solution ### Setup Convert the standard deviations and correlation into variances and covariance. LaTeX: \operatorname{Var}(X)=4,\qquad \operatorname{Var}(Y)=9 LaTeX: \operatorname{Cov}(X,Y)=0.5(2)(3)=3 ### Model For a linear combination, include both squared coefficients and the covariance cross term. LaTeX: \operatorname{Var}(aX+bY)=a^2\operatorname{Var}(X)+b^2\operatorname{Var}(Y)+2ab\operatorname{Cov}(X,Y) ### Compute Substitute a=2 and b=-1 into the linear-combination formula. LaTeX: \operatorname{Var}(Z)=4(4)+9-4(3)=13 ### Answer The correction score has variance 13. LaTeX: \boxed{13\quad\text{(C)}} ### Verification record - Main value: 13 - Main method: Convert correlation to covariance and apply the variance formula for the linear combination 2X-Y. - Independent check: The covariance matrix is [[4,3],[3,9]]. Its quadratic form with coefficient vector (2,-1) equals 13. - Independent value: 13 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #588 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-588/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Gamma Distribution; Sums of Independent Random Variables This is a sum of five independent exponential waiting times. Each has variance 1000^2, so the total variance is 5,000,000 and its standard deviation is 1000 sqrt(5), approximately 2236, making choice D correct. ### Independent solution ### Setup Let each successive lifetime be an independent exponential variable with mean 1000. The requested time is their five-term sum. LaTeX: T=X_1+X_2+X_3+X_4+X_5 LaTeX: E[X_i]=1000,\qquad \operatorname{Var}(X_i)=1000^2 ### Model Independence makes the variance of the sum equal to the sum of the five component variances. LaTeX: \operatorname{Var}(T)=\sum_{i=1}^{5}\operatorname{Var}(X_i) ### Compute Add the variances and then take one square root because the question asks for standard deviation. LaTeX: \operatorname{Var}(T)=5(1000^2)=5{,}000{,}000 LaTeX: \operatorname{SD}(T)=1000\sqrt{5}=2236.067977\ldots ### Answer The standard deviation rounds to 2236 hours. LaTeX: \boxed{2236\quad\text{(D)}} ### Distractor analysis - A: This is approximately sqrt(5(1000)); it treats the exponential mean as a variance rather than squaring it. - B: This also comes from applying a square root to an incorrectly scaled component variance before combining the five lifetimes. - C: This is the standard deviation of one exponential lifetime and ignores variability accumulated over the other four lifetimes. - E: This is the total mean 5(1000), not the standard deviation of the total time. ### Verification record - Main value: 2236.06797749979 - Main method: Add the variances of five independent exponential variables with common mean and standard deviation 1000, then take the square root. - Independent check: The five-term sum has a gamma distribution with shape 5 and scale 1000, whose variance is shape times scale squared, or 5,000,000. - Independent value: 2236.06797749979 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: random-stage calibration A calibration job requires either one stage or three stages, each with probability 0.5. Stage durations are independent exponential random variables with mean 2 minutes and are independent of the number of stages. Calculate the variance of the total calibration time. - A: 4 - B: 8 - C: 12 - D: 16 - E: 28 The random stage count has mean 2 and variance 1. Total variance gives 4E[N]+4Var(N)=8+4=12, so choice C is correct. #### Variant solution ### Setup Let N be the stage count and let S be the sum of the N independent stage durations. LaTeX: \Pr(N=1)=\Pr(N=3)=\frac12 LaTeX: E[N]=2,\qquad \operatorname{Var}(N)=1 ### Model Conditioned on N, the exponential-stage sum has mean 2N and variance 4N. LaTeX: E[S\mid N]=2N,\qquad \operatorname{Var}(S\mid N)=4N ### Compute Apply the law of total variance to include both stage-duration randomness and stage-count randomness. LaTeX: \operatorname{Var}(S)=E[4N]+\operatorname{Var}(2N) LaTeX: \operatorname{Var}(S)=4(2)+4(1)=12 ### Answer The total calibration time has variance 12 square minutes. LaTeX: \boxed{12\quad\text{(C)}} ### Verification record - Main value: 12 - Main method: Use conditional mean 2N, conditional variance 4N, and the law of total variance. - Independent check: For exponential stages, E[S^2 given N]=4N+4N^2. Averaging over N gives E[S^2]=28; since E[S]=4, the variance is 28-16=12. - Independent value: 12 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #589 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-589/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Compound Poisson Distribution; Deductible Payments; Expected Aggregate Loss This is a compound-Poisson expectation with an ordinary deductible. The expected payment per event is the tail integral from 2 to 8, equal to 1.125, so multiplying by the expected count 12 gives 13.5 and choice B is correct. ### Independent solution ### Setup Let Y be the payment from one event. Under a deductible of 2, Y is the positive part of the loss above 2. LaTeX: Y=(X-2)_+ ### Model Integrating the stated density gives a simple survival function on the loss support. The tail-integral identity then gives the mean payment. LaTeX: \Pr(X>x)=\frac{(8-x)^2}{64},\qquad 0\le x\le 8 LaTeX: E[Y]=\int_2^8\Pr(X>x)\,dx ### Compute Evaluate the per-event mean, then apply the compound-sum mean formula with expected event count 12. LaTeX: E[Y]=\int_2^8\frac{(8-x)^2}{64}\,dx=\frac{6^3}{3(64)}=1.125 LaTeX: E\!\left[\sum_{i=1}^{N}Y_i\right]=E[N]E[Y]=12(1.125)=13.5 ### Answer The expected aggregate payment is 13.5. LaTeX: \boxed{13.5\quad\text{(B)}} ### Distractor analysis - A: This uses E[(X-2)+]=E[X]-2, giving 12(8/3-2)=8, and incorrectly permits negative payments below the deductible. - C: This can arise from integrating the excess against an incorrectly normalized tail rather than the survival function implied by the density. - D: This overstates the conditional excess and then applies it to every event instead of weighting by the chance that the deductible is exceeded. - E: This is 12E[X]=12(8/3)=32 and ignores the deductible entirely. ### Verification record - Main value: 13.5 - Main method: Integrate the loss survival function above the deductible to obtain the mean payment per event, then multiply by the Poisson mean. - Independent check: Directly integrate (x-2)(8-x)/32 from 2 to 8 to obtain 1.125 per event; conditioning on the event count then gives 12(1.125)=13.5. - Independent value: 13.5 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: pop-up service expenses The number of technical-support incidents during a pop-up event is Poisson with mean 5. Each incident costs 2 units with probability 0.6 and 5 units with probability 0.4, independently of all other incidents and of the count. Calculate the variance of the total incident cost. - A: 10.8 - B: 16.0 - C: 51.2 - D: 62.0 - E: 256.0 A single incident has second moment 0.6(2^2)+0.4(5^2)=12.4. A compound Poisson sum has variance lambda E[Y^2], so the aggregate variance is 5(12.4)=62 and choice D is correct. #### Variant solution ### Setup Let Y denote one incident cost and S the random sum of costs. LaTeX: S=\sum_{i=1}^{N}Y_i,\qquad N\sim\operatorname{Poisson}(5) ### Model For a compound Poisson sum, the aggregate variance is the Poisson mean times the severity second moment. LaTeX: \operatorname{Var}(S)=5E[Y^2] ### Compute Calculate the discrete severity second moment and multiply by the frequency mean. LaTeX: E[Y^2]=0.6(2^2)+0.4(5^2)=12.4 LaTeX: \operatorname{Var}(S)=5(12.4)=62 ### Answer The total incident cost has variance 62. LaTeX: \boxed{62.0\quad\text{(D)}} ### Verification record - Main value: 62 - Main method: Compute the incident-cost second moment and apply the compound Poisson identity Var(S)=lambda E[Y^2]. - Independent check: Total variance gives E[N Var(Y)]+Var(N E[Y])=5(2.16)+5(3.2^2)=10.8+51.2=62. - Independent value: 62 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #590 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-590/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Geometric Distribution; Independence This is a geometric stopping-time calculation. A round continues with probability (1/4)^3+(3/4)^3=7/16, so exactly five rounds requires four continuations followed by a stop: (7/16)^4(9/16)=0.020608, making choice C correct. ### Independent solution ### Setup In one round, continuation occurs only when all three outcomes match. Add the all-first-outcome and all-second-outcome probabilities. LaTeX: r=\left(\frac14\right)^3+\left(\frac34\right)^3=\frac7{16} ### Model Exactly five rounds means that each of the first four rounds continues and the fifth round stops. LaTeX: \Pr(T=5)=r^4(1-r) ### Compute Substitute the one-round continuation and stopping probabilities. LaTeX: \Pr(T=5)=\left(\frac7{16}\right)^4\left(\frac9{16}\right) LaTeX: \Pr(T=5)=\frac{21609}{1048576}=0.0206079483\ldots ### Answer The probability rounds to 0.0206. LaTeX: \boxed{0.0206\quad\text{(C)}} ### Distractor analysis - A: This treats continuation as a prescribed rare matching pattern rather than allowing both possible all-matching outcomes. - B: This over-multiplies the rare all-first-outcome event instead of using the full one-round continuation probability 7/16. - D: This is close to an off-by-one calculation with only three continuation rounds before the stopping round. - E: This is close to r^2(1-r), the probability of stopping on round three rather than round five. ### Verification record - Main value: 0.020607948303222656 - Main method: Compute the one-round continuation probability and apply the geometric waiting-time probability r^4(1-r). - Independent check: Enumerate the required event pattern C,C,C,C,S. Independence across rounds gives four factors of 7/16 and one factor of 9/16. - Independent value: 0.020607948303222656 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: two-state incident response An incident starts at level 0. During the first response cycle it is resolved with probability 0.3 or advances to level 1 with probability 0.7. Once at level 1, each later cycle resolves it with probability 0.6 or leaves it at level 1 with probability 0.4. Calculate the probability that the incident is resolved exactly on cycle 4. - A: 0.0288 - B: 0.0672 - C: 0.0960 - D: 0.1680 - E: 0.2800 Resolution on cycle 4 requires advancement on cycle 1, two further unresolved level-1 cycles, and resolution on cycle 4. The probability is 0.7(0.4)^2(0.6)=0.0672, so choice B is correct. #### Variant solution ### Setup Identify the unique state path that permits resolution on the fourth cycle. LaTeX: 0\longrightarrow 1\longrightarrow 1\longrightarrow 1\longrightarrow \text{resolved} ### Model The first transition has probability 0.7, the next two cycles must remain unresolved, and the final cycle must resolve. LaTeX: \Pr(T=4)=0.7(0.4)(0.4)(0.6) ### Compute Multiply the transition probabilities along the required path. LaTeX: \Pr(T=4)=0.7(0.4)^2(0.6)=0.0672 ### Answer The incident resolves on cycle 4 with probability 0.0672. LaTeX: \boxed{0.0672\quad\text{(B)}} ### Verification record - Main value: 0.0672 - Main method: Multiply the probabilities along the only path that first resolves the incident on cycle 4. - Independent check: After the compulsory level-0 transition, the number of level-1 cycles to resolution is geometric with success probability 0.6; 0.7 times its third-trial mass 0.4^2(0.6) equals 0.0672. - Independent value: 0.0672 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #591 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-591/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Normal Distribution; Central Probability; Quantiles This is a symmetric normal-quantile calculation. A central probability of 0.25 leaves cumulative probability 0.625 at the upper endpoint, whose z-score is 0.31864; multiplying by the standard deviation 4 gives a half-width of 1.2746 and the interval (18.7,21.3), so choice E is correct. ### Independent solution ### Setup The variance is 16, so the standard deviation is 4. Write the centered interval as the mean plus or minus a multiple of this standard deviation. LaTeX: \sigma=\sqrt{16}=4 LaTeX: I=(20-4z,\,20+4z) ### Model Symmetry splits the remaining probability equally between the two tails. Therefore the upper standardized endpoint has cumulative probability 0.5+0.25/2. LaTeX: \Pr(-zc)=0.10 LaTeX: \Pr\!\left(Z\le \frac{c-6}{5}\right)=0.90 ### Compute Use the standard-normal 90th percentile and transform back to the original scale. LaTeX: \Phi^{-1}(0.90)=1.281551566\ldots LaTeX: c=6+5(1.281551566\ldots)=12.40775783\ldots ### Answer The required threshold is approximately 12.4078. LaTeX: \boxed{c=12.4078\quad\text{(E)}} ### Verification record - Main value: 12.407757827723001 - Main method: Derive X-Y as normal with mean 6 and standard deviation 5, then transform the standard-normal 90th percentile. - Independent check: Substitution gives (c-6)/5=1.281551566, whose standard-normal survival probability is 0.10. - Independent value: 0.1 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ## SOA Exam P Sample Question #592 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-592/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Binomial Distribution; Modes The count is binomial with n=18 and p=0.15. Since (n+1)p=2.85 is not an integer, its unique mode is floor(2.85)=2, so choice C is correct. ### Independent solution ### Setup Model the number of positive results as a binomial count. LaTeX: X\sim\operatorname{Binomial}(18,0.15) ### Model A binomial distribution has unique mode floor((n+1)p) when (n+1)p is not an integer. LaTeX: (n+1)p=19(0.15)=2.85 ### Compute Take the integer below 2.85. LaTeX: \operatorname{mode}(X)=\lfloor2.85\rfloor=2 ### Answer The most likely count is 2. LaTeX: \boxed{2\quad\text{(C)}} ### Distractor analysis - A: This treats the low probability for one trial as implying a zero modal count and ignores the accumulation across eighteen trials. - B: This uses an incorrect mode rule such as floor(np)-1, moving one count below the actual peak. - D: This rounds the mean np=2.7 to 3; a binomial mode is determined by (n+1)p rather than ordinary rounding of the mean. - E: This moves farther into the decreasing upper side of the binomial mass after the peak near the mean. ### Verification record - Main value: 2 - Main method: Apply the unique binomial mode formula floor((n+1)p) to n=18 and p=0.15. - Independent check: Adjacent probability ratios give P(X=2)/P(X=1)=1.5>1 and P(X=3)/P(X=2)=16/17<1, confirming that the mass peaks at 2. - Independent value: 2 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: inferred inspection rate The number X of flagged items in ten independent inspections is binomial with an unknown flag probability. The probability of exactly four flags is 1.5 times the probability of exactly three flags. Calculate the mode of X. - A: 3 - B: 4 - C: 5 - D: 6 - E: 10 The adjacent-mass ratio gives 7p/[4(1-p)]=1.5, so p=6/13. Then (10+1)p=66/13 lies between 5 and 6, making 5 the unique mode and choice C correct. #### Variant solution ### Setup Use the ratio of adjacent binomial masses to recover the unknown probability. LaTeX: \frac{\Pr(X=4)}{\Pr(X=3)}=\frac{10-3}{4}\frac{p}{1-p}=1.5 ### Model Solve the ratio equation for p. LaTeX: \frac{7p}{4(1-p)}=\frac32 LaTeX: p=\frac6{13} ### Compute Apply the binomial mode rule after inferring the parameter. LaTeX: (10+1)p=\frac{66}{13}=5.076923\ldots LaTeX: \operatorname{mode}(X)=5 ### Answer The most likely number of flagged items is 5. LaTeX: \boxed{5\quad\text{(C)}} ### Verification record - Main value: 5 - Main method: Recover p from the adjacent binomial mass ratio and apply the unique-mode formula. - Independent check: At p=6/13, P(X=5)/P(X=4)=36/35>1 while P(X=6)/P(X=5)=5/7<1, so the maximum occurs at 5. - Independent value: 5 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #593 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-593/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Normal Distribution; Percentiles and Quantiles The 28th-percentile z-score is -0.582842, so the 18-unit gap below the mean determines standard deviation 30.8832. A value of 3 is 1.165683 standard deviations below the mean, with cumulative probability 0.121871, so choice E is correct. ### Independent solution ### Setup Translate the known percentile into a standard-normal equation. LaTeX: \frac{21-39}{\sigma}=\Phi^{-1}(0.28)=-0.5828415073\ldots ### Model Solve for the normal standard deviation. LaTeX: \sigma=\frac{-18}{-0.5828415073}=30.8831814060\ldots ### Compute Standardize the target value and evaluate its cumulative probability. LaTeX: z=\frac{3-39}{30.8831814060}=-1.1656830145\ldots LaTeX: \Pr(X\le3)=\Phi(z)=0.1218713150\ldots ### Answer A cumulative probability near 12.2% corresponds to the listed 12th percentile. LaTeX: \boxed{\text{12th percentile}\quad\text{(E)}} ### Distractor analysis - A: This substitutes the 20th-percentile z-score, about -0.842, for the stated 28th percentile; doubling it gives about -1.68 and a lower tail near 4.6%. - B: This treats probability as linear in distance from the mean: 0.50-2(0.50-0.28)=0.06, even though normal CDF probabilities do not scale linearly. - C: This squares the known percentile probability, 0.28^2=0.0784, and rounds it to the 8th percentile instead of doubling the standardized distance. - D: This subtracts the 18-unit drop from 21 to 3 directly from the percentile number, 28-18=10, incorrectly treating profit units as percentile points. ### Verification record - Main value: 0.12187131503350812 - Main method: Use the known normal percentile to infer sigma, then standardize the target value and evaluate its CDF. - Independent check: The target is twice as far below mean 39 as the known value 21, so its z-score is exactly 2Phi^(-1)(0.28); evaluating Phi at that score gives 0.1218713150. - Independent value: 0.12187131503350812 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: spread from paired quantiles A normally distributed sensor reading has 10th percentile 40 and 90th percentile 60. Calculate the variance of the reading. - A: 7.80 - B: 10.00 - C: 60.89 - D: 100.00 - E: 400.00 Normal symmetry places the mean at 50. The half-width 10 equals Phi inverse 0.90 times sigma, giving sigma=7.80304 and variance 60.88746, so choice C is correct. #### Variant solution ### Setup Use symmetry of the paired normal percentiles to find the center. LaTeX: \mu=\frac{40+60}{2}=50 ### Model Relate the upper percentile's distance from the mean to its standard-normal quantile. LaTeX: \frac{60-50}{\sigma}=\Phi^{-1}(0.90)=1.2815515655\ldots ### Compute Solve for the standard deviation and square it. LaTeX: \sigma=\frac{10}{1.2815515655}=7.8030414607\ldots LaTeX: \sigma^2=60.8874560378\ldots ### Answer The variance rounds to 60.89. LaTeX: \boxed{60.89\quad\text{(C)}} ### Verification record - Main value: 60.88745603777448 - Main method: Use symmetry to find the mean, divide the upper half-width by the 90th-percentile z-score, and square the resulting standard deviation. - Independent check: With mean 50 and variance 60.88745604, standardizing 40 and 60 gives -1.281551566 and 1.281551566, whose CDF values are 0.10 and 0.90. - Independent value: 60.88745603777448 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #594 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-594/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Sampling Without Replacement; Hypergeometric Distribution Favorable hands contain two, three, or four kings, no cards from the two excluded ranks, and all remaining cards from the 40 neutral cards. Summing those combinations over all five-card hands gives probability 0.02402499, so choice C is correct. ### Independent solution ### Setup Partition the deck into four target cards, eight excluded cards, and forty neutral cards. LaTeX: 4+8+40=52 ### Model Count favorable hands separately for two, three, and four target cards while choosing zero excluded cards. LaTeX: N_f=\sum_{k=2}^{4}\binom4k\binom80\binom{40}{5-k} ### Compute Divide the favorable count by the number of unordered five-card hands. LaTeX: \Pr=\frac{\binom42\binom{40}3+\binom43\binom{40}2+\binom44\binom{40}1}{\binom{52}5} LaTeX: \Pr=0.0240249946\ldots ### Answer The probability rounds to 0.0240. LaTeX: \boxed{0.0240\quad\text{(C)}} ### Distractor analysis - A: This excludes an additional four-card rank along with the two prohibited ranks, leaving only 36 neutral cards and producing about 0.0174. - B: This counts exactly two kings and no prohibited cards but omits favorable hands containing three or four kings. - D: This treats the five selections as independent draws with replacement, producing about 0.0298. - E: This counts hands with at least two kings but ignores the restriction excluding aces and jacks. ### Verification record - Main value: 0.02402499461322991 - Main method: Count unordered hands with two through four kings, zero excluded-rank cards, and neutral cards filling the remaining positions. - Independent check: Factor as P(no excluded rank)=C(44,5)/C(52,5) times the conditional hypergeometric probability of at least two kings among five draws from four kings and forty neutral cards; the product is 0.0240249946. - Independent value: 0.02402499461322991 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: conditioned archive tokens An archive contains 6 bronze, 5 silver, and 4 gold tokens. Five tokens are selected without replacement. Given that exactly two selected tokens are bronze, calculate the probability that at least one selected token is gold. - A: 0.119 - B: 0.444 - C: 0.476 - D: 0.881 - E: 1.000 After conditioning on two bronze tokens, the other three are drawn from five silver and four gold tokens. The probability that all three are silver is 10/84, so the probability of at least one gold is 37/42=0.880952 and choice D is correct. #### Variant solution ### Setup Conditioning fixes two bronze selections, leaving three selections from the nine nonbronze tokens. LaTeX: 5\text{ silver}+4\text{ gold}=9\text{ nonbronze} ### Model Use the complement event that all three remaining selections are silver. LaTeX: \Pr(\text{at least one gold}\mid\text{two bronze}) LaTeX: =1-\frac{\binom53}{\binom93} ### Compute Evaluate the combination ratio. LaTeX: 1-\frac{10}{84}=\frac{74}{84}=\frac{37}{42} LaTeX: \frac{37}{42}=0.8809523810\ldots ### Answer The conditional probability rounds to 0.881. LaTeX: \boxed{0.881\quad\text{(D)}} ### Verification record - Main value: 0.8809523809523809 - Main method: Condition on the two bronze tokens and complement the hypergeometric probability that all three remaining selections are silver. - Independent check: Enumerating one, two, or three gold tokens gives [C(4,1)C(5,2)+C(4,2)C(5,1)+C(4,3)]/C(9,3)=74/84=37/42. - Independent value: 0.8809523809523809 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #595 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-595/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Uniform Distribution; Percentiles and Quantiles The smallest threshold exceeded by no more than 5% of observations is the 95th percentile. For a uniform interval from 1.1 to 8.6, this is 1.1+0.95(7.5)=8.225, so choice D is correct. ### Independent solution ### Setup Translate the upper-tail requirement into a cumulative probability. LaTeX: \Pr(X>y)\le0.05 LaTeX: F(y)\ge0.95 ### Model At the minimum admissible threshold, equality holds in the continuous uniform CDF. LaTeX: F(y)=\frac{y-1.1}{8.6-1.1}=0.95 ### Compute Solve the percentile equation. LaTeX: y=1.1+0.95(7.5)=8.225 ### Answer The minimum threshold is 8.225 minutes. LaTeX: \boxed{8.225\quad\text{(D)}} ### Distractor analysis - A: This calculates the 5th percentile, 1.1+0.05(7.5), instead of the 95th percentile required by an upper 5% tail. - B: This multiplies the interval width by 0.95 but forgets to add the lower endpoint 1.1. - C: This computes 0.95(8.6), treating the interval as though it began at zero. - E: This subtracts 5% of the lower endpoint from 8.6 instead of subtracting 5% of the full interval width. ### Verification record - Main value: 8.225 - Main method: Set the continuous uniform CDF equal to 0.95 and solve for the minimum upper-tail threshold. - Independent check: At y=8.225, the upper interval length is 8.6-8.225=0.375, and 0.375/7.5=0.05; every smaller y produces an upper tail greater than 0.05. - Independent value: 8.225 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: spread from uniform quartiles A sensor latency is uniformly distributed on an unknown interval. Its 25th percentile is 14 milliseconds and its 75th percentile is 26 milliseconds. Calculate the standard deviation of the latency. - A: 6.000 - B: 6.928 - C: 12.000 - D: 13.856 - E: 24.000 For a uniform distribution, the interquartile range equals one-half of the support width. The 12-millisecond interquartile range therefore gives width 24 and standard deviation 24/sqrt(12)=6.9282, so choice B is correct. #### Variant solution ### Setup Relate uniform percentiles to the unknown support width w. LaTeX: q_{0.75}-q_{0.25}=(0.75-0.25)w ### Model Use the observed interquartile range to recover the full width. LaTeX: 26-14=0.5w LaTeX: w=24 ### Compute Apply the continuous uniform standard-deviation formula. LaTeX: \operatorname{SD}(X)=\frac{w}{\sqrt{12}} LaTeX: \operatorname{SD}(X)=\frac{24}{\sqrt{12}}=6.9282032303\ldots ### Answer The latency standard deviation rounds to 6.928 milliseconds. LaTeX: \boxed{6.928\quad\text{(B)}} ### Verification record - Main value: 6.92820323027551 - Main method: Use the uniform interquartile range to recover width 24, then divide by sqrt(12). - Independent check: The implied endpoints are 8 and 32 because 8+0.25(24)=14 and 8+0.75(24)=26; their uniform variance is 24^2/12=48 and SD=sqrt(48). - Independent value: 6.92820323027551 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #596 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-596/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Normal Distribution; Percentiles and Quantiles The two upper-tail probabilities correspond to z-scores 0.449876 and 0.599859. Their 0.149984 difference spans 0.12 units, giving standard deviation 0.800088 and variance 0.640141, so choice B is correct. ### Independent solution ### Setup Convert each upper-tail probability to its cumulative standard-normal z-score. LaTeX: z_1=\Phi^{-1}(1-0.3264)=0.4498757972\ldots LaTeX: z_2=\Phi^{-1}(1-0.2743)=0.5998593133\ldots ### Model The difference between two standardized thresholds eliminates the unknown mean. LaTeX: z_2-z_1=\frac{3.62-3.50}{\sigma} ### Compute Solve for the standard deviation and square it. LaTeX: \sigma=\frac{0.12}{0.5998593133-0.4498757972}=0.8000879238\ldots LaTeX: \sigma^2=0.6401406858\ldots ### Answer The variance rounds to 0.64. LaTeX: \boxed{0.64\quad\text{(B)}} ### Distractor analysis - A: This results from an incorrect normal-table conversion that makes the z-score gap too large and therefore understates the variance. - C: This reports the standard deviation, approximately 0.80, rather than the requested variance. - D: This reports approximately the reciprocal of the standard deviation, 1/0.80=1.25. - E: This reports approximately the reciprocal of the variance, 1/0.64=1.56. ### Verification record - Main value: 0.6401406857843783 - Main method: Convert both upper-tail probabilities to z-scores, use the threshold difference to solve for sigma, and square it. - Independent check: The resulting sigma=0.8000879238 and mu=3.1400598075 reproduce upper tails 0.3264 at 3.50 and 0.2743 at 3.62. - Independent value: 0.6401406857843783 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: demand mean from CV A daily demand X is normally distributed with positive mean. Its coefficient of variation is 0.25, and its 90th percentile is 132.038789138615. Calculate the mean of X. - A: 25.000 - B: 99.029 - C: 100.000 - D: 105.631 - E: 132.039 CV 0.25 gives sigma=0.25mu. The 90th percentile is therefore mu[1+0.25Phi inverse 0.90]. Equating this to 132.03878914 gives mu=100, so choice C is correct. #### Variant solution ### Setup Translate the coefficient of variation into a relation between standard deviation and mean. LaTeX: \frac{\sigma}{\mu}=0.25 LaTeX: \sigma=0.25\mu ### Model Write the normal 90th percentile using its standard-normal quantile. LaTeX: q_{0.90}=\mu+\Phi^{-1}(0.90)\sigma LaTeX: 132.038789138615=\mu\left[1+0.25(1.2815515655)\right] ### Compute Solve the linear equation for the positive mean. LaTeX: 132.038789138615=1.3203878914\mu LaTeX: \mu=100 ### Answer The mean daily demand is 100. LaTeX: \boxed{100.000\quad\text{(C)}} ### Verification record - Main value: 100 - Main method: Express sigma as 0.25mu, substitute into the normal 90th-percentile equation, and solve for mu. - Independent check: At mu=100 and sigma=25, standardizing 132.038789138615 gives z=1.2815515655, whose normal CDF is 0.90; sigma/mu is also 0.25. - Independent value: 100 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #597 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-597/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Uniform Distribution; Deductible Payments The no-deductible mean identifies the uniform support as [1,19]. A deductible of 4 leaves a positive excess over a 15-unit interval; integrating that excess against density 1/18 gives 225/36=6.25, so choice B is correct. ### Independent solution ### Setup Use the no-deductible expected payment to recover the lower endpoint of the uniform loss distribution. LaTeX: \frac{a+19}{2}=10 LaTeX: a=1 ### Model The loss therefore has constant density 1/18 on [1,19], and the second policy pays only the positive excess above 4. LaTeX: X\sim\operatorname{Uniform}(1,19) LaTeX: Y=(X-4)_+ ### Compute Integrate the payment over the part of the support above the deductible. LaTeX: E[Y]=\int_{4}^{19}(x-4)\frac{1}{18}\,dx LaTeX: E[Y]=\frac{(19-4)^2}{36}=6.25 ### Answer The expected payment under the deductible policy is 6.25. LaTeX: \boxed{6.25\quad\text{(B)}} ### Distractor analysis - A: This subtracts the deductible directly from the original mean, 10-4=6, and allows negative payments for losses below 4. - C: This is the mean excess conditional on a payment, (19-4)/2=7.5, and omits the probability that a payment occurs. - D: This scales the original mean by the fraction (19-4)/19, treating the support as if it began at zero instead of integrating the excess. - E: This multiplies the original mean 10 by P(X>4)=15/18 but does not subtract the deductible from paid losses. ### Verification record - Main value: 6.25 - Main method: Infer the lower uniform endpoint from its mean and integrate the positive excess above the deductible. - Independent check: A payment occurs with probability 15/18; conditional on payment, the excess is uniform on [0,15] with mean 7.5, and (15/18)(7.5)=6.25. - Independent value: 6.25 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: research-vessel repair threshold A research vessel's repair-cost index X is uniformly distributed from 0 to 24. A reserve pays the amount of X above an unknown activation threshold d and pays zero otherwise. The expected reserve payment is 3. Determine d. - A: 3 - B: 9 - C: 12 - D: 18 - E: 21 For a uniform index on [0,24], the expected positive excess over d is (24-d)^2/48. Equating this to 3 gives 24-d=12 and hence d=12, so choice C is correct. #### Variant solution ### Setup Represent the reserve payment as the positive part above the unknown threshold. LaTeX: X\sim\operatorname{Uniform}(0,24) LaTeX: R=(X-d)_+ ### Model Integrate the excess payment using the constant uniform density. LaTeX: E[R]=\int_d^{24}(x-d)\frac1{24}\,dx LaTeX: E[R]=\frac{(24-d)^2}{48} ### Compute Set the expected payment equal to its specified value and retain the threshold within the support. LaTeX: \frac{(24-d)^2}{48}=3 LaTeX: (24-d)^2=144 LaTeX: d=12 ### Answer The reserve activates at a cost index of 12. LaTeX: \boxed{d=12\quad\text{(C)}} ### Verification record - Main value: 12 - Main method: Integrate the uniform positive excess, equate (24-d)^2/48 to 3, and solve for the admissible threshold. - Independent check: At d=12, the threshold is exceeded with probability 1/2 and the conditional excess is uniform on [0,12] with mean 6, so the expected payment is (1/2)(6)=3. - Independent value: 12 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #598 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-598/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Normal Distribution; Raw Moments The probability condition gives a mean-to-standard-deviation ratio of Phi inverse 0.92, or 1.405072. Combining this ratio with E[X^2]=mu^2+sigma^2=74 yields variance 24.8804, so choice A is correct. ### Independent solution ### Setup Write the normal variable in terms of its mean and standard deviation and translate the positive-outcome probability into a standard-normal quantile. LaTeX: X\sim N(\mu,\sigma^2) LaTeX: \Pr(X>0)=\Phi\!\left(\frac{\mu}{\sigma}\right)=0.92 ### Model The quantile fixes the ratio of the mean to the standard deviation, while the supplied second raw moment fixes their squared sum. LaTeX: \frac{\mu}{\sigma}=\Phi^{-1}(0.92)=1.4050715603\ldots LaTeX: E[X^2]=\mu^2+\sigma^2=74 ### Compute Substitute the quantile ratio into the second-moment equation and solve for the variance. LaTeX: \left(1.4050715603^2+1\right)\sigma^2=74 LaTeX: \sigma^2=24.8804219219\ldots ### Answer The variance rounds to 24.88. LaTeX: \boxed{24.88\quad\text{(A)}} ### Distractor analysis - B: This uses mu squared equal to z times sigma squared instead of z squared times sigma squared, producing 74/(1+1.40507)=30.77. - C: This writes the second-moment identity as mu squared plus sigma rather than mu squared plus sigma squared, leading to a reported variance near 34.50. - D: This corresponds to using an incorrect standardized distance of about 0.82 in place of the 92nd-percentile value 1.40507. - E: This writes the second moment as mu plus sigma squared, omitting the square on the mean and producing variance about 62.86. ### Verification record - Main value: 24.880421921851084 - Main method: Convert the positive-outcome probability to mu/sigma, substitute into E[X^2]=mu^2+sigma^2, and solve for sigma squared. - Independent check: A direct numerical root of Phi(sqrt((74-v)/v))=0.92 over 0t)=e^{-3\lambda t} ### Compute At a 75th percentile, the survival probability is one-quarter. LaTeX: e^{-3(\ln2/12)t}=0.25 LaTeX: t=\frac{\ln4}{3\ln2/12}=8 ### Answer The 75th percentile of the earliest signal time is 8 minutes. LaTeX: \boxed{8\text{ minutes}\quad\text{(B)}} ### Verification record - Main value: 8 - Main method: Infer each beacon rate from its median, add the three rates for the minimum, and invert the minimum's survival function at 0.25. - Independent check: At t=8, each beacon survives with probability exp[-(ln2/12)8]=2^(-2/3); cubing for three independent beacons gives 1/4, hence the minimum CDF is 3/4. - Independent value: 8 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #600 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-600/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Cumulative Probabilities; Geometric Distribution Evaluating the loss CDF at the reimbursement limit gives probability 0.65 of full reimbursement and 0.35 of a partial shortfall. Independence makes the probability of two full reimbursements followed by a shortfall equal to 0.65 squared times 0.35, or 0.147875, so choice D is correct. ### Independent solution ### Setup A loss is fully reimbursed exactly when it does not exceed the payment limit. LaTeX: p=\Pr(X\le0.80)=F(0.80) ### Model Evaluate the supplied CDF and use its complement for a partially unreimbursed loss. LaTeX: p=\frac{0.80+15(0.80)^2}{16}=0.65 LaTeX: q=1-p=0.35 ### Compute The first shortfall occurs on the third event only when the first two are fully reimbursed and the third exceeds the limit. LaTeX: \Pr(T=3)=p^2q LaTeX: \Pr(T=3)=(0.65)^2(0.35)=0.147875 ### Answer The required probability rounds to 0.148. LaTeX: \boxed{0.148\quad\text{(D)}} ### Distractor analysis - A: This uses the four-event pattern 0.65(0.35)^3, inserting three shortfalls after one fully reimbursed loss instead of locating the first shortfall on event three. - B: This computes 0.65(0.35)^2, requiring one full reimbursement followed by two shortfalls rather than two full reimbursements followed by one shortfall. - C: Omitting the square on 0.80 inside the CDF gives p=0.80 and then p squared times (1-p)=0.128. - E: This is the probability 0.35 that one loss exceeds the limit and ignores the required position of the first such loss. ### Verification record - Main value: 0.147875 - Main method: Evaluate the CDF at the reimbursement limit and multiply the independent sequence probabilities for full, full, and partial reimbursement. - Independent check: The index of the first loss above the limit is geometric with event probability 0.35; its mass at 3 is (1-0.35)^2(0.35)=0.147875. - Independent value: 0.147875 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: polymer curing flags Four independent polymer samples have curing index Y with distribution function F(y)=y^2 for 0<=y<=1. A scanner flags a sample when its curing index exceeds 0.50. Calculate the probability that exactly two of the four samples are flagged. - A: 0.035 - B: 0.211 - C: 0.262 - D: 0.422 - E: 0.750 A sample is flagged with probability 1-F(0.5)=0.75. The number flagged among four independent samples is binomial, and its mass at two is 0.2109375, which rounds to choice B. #### Variant solution ### Setup Use the distribution function to find the flag probability for one sample. LaTeX: q=\Pr(Y>0.50)=1-F(0.50)=1-(0.50)^2=0.75 ### Model Independence makes the number of flagged samples a binomial count. LaTeX: K\sim\operatorname{Binomial}(4,0.75) ### Compute Evaluate the binomial mass for exactly two flags. LaTeX: \Pr(K=2)=\binom42(0.75)^2(0.25)^2 LaTeX: \Pr(K=2)=0.2109375 ### Answer The probability rounds to 0.211. LaTeX: \boxed{0.211\quad\text{(B)}} ### Verification record - Main value: 0.2109375 - Main method: Evaluate the CDF complement at 0.50 and use the binomial mass for two successes in four independent trials. - Independent check: Each named pair of flagged samples with the other two unflagged has probability (0.75)^2(0.25)^2=0.03515625; summing the six disjoint pairs gives 0.2109375. - Independent value: 0.2109375 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #601 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-601/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Insurance Payment Variables; Tail Probabilities The density corresponds to a Pareto tail S(x)=(1000/x)^3 above 1000. The expected positive excess over 1100 is the integral of this survival function, equal to 1000^3 divided by 2(1100)^2, or 413.2231, so choice C is correct. ### Independent solution ### Setup Integrate the density once to express the loss survival function above its lower endpoint. LaTeX: S(x)=\Pr(X>x)=\left(\frac{1000}{x}\right)^3,\qquad x\ge1000 ### Model For an ordinary deductible, the expected positive excess equals the tail integral beginning at the deductible. LaTeX: E[(X-1100)_+]=\int_{1100}^{\infty}S(x)\,dx ### Compute Evaluate the power-tail integral. LaTeX: E[(X-1100)_+]=1000^3\int_{1100}^{\infty}x^{-3}\,dx LaTeX: E[(X-1100)_+]=\frac{1000^3}{2(1100)^2}=413.2231405\ldots ### Answer The expected payout rounds to the listed value 413. LaTeX: \boxed{413\quad\text{(C)}} ### Distractor analysis - A: Omitting the leading factor 3 from the Pareto density divides the correct integral by three, giving 137.74 and the nearby listed value 140. - B: This subtracts the deductible from the unconditional mean, 1500-1100=400, allowing negative payments from losses below the deductible. - D: This is 1100P(X>1100)=826.45, the expected deductible amount removed on payment-producing losses rather than the remaining payout. - E: This is E[X 1{X>1100}]=1239.67, the loss contribution above the threshold before subtracting the deductible on those losses. ### Verification record - Main value: 413.22314049586777 - Main method: Derive the Pareto survival function and integrate it from the deductible to infinity. - Independent check: The exceedance probability is (1000/1100)^3; conditional on exceedance, the Pareto mean excess is 1100/(3-1)=550, and their product is 413.2231405. - Independent value: 413.22314049586777 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: capped endurance reading A battery endurance X has survival function 1 for 0<=x<200 and (200/x)^2 for x>=200. A test rig records the smaller of X and an upper display limit L. If the expected displayed reading is 360, determine L. - A: 360 - B: 400 - C: 500 - D: 1000 - E: 2000 The limited expectation is the survival integral through L, equal to 400-40000/L. Setting this expression to 360 gives L=1000, so choice D is correct. #### Variant solution ### Setup Express a capped nonnegative expectation as an integral of the survival function up to the cap. LaTeX: E[\min(X,L)]=\int_0^L\Pr(X>x)\,dx ### Model Split the integral at the Pareto lower endpoint. LaTeX: E[\min(X,L)]=\int_0^{200}1\,dx+\int_{200}^{L}\left(\frac{200}{x}\right)^2dx LaTeX: E[\min(X,L)]=400-\frac{40000}{L} ### Compute Equate the limited expectation to the observed display mean and solve for the upper limit. LaTeX: 400-\frac{40000}{L}=360 LaTeX: \frac{40000}{L}=40 LaTeX: L=1000 ### Answer The display limit is 1000 endurance units. LaTeX: \boxed{L=1000\quad\text{(D)}} ### Verification record - Main value: 1000 - Main method: Integrate the survival function only up to the unknown cap, set the limited expectation equal to 360, and solve for the cap. - Independent check: At L=1000, E[X 1{X<=L}]=400-80000/1000=320 and L P(X>L)=1000(200/1000)^2=40; their sum is 360. - Independent value: 1000 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #602 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-602/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Discrete Random Variables; Cumulative Probabilities; Probability Ratios The four relevant count probabilities are proportional to 60, 20, 5, and 1. Their total weight 86 represents probability 0.95, while the first three weights total 85, giving 0.95(85/86)=0.938953 and choice D. ### Independent solution ### Setup Let q be the probability assigned to count three. Propagate the supplied ratios downward through counts two, one, and zero. LaTeX: p_3=q,\qquad p_2=5q LaTeX: p_1=4p_2=20q,\qquad p_0=3p_1=60q ### Model The cumulative probability through count three fixes the common scale q. LaTeX: p_0+p_1+p_2+p_3=(60+20+5+1)q=86q LaTeX: 86q=0.95,\qquad q=\frac{0.95}{86} ### Compute For a count no greater than two, retain the first three probability weights. LaTeX: \Pr(X\le2)=(60+20+5)q LaTeX: \Pr(X\le2)=0.95\left(\frac{85}{86}\right)=0.9389534884\ldots ### Answer The probability rounds to 0.94. LaTeX: \boxed{0.94\quad\text{(D)}} ### Distractor analysis - A: This is much too small because it effectively retains only the dominant low-count mass and discards material probability from the other allowed counts. - B: This can result from reversing or truncating one of the probability-ratio links. Every ratio must be propagated from count three before normalization. - C: Omitting the zero-count category gives weights 20, 5, and 1; normalizing those to 0.95 and then excluding count three produces about 0.91. - E: A probability through count two cannot exceed the given probability through count three. This choice violates nesting of cumulative events. ### Verification record - Main value: 0.938953488372093 - Main method: Express the probabilities at counts zero through three as q times weights 60, 20, 5, and 1, scale their sum to 0.95, and retain weights totaling 85. - Independent check: Conditional on the count being at most three, the relative distribution has total weight 86 and assigns weight 85 to counts at most two. Thus P(X at most 2)=0.95(85/86). - Independent value: 0.938953488372093 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: even-ranked search result A retrieval system returns a result at rank R in {1,2,3,4}. Before normalization, rank r receives probability weight r(5-r). An analyst is told that the returned rank is even. Calculate the conditional expected rank. - A: 0.400 - B: 0.600 - C: 2.500 - D: 2.800 - E: 4.000 The four rank weights are 4, 6, 6, and 4. Conditioning on an even rank leaves weights 6 and 4 at ranks two and four, so the conditional mean is 2(0.6)+4(0.4)=2.8 and choice D. #### Variant solution ### Setup Evaluate the unnormalized weight at each possible rank. LaTeX: \bigl(w_1,w_2,w_3,w_4\bigr)=\bigl(1(4),2(3),3(2),4(1)\bigr)=(4,6,6,4) ### Model After the even-rank condition, only ranks two and four remain. Normalize their two weights. LaTeX: \Pr(R=2\mid R\text{ even})=\frac6{6+4}=0.6 LaTeX: \Pr(R=4\mid R\text{ even})=\frac4{6+4}=0.4 ### Compute Weight the two remaining ranks by their conditional probabilities. LaTeX: \operatorname{E}[R\mid R\text{ even}]=2(0.6)+4(0.4)=2.8 ### Answer The conditional expected rank is 2.800. LaTeX: \boxed{2.800\quad\text{(D)}} ### Verification record - Main value: 2.8 - Main method: Evaluate the proportional weights and normalize the two even-rank weights before calculating their weighted average. - Independent check: View the even-conditioned population as ten equal-weight tickets: six carry rank two and four carry rank four. Their total rank is 28, so their average is 28/10. - Independent value: 2.8 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #603 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-603/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Bayes' Theorem; Poisson Distribution; Mixture Distributions The zero-count relationship determines the first class's Poisson mean as 3-ln(3)=1.901388. Combining the two exact-count likelihoods with prior weights 0.40 and 0.60 gives posterior probability 0.445490, so choice D is correct. ### Independent solution ### Setup Let S denote the first class and N the second. Their prior probabilities are 0.40 and 0.60, and the second class has Poisson mean three. LaTeX: \Pr(S)=0.40,\qquad \Pr(N)=0.60 LaTeX: \lambda_N=3 ### Model Translate the zero-count comparison into an equation for the unknown mean of the first class. LaTeX: e^{-\lambda_S}=3e^{-3} LaTeX: \lambda_S=3-\ln 3=1.9013877113\ldots ### Compute Evaluate the two likelihoods at count two and apply Bayes' theorem. LaTeX: \Pr(X=2\mid S)=e^{-\lambda_S}\frac{\lambda_S^2}{2!}=0.2699909325\ldots LaTeX: \Pr(X=2\mid N)=e^{-3}\frac{3^2}{2!}=0.2240418077\ldots LaTeX: \Pr(S\mid X=2)=\frac{0.40(0.2699909325)}{0.40(0.2699909325)+0.60(0.2240418077)} LaTeX: \Pr(S\mid X=2)=0.4454901561\ldots ### Answer The posterior probability rounds to 0.45. LaTeX: \boxed{0.45\quad\text{(D)}} ### Distractor analysis - A: This is the joint probability Pr(S and X=2), approximately 0.108. It is the Bayes numerator before normalization. - B: This is the likelihood Pr(X=2 given S), approximately 0.270. It does not incorporate either prior class weight. - C: This is the prior probability of class S. Observing the count changes that probability because the two Poisson likelihoods differ. - E: This is the posterior complement, the probability of the other class after the count is observed. ### Verification record - Main value: 0.4454901561406148 - Main method: Infer the unknown Poisson mean from the zero-count ratio, evaluate both count-two likelihoods, and normalize their prior-weighted values. - Independent check: Posterior odds equal prior odds 2/3 times the likelihood ratio P(X=2|S)/P(X=2|N). Converting those odds back to a probability gives 0.4454901561. - Independent value: 0.4454901561406148 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: source allocation after a combined count Two independent particle counters register Poisson counts with means 2 and 3 during the same interval. The combined display reports exactly four particles. Conditional on that total, calculate the probability that the first counter registered at least two particles. - A: 0.1296 - B: 0.3456 - C: 0.4752 - D: 0.5248 - E: 0.8704 Given the combined total, the first source count is binomial with four trials and success probability 2/5. Its probability of at least two is 1-0.1296-0.3456=0.5248, selecting choice D. #### Variant solution ### Setup Let X and Y be the independent source counts and T their observed sum. LaTeX: X\sim\operatorname{Poisson}(2),\qquad Y\sim\operatorname{Poisson}(3) LaTeX: T=X+Y=4 ### Model Independent Poisson sources split a fixed combined total in proportion to their means. LaTeX: X\mid(T=4)\sim\operatorname{Binomial}\!\left(4,\frac{2}{2+3}\right) ### Compute Use the complement of zero or one particle from the first counter. LaTeX: \Pr(X\ge2\mid T=4)=1-\left(\frac35\right)^4-4\left(\frac25\right)\left(\frac35\right)^3 LaTeX: \Pr(X\ge2\mid T=4)=1-0.1296-0.3456=0.5248 ### Answer The conditional allocation probability is 0.5248. LaTeX: \boxed{0.5248\quad\text{(D)}} ### Verification record - Main value: 0.5248 - Main method: Use the Poisson-splitting conditional binomial distribution and subtract its zero- and one-count probabilities. - Independent check: Directly sum P(X=x)P(Y=4-x) for x=2,3,4 and divide by the Poisson(5) mass at four; the ratio simplifies to 0.5248. - Independent value: 0.5248 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #604 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-604/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Independence; Independent Increments A zero count across six equal independent periods has probability exp(-6 lambda)=0.008. The zero-count probability across the final two periods is therefore the cube root of 0.008, or 0.20; the observation in a disjoint period does not alter it, so choice A is correct. ### Independent solution ### Setup Let lambda be the common Poisson mean for one active month. A six-month total then has mean 6 lambda. LaTeX: N_6\sim\operatorname{Poisson}(6\lambda) LaTeX: \Pr(N_6=0)=e^{-6\lambda}=0.008 ### Model The two target months form a disjoint two-month count with mean 2 lambda. Their count is independent of what happened in the earlier observed month. LaTeX: N_2\sim\operatorname{Poisson}(2\lambda) LaTeX: \Pr(N_2=0\mid\text{June observation})=\Pr(N_2=0)=e^{-2\lambda} ### Compute Relate the two-month survival factor to the supplied six-month factor. LaTeX: e^{-2\lambda}=\left(e^{-6\lambda}\right)^{1/3} LaTeX: e^{-2\lambda}=(0.008)^{1/3}=0.20 ### Answer The conditional probability is 0.20. LaTeX: \boxed{0.20\quad\text{(A)}} ### Distractor analysis - B: This is a numerical cube-root error. Since 0.20 cubed is exactly 0.008, no approximation above 0.20 is needed. - C: This is close to the probability that both target months are nonempty, obtained from the one-month nonzero probability. The requested event has both counts equal to zero. - D: This is close to the probability that a single active month is nonempty. It changes both the event direction and the two-month exposure length. - E: Multiplying the two-month probability 0.20 by the three two-month blocks in six months treats intersecting zero-count requirements as a disjoint union. Independent block probabilities multiply. ### Verification record - Main value: 0.2 - Main method: Use the Poisson zero-count formula and take the cube root that converts six equal months of exposure to two months; independence removes the earlier-month condition. - Independent check: Solving explicitly gives lambda=-ln(0.008)/6=0.8047189562. Substitution into exp(-2 lambda) returns 0.2000000000. - Independent value: 0.2 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: event locations within a shift Signals arrive according to a homogeneous Poisson process during an eight-hour research shift. Exactly three signals were observed over the full shift. Conditional on that total, calculate the probability that at least two signals arrived during the first two hours. - A: 0.0156 - B: 0.1406 - C: 0.1563 - D: 0.4219 - E: 0.5781 Given the three-event total, each event independently falls in the first quarter of the shift with probability 1/4. A Binomial(3,1/4) count is at least two with probability 10/64=0.15625, so choice C is correct. #### Variant solution ### Setup Let K count signals in the first two hours, given that the eight-hour total equals three. LaTeX: \frac{\text{early exposure}}{\text{total exposure}}=\frac28=\frac14 ### Model Conditional on a homogeneous Poisson total, event locations are independent and uniform across the observation window. LaTeX: K\mid(N_8=3)\sim\operatorname{Binomial}\!\left(3,\frac14\right) ### Compute Add the probabilities of exactly two and exactly three early signals. LaTeX: \Pr(K\ge2\mid N_8=3)=\binom32\left(\frac14\right)^2\left(\frac34\right)+\left(\frac14\right)^3 LaTeX: \Pr(K\ge2\mid N_8=3)=\frac9{64}+\frac1{64}=\frac5{32}=0.15625 ### Answer The conditional probability rounds to 0.1563. LaTeX: \boxed{0.1563\quad\text{(C)}} ### Verification record - Main value: 0.15625 - Main method: Use the conditional uniform-location property of a Poisson process and evaluate the upper tail of a Binomial(3,1/4) variable. - Independent check: Split the shift into independent Poisson counts with means 2r and 6r. Dividing the joint probabilities for early counts two and three with total three by the Poisson(8r) mass at three simplifies to 5/32 for every positive r. - Independent value: 0.15625 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #605 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-605/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Bayes' Theorem; Exponential Distribution; Mixture Distributions The three-year failure probabilities are 1-exp(-3/2) for type A and 1-exp(-3/4) for type B. Bayes' theorem with prior weights 0.10 and 0.90 gives posterior type-A probability 0.140595, so choice B is correct. ### Independent solution ### Setup Let A and B identify the two product types, and let F be failure before year three. LaTeX: \Pr(A)=0.10,\qquad \Pr(B)=0.90 LaTeX: F=\{T<3\} ### Model Use each exponential survival function to calculate the likelihood of the observed early failure. LaTeX: \Pr(F\mid A)=1-e^{-3/2}=0.7768698399\ldots LaTeX: \Pr(F\mid B)=1-e^{-3/4}=0.5276334473\ldots ### Compute Weight the two likelihoods by their prior probabilities and normalize. LaTeX: \Pr(A\mid F)=\frac{0.10(1-e^{-3/2})}{0.10(1-e^{-3/2})+0.90(1-e^{-3/4})} LaTeX: \Pr(A\mid F)=0.1405953989\ldots ### Answer The posterior probability rounds to 0.14. LaTeX: \boxed{0.14\quad\text{(B)}} ### Distractor analysis - A: This is the joint probability of type A and early failure, about 0.0777. It is the Bayes numerator before division by the total early-failure probability. - C: This is the probability that a type-B item survives three years, exp(-3/4). It reverses the observed event and omits Bayes' theorem. - D: This is close to the overall early-failure probability, the denominator that normalizes the two weighted likelihoods. It is not a type posterior. - E: This is the likelihood of early failure for type A, 1-exp(-3/2), without accounting for the small prior share of that type. ### Verification record - Main value: 0.140595398883898 - Main method: Evaluate both exponential CDFs at three years and apply Bayes' theorem with prior type probabilities 0.10 and 0.90. - Independent check: Posterior odds equal prior odds 1/9 multiplied by likelihood ratio (1-exp(-3/2))/(1-exp(-3/4)); converting the resulting odds to probability gives 0.1405953989. - Independent value: 0.140595398883898 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: one module failure by inspection A research rack contains four independent modules whose lifetimes are exponential with mean 5 years. At an inspection two years after installation, calculate the probability that exactly one module has failed. - A: 0.2019 - B: 0.3297 - C: 0.3972 - D: 0.7981 - E: 1.3187 Each module fails by year two with probability 1-exp(-0.4). The number failed is binomial with four trials, giving exactly-one probability 4(1-exp(-0.4))exp(-1.2)=0.397191 and choice C. #### Variant solution ### Setup Compute the failure and survival probabilities for one module at the inspection time. LaTeX: p=\Pr(T\le2)=1-e^{-2/5}=1-e^{-0.4} LaTeX: 1-p=e^{-0.4} ### Model Independence makes the number of failed modules at year two binomial. LaTeX: K\sim\operatorname{Binomial}\!\left(4,1-e^{-0.4}\right) ### Compute Choose which one module failed and require the other three to survive. LaTeX: \Pr(K=1)=\binom41(1-e^{-0.4})(e^{-0.4})^3 LaTeX: \Pr(K=1)=4(1-e^{-0.4})e^{-1.2}=0.3971907757\ldots ### Answer The exactly-one probability rounds to 0.3972. LaTeX: \boxed{0.3972\quad\text{(C)}} ### Verification record - Main value: 0.3971907756701868 - Main method: Convert one exponential lifetime to a two-year failure probability and evaluate the exactly-one mass of a four-trial binomial variable. - Independent check: Integrate over the first failure time t from zero to two: the density that one of four modules fails first at t is (4/5)exp(-4t/5), and the remaining three survive to year two with factor exp(-3(2-t)/5). The integral is 0.3971907757. - Independent value: 0.3971907756701868 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #606 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-606/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Joint Distributions; Conditional Distributions; Variance Fixing the second count at one gives relative masses 12, 20, and 8 for first-count values zero, one, and two. After normalization their mean is 0.9 and second moment is 1.3, so the conditional variance is 1.3-0.9^2=0.49 and choice B. ### Independent solution ### Setup Evaluate the joint probability rule at the conditioning value. Common factors cancel, leaving three relative masses. LaTeX: p(0,1):p(1,1):p(2,1)=12:20:8 ### Model Normalize the three masses by their total of forty to obtain the conditional distribution. LaTeX: \bigl(\Pr(X=0\mid Y=1),\Pr(X=1\mid Y=1),\Pr(X=2\mid Y=1)\bigr)=\left(\frac{12}{40},\frac{20}{40},\frac8{40}\right) LaTeX: \bigl(\Pr(X=0\mid Y=1),\Pr(X=1\mid Y=1),\Pr(X=2\mid Y=1)\bigr)=(0.3,0.5,0.2) ### Compute Calculate the conditional first and second moments, then subtract the squared mean. LaTeX: \operatorname{E}[X\mid Y=1]=0(0.3)+1(0.5)+2(0.2)=0.9 LaTeX: \operatorname{E}[X^2\mid Y=1]=0^2(0.3)+1^2(0.5)+2^2(0.2)=1.3 LaTeX: \operatorname{Var}(X\mid Y=1)=1.3-(0.9)^2=0.49 ### Answer The conditional variance is 49/100. LaTeX: \boxed{\frac{49}{100}\quad\text{(B)}} ### Distractor analysis - A: This can result from losing a conditional mass or using an incorrect normalizing total before calculating moments. The three retained weights must sum to forty. - C: This is the variance of a discrete uniform variable on zero, one, and two. The actual conditional masses are not equal. - D: The value 0.70 is the conditional standard deviation, the square root of 0.49, rather than the variance. - E: This treats a rounded moment or scale value as the variance. Variance requires the second moment minus the square of the full conditional mean. ### Verification record - Main value: 0.49 - Main method: Substitute the conditioning value into the joint rule, normalize weights 12, 20, and 8, and apply the first-two-moments variance identity. - Independent check: Compute the centered conditional sum directly: 0.3(0-0.9)^2+0.5(1-0.9)^2+0.2(2-0.9)^2=0.49. - Independent value: 0.49 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: horizontal error on a disk chord A robotic alignment error (X,Y) is uniformly distributed over the unit disk x^2+y^2<1. On a particular run, the vertical error is observed to be Y=3/5. Calculate the conditional variance of the horizontal error X. - A: 4/75 - B: 16/75 - C: 1/4 - D: 16/25 - E: 4/5 At vertical coordinate 3/5, the disk's horizontal chord runs from -4/5 to 4/5. The conditional horizontal coordinate is uniform on that chord, whose variance is (8/5)^2/12=16/75, selecting choice B. #### Variant solution ### Setup Intersect the unit disk with the observed vertical coordinate to find the allowable horizontal range. LaTeX: x^2+\left(\frac35\right)^2<1 LaTeX: |x|<\sqrt{1-\frac9{25}}=\frac45 ### Model The joint density is constant over the disk. Along a fixed horizontal chord, the conditional density is therefore uniform. LaTeX: X\mid\left(Y=\frac35\right)\sim\operatorname{Uniform}\!\left(-\frac45,\frac45\right) ### Compute Use the uniform variance formula with full chord width eight-fifths. LaTeX: \operatorname{Var}\!\left(X\mid Y=\frac35\right)=\frac{\left(\frac85\right)^2}{12} LaTeX: \operatorname{Var}\!\left(X\mid Y=\frac35\right)=\frac{16}{75}=0.213333\ldots ### Answer The conditional variance is 16/75. LaTeX: \boxed{\frac{16}{75}\quad\text{(B)}} ### Verification record - Main value: 0.21333333333333335 - Main method: Condition the uniform disk density on Y=3/5, identify the horizontal chord [-4/5,4/5], and use the uniform variance formula. - Independent check: The conditional density is 5/8 on [-4/5,4/5]. Direct integration of x squared against this density gives 16/75, while symmetry gives conditional mean zero. - Independent value: 0.21333333333333335 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #607 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-607/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Disjoint Events; Interval Probabilities This is a cancellation problem involving five overlapping two-unit intervals. Alternating their probabilities cancels every interior one-unit bin and leaves the two requested edge bins, whose combined probability is 0.18, so choice B is correct. ### Independent solution ### Setup Let q_k be the probability of a duration in the one-unit interval beginning at k, for k from two through seven. LaTeX: q_k=\Pr(k\le T3.2)=1-F(3.2) ### Compute Evaluate the distribution function at the deductible and take its complement. LaTeX: F(3.2)=\frac{3.2^2+3.2}{30}=\frac{13.44}{30}=0.448 LaTeX: \Pr(X>3.2)=1-0.448=0.552 ### Answer The insurer makes a positive payment with probability 0.552. LaTeX: \boxed{0.552\quad\text{(D)}} ### Distractor analysis - A: Treating the supplied polynomial CDF as though it were a density and normalizing its integral gives an erroneous lower-tail probability near 0.296. - B: This treats the loss as uniform on the support and uses the remaining length (5-3.2)/5=0.36, ignoring the stated distribution function. - C: This is F(3.2), the probability that the loss does not exceed the deductible. It is the complement of the positive-payment event. - E: This is the complement of choice A, obtained by treating the CDF expression as a density before forming the upper tail. ### Verification record - Main value: 0.552 - Main method: Normalize the CDF at x=5, evaluate it at the deductible 3.2, and take the upper-tail complement. - Independent check: Differentiating gives density (2x+1)/30 on [0,5]. Direct integration from 3.2 to 5 equals [(x^2+x)/30] from 3.2 to 5, which is 0.552. - Independent value: 0.552 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: transformed polymer response A normalized polymer response X has density f(x)=k x(2-x) for 0m) LaTeX: \operatorname{E}[Y]=\frac{m^2}{4000}+m\left(1-\frac{m}{2000}\right)=m-\frac{m^2}{4000} ### Compute Set the limited expectation equal to the supplied value and solve the quadratic. LaTeX: m-\frac{m^2}{4000}=910 LaTeX: m^2-4000m+3640000=0 LaTeX: m=1400\ \text{or}\ 2600 ### Answer A policy maximum above 2000 would be outside the stated loss range, leaving 1400 as the admissible solution. LaTeX: \boxed{m=1400\quad\text{(C)}} ### Distractor analysis - A: This copies the expected payment as the policy maximum. Because some losses exceed the maximum, the payment mean is not generally equal to the maximum. - B: Substitution gives 1150-1150^2/4000=819.375, so this proposed maximum does not reproduce the stated expectation. - D: Substitution gives 1600-1600^2/4000=960, which is larger than the stated expected payment. - E: This sets m/2=910 as though every payment were uniform on [0,m]. Losses above m instead create a probability mass at the maximum. ### Verification record - Main value: 1400 - Main method: Integrate the uniform limited payment, solve m-m^2/4000=910, and retain the root inside [0,2000]. - Independent check: The tail-integral identity gives E[min(X,m)]=integral from 0 to m of (1-t/2000) dt. At m=1400 this equals 1400-1400^2/4000=910. - Independent value: 1400 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: coinsured repair layer Equipment repair loss X is uniform from 0 to 1000. A contract pays 80% of the portion of a loss between 200 and 700, so its payment is Y=0.8 min((X-200)+,500). Calculate the variance of Y. - A: 162.07 - B: 26,266.67 - C: 41,041.67 - D: 48,400.00 - E: 74,666.67 Before coinsurance, the layer payment has first moment 275 and second moment 350000/3. Its variance is 123125/3, and multiplying by 0.8^2 gives Var(Y)=78800/3=26266.67, so choice B is correct. #### Variant solution ### Setup Separate the three-part contract design into an underlying layer amount Z and the 80% coinsurance factor. LaTeX: Z=\min\!\left((X-200)_+,500\right) LaTeX: Y=0.8Z ### Model For losses between 200 and 700, Z increases continuously from zero to 500. Losses at or above 700 produce a point mass at the layer maximum. LaTeX: \Pr(X\ge700)=0.30 ### Compute Calculate the first two layer moments, form its variance, and then square the coinsurance factor. LaTeX: \operatorname{E}[Z]=\int_0^{500}\frac{z}{1000}\,dz+500(0.30)=275 LaTeX: \operatorname{E}[Z^2]=\int_0^{500}\frac{z^2}{1000}\,dz+500^2(0.30)=\frac{350000}{3} LaTeX: \operatorname{Var}(Y)=0.8^2\left(\frac{350000}{3}-275^2\right)=\frac{78800}{3}=26266.6667\ldots ### Answer The contract-payment variance is approximately 26,266.67 square currency units. LaTeX: \boxed{26266.67\quad\text{(B)}} ### Verification record - Main value: 26266.666666666668 - Main method: Compute the first two moments of the deductible-and-cap layer, subtract the squared mean, and apply the squared coinsurance factor. - Independent check: The tail identities give E[Z]=integral from 0 to 500 of (800-t)/1000 dt=275 and E[Z^2]=integral from 0 to 500 of 2t(800-t)/1000 dt=350000/3, confirming the same variance. - Independent value: 26266.666666666668 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #611 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-611/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Continuous Random Variables; Piecewise Density; Percentiles and Quantiles This is a percentile calculation under a symmetric triangular density. Normalization gives c=1/9, and solving (x-5)^2/18=0.30 on the rising half gives x=5+sqrt(5.4)=7.32379, so choice E is correct. ### Independent solution ### Setup Normalize the two linear pieces. Geometrically they form two congruent triangles, each with base three and height 3c. LaTeX: 1=2\left(\frac12\right)(3)(3c)=9c LaTeX: c=\frac19 ### Model Half of the total density lies below the center at eight, so the 30th percentile belongs to the rising left branch. LaTeX: F(8)=\frac12 LaTeX: F(x)=\int_5^x\frac{t-5}{9}\,dt=\frac{(x-5)^2}{18},\qquad 5\le x\le8 ### Compute Set the left-branch cumulative probability equal to 0.30 and use the positive distance from the lower endpoint. LaTeX: \frac{(x-5)^2}{18}=0.30 LaTeX: (x-5)^2=5.4 LaTeX: x=5+\sqrt{5.4}=7.3237900077\ldots ### Answer The 30th percentile rounds to 7.32. LaTeX: \boxed{7.32\quad\text{(E)}} ### Distractor analysis - A: This places the percentile almost at the lower endpoint, where the triangular density is near zero and too little probability has accumulated. - B: This also understates the quantile by treating the rising density too much like a constant height instead of integrating its triangular area. - C: This is the 30%-of-range location 5+0.30(11-5)=6.80, which would apply to a uniform density rather than the stated triangular density. - D: This results from underestimating the area needed on the rising branch; the correct CDF is quadratic in the distance x-5, not linear. ### Verification record - Main value: 7.32379000772445 - Main method: Normalize the symmetric triangular density, integrate its rising branch, and invert the CDF at probability 0.30. - Independent check: A triangle from 5 to x has area one-half times base (x-5) times height (x-5)/9. At x=5+sqrt(5.4), this geometric area is 5.4/18=0.30. - Independent value: 7.32379000772445 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: maximum strength-ratio median Three independently manufactured fibers have strength ratios X1, X2, and X3, each with density f(x)=2x for 010)=e^{-10c}=0.5 LaTeX: c=\frac{\ln2}{10} ### Model Separate the two payment levels into disjoint time intervals. LaTeX: E[W]=35\Pr(T\le5)+25\Pr(55), which reverses failure and survival at the first warranty boundary and ignores the second payment tier. - E: This extends the second payment beyond the end of the warranty instead of assigning zero after 7.5 years. ### Verification record - Main value: 13.063843250600513 - Main method: Infer the exponential rate from its median and sum payment times probability over the two disjoint covered intervals. - Independent check: Direct integration of the piecewise payment against c exp(-ct) over [0,5] and (5,7.5] gives 13.0638432506. - Independent value: 13.063843250600513 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: capped operating credits A field instrument remains operational for an exponential time T with mean 4 years. A research program earns 12 credits for each year of operation, but only during the instrument's first three years. The credited amount is proportional for partial years. Calculate the expected total credit. - A: 5.668 - B: 6.332 - C: 25.326 - D: 36.000 - E: 48.000 The credited operating time is min(T,3). Integrating its survival function gives 4(1-e^(-0.75)) years, so the expected credit is 48(1-e^(-0.75))=25.32640547 and choice C is correct. #### Variant solution ### Setup Represent the credited duration by the operating lifetime capped at three years. LaTeX: T\sim\operatorname{Exponential}\!\left(\text{mean }4\right) LaTeX: C=12\min(T,3) ### Model Use the survival-integral identity for a capped nonnegative duration. LaTeX: E[\min(T,3)]=\int_0^3\Pr(T>t)\,dt LaTeX: \Pr(T>t)=e^{-t/4} ### Compute Evaluate the capped lifetime integral and multiply by the annual credit rate. LaTeX: E[C]=12\int_0^3e^{-t/4}\,dt LaTeX: E[C]=48(1-e^{-0.75})=25.3264054684\ldots ### Answer The expected total credit rounds to 25.326. LaTeX: \boxed{25.326\quad\text{(C)}} ### Verification record - Main value: 25.326405468431297 - Main method: Integrate the exponential survival function through the three-year cap and multiply by 12. - Independent check: Decompose min(T,3) as T on T<=3 and 3 on T>3; evaluating E[T 1(T<=3)]+3P(T>3) and multiplying by 12 gives 25.3264054684. - Independent value: 25.326405468431297 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #613 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-613/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Variance and Standard Deviation The first Poisson mean is the square of its standard deviation, or 2.25. The ratio of zero-count probabilities gives the second mean as 2.25-ln(1.1), whose square root is 1.467886, so choice D is correct. ### Independent solution ### Setup For a Poisson count, the mean and variance are equal. Convert the first standard deviation into its mean. LaTeX: \lambda_1=(1.50)^2=2.25 ### Model Express the stated zero-count probability ratio with the Poisson mass at zero. LaTeX: e^{-\lambda_2}=1.1e^{-\lambda_1} LaTeX: \lambda_2=\lambda_1-\ln(1.1) ### Compute Calculate the second mean and then take its square root to obtain the standard deviation. LaTeX: \lambda_2=2.25-\ln(1.1)=2.1546898202\ldots LaTeX: \operatorname{SD}(N_2)=\sqrt{\lambda_2}=1.4678861741\ldots ### Answer The second standard deviation rounds to 1.47. LaTeX: \boxed{1.47\quad\text{(D)}} ### Distractor analysis - A: This reduces the second Poisson mean far too much by treating the 1.1 probability multiplier as a direct linear change instead of taking its logarithm. - B: This similarly applies the zero-probability ratio to the standard deviation scale rather than to the exponential zero-count probability. - C: This subtracts ln(1.1) directly from the first standard deviation, giving 1.4047, instead of subtracting it from the Poisson mean. - E: This increases the standard deviation even though a larger zero-count probability requires a smaller Poisson mean. ### Verification record - Main value: 1.4678861741278426 - Main method: Square the first standard deviation to get its Poisson mean, take logarithms of the zero-count probability ratio, and square-root the second mean. - Independent check: At lambda2=2.25-ln(1.1), the ratio exp(-lambda2)/exp(-2.25)=exp(ln1.1)=1.1, and lambda2 equals 1.4678861741 squared. - Independent value: 1.4678861741278426 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: sampled fault reports A server's number N of fault reports in one cycle is Poisson. The coefficient of variation of N is 0.40. Each report is independently selected for a detailed audit with probability 0.25; let M be the selected count. Calculate the standard deviation of M. - A: 0.40 - B: 0.63 - C: 0.79 - D: 1.25 - E: 2.50 For a Poisson count, CV=1/sqrt(lambda), so CV 0.40 gives lambda=6.25. Independent thinning makes M Poisson with mean 0.25(6.25)=1.5625 and standard deviation 1.25, so choice D is correct. #### Variant solution ### Setup Recover the original Poisson mean from its coefficient of variation. LaTeX: \operatorname{CV}(N)=\frac{\sqrt\lambda}{\lambda}=\frac1{\sqrt\lambda}=0.40 LaTeX: \lambda=\frac1{0.40^2}=6.25 ### Model Independent selection thins a Poisson count to another Poisson count. LaTeX: M\sim\operatorname{Poisson}(0.25\lambda) LaTeX: E[M]=\operatorname{Var}(M)=1.5625 ### Compute Take the square root of the thinned-count variance. LaTeX: \operatorname{SD}(M)=\sqrt{1.5625}=1.25 ### Answer The selected-count standard deviation is 1.25. LaTeX: \boxed{1.25\quad\text{(D)}} ### Verification record - Main value: 1.25 - Main method: Use the Poisson coefficient of variation to recover lambda, thin the count by 0.25, and take the square root of the thinned mean. - Independent check: Conditionally M|N is Binomial(N,0.25). Total variance gives 0.25(0.75)E[N]+0.25^2 Var(N)=1.5625, so SD(M)=1.25. - Independent value: 1.25 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #614 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-614/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Normal Distribution; Linear Combinations The independent two-period total is normal with standard deviation sqrt(1100^2+2640^2)=2860. The positive-total probability corresponds to z=1.099844, so the unknown second-period mean is 2860z-660=2485.55 and choice C is correct. ### Independent solution ### Setup Form the distribution of the sum of the two independent normal results. LaTeX: S=X_1+X_2 LaTeX: E[S]=660+\mu ### Model Independence makes the variances additive. LaTeX: \operatorname{SD}(S)=\sqrt{1100^2+2640^2}=2860 LaTeX: S\sim N(660+\mu,2860^2) ### Compute Translate the positive-total probability into a standard-normal quantile and solve for the unknown mean. LaTeX: \Phi\!\left(\frac{660+\mu}{2860}\right)=0.8643 LaTeX: \frac{660+\mu}{2860}=\Phi^{-1}(0.8643)=1.099844224\ldots LaTeX: \mu=2485.5544806\ldots ### Answer The unknown mean rounds to the listed value 2486. LaTeX: \boxed{2486\quad\text{(C)}} ### Distractor analysis - A: This understates the mean needed for an 86.43% positive-total probability by using an inadequately small spread for the two-period sum. - B: This results from using the stated probability itself in place of the required standard-normal quantile and then applying an inconsistent mean adjustment. - D: This uses a shifted or rounded cutoff equation instead of standardizing zero against the full total mean 660+mu. - E: This adds the two standard deviations to get 3740 instead of adding variances; 3740(1.099844)-660 is approximately 3454. ### Verification record - Main value: 2485.55448057018 - Main method: Add the independent normal variances, convert the positive-total probability to a z-quantile, and solve the total-mean equation. - Independent check: Substituting mu=2485.5544806 gives total mean 3145.5544806 and z=3145.5544806/2860=1.099844224; the normal CDF at this z is 0.8643. - Independent value: 2485.55448057018 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: combined production cost Three independent production-stage costs are normally distributed. Their respective means are 100, 150, and 250, and their standard deviations are 10, 20, and 30. Calculate the 95th percentile of the total production cost. - A: 438.5 - B: 500.0 - C: 537.4 - D: 561.5 - E: 598.7 The total cost is normal with mean 500 and variance 1400. Adding 1.6448536 standard deviations to the mean gives the 95th percentile 561.5448, so choice D is correct. #### Variant solution ### Setup Add the three stage means to obtain the center of the total cost. LaTeX: E[T]=100+150+250=500 ### Model The sum is normal, and independence makes the stage variances additive. LaTeX: \operatorname{Var}(T)=10^2+20^2+30^2=1400 LaTeX: T\sim N(500,1400) ### Compute Apply the 95th-percentile standard-normal quantile to the total standard deviation. LaTeX: q_{0.95}=500+\Phi^{-1}(0.95)\sqrt{1400} LaTeX: q_{0.95}=500+1.644853627(\sqrt{1400}) LaTeX: q_{0.95}=561.5447872\ldots ### Answer The 95th percentile rounds to 561.5. LaTeX: \boxed{561.5\quad\text{(D)}} ### Verification record - Main value: 561.5447872344489 - Main method: Add the three independent normal means and variances, then apply the 0.95 standard-normal quantile. - Independent check: Standardizing 561.544787 with mean 500 and standard deviation sqrt(1400) gives z=1.644853627, whose normal CDF is 0.95. - Independent value: 561.5447872344489 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #615 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-615/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Cumulative Probabilities The weekly mean gives a daily Poisson mean of 9 and standard deviation 3. A count at least two standard deviations below the mean is at most 3, whose Poisson cumulative probability is 0.02122649, so choice B is correct. ### Independent solution ### Setup Convert the weekly expected count to a daily Poisson mean. LaTeX: \lambda=\frac{63}{7}=9 LaTeX: D\sim\operatorname{Poisson}(9) ### Model A Poisson count has variance equal to its mean, so determine the requested lower cutoff. LaTeX: \operatorname{SD}(D)=\sqrt9=3 LaTeX: 9-2(3)=3 ### Compute Sum the Poisson masses through the integer cutoff. LaTeX: \Pr(D\le3)=e^{-9}\sum_{k=0}^{3}\frac{9^k}{k!} LaTeX: \Pr(D\le3)=0.0212264863\ldots ### Answer The probability rounds to 0.021. LaTeX: \boxed{0.021\quad\text{(B)}} ### Distractor analysis - A: This uses a strict cutoff below 3 and computes P(D<=2)=0.006232 instead of including count 3. - C: This uses the higher cutoff 5, giving P(D<=5)=0.115691 rather than moving two full standard deviations below the mean. - D: This results from an inappropriate approximation or a noninteger cutoff rather than evaluating the exact Poisson lower tail at 3. - E: This subtracts only one standard deviation, using cutoff 6 and obtaining P(D<=6)=0.206781. ### Verification record - Main value: 0.0212264863029089 - Main method: Scale the weekly Poisson mean to one day, compute its standard deviation, and sum the exact Poisson masses through the two-standard-deviation cutoff. - Independent check: Starting with p0=exp(-9) and applying the recurrence p(k+1)=9p(k)/(k+1) through k=3 gives cumulative probability 0.0212264863. - Independent value: 0.0212264863029089 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: transformed incident score A cycle's incident count N is Poisson. An operational score is defined by Y=2N+5, and Var(Y)=36. Calculate the probability that Y is at least 29. - A: 0.073 - B: 0.197 - C: 0.294 - D: 0.413 - E: 0.803 Because Var(2N+5)=4Var(N), the Poisson mean is 9. The event Y>=29 is equivalent to N>=12, whose Poisson upper-tail probability is 0.19699162, so choice B is correct. #### Variant solution ### Setup Use the variance of the transformed score to recover the Poisson parameter. LaTeX: \operatorname{Var}(Y)=2^2\operatorname{Var}(N)=4\lambda=36 LaTeX: \lambda=9 ### Model Translate the score threshold into an integer count threshold. LaTeX: 2N+5\ge29 LaTeX: N\ge12 ### Compute Evaluate the Poisson upper tail by complementing the cumulative probability through 11. LaTeX: \Pr(Y\ge29)=\Pr(N\ge12) LaTeX: \Pr(N\ge12)=1-e^{-9}\sum_{k=0}^{11}\frac{9^k}{k!} LaTeX: \Pr(N\ge12)=0.1969916175\ldots ### Answer The upper-tail probability rounds to 0.197. LaTeX: \boxed{0.197\quad\text{(B)}} ### Verification record - Main value: 0.19699161747065808 - Main method: Recover the Poisson mean from the transformed variance, convert the score event to N>=12, and evaluate the upper tail. - Independent check: Generate Poisson(9) masses recursively from p0=exp(-9); their sum through 11 is 0.8030083825, leaving complement 0.1969916175. - Independent value: 0.19699161747065808 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #616 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-616/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Uniform Distribution; Variance and Standard Deviation The continuous uniform standard deviation is 10/sqrt(12)=2.886751. The discrete uniform distribution on eleven consecutive integers has variance 10 and standard deviation 3.162278. Their absolute difference is 0.275526, which rounds to choice C. ### Independent solution ### Setup Calculate the variance of the continuous uniform selection from its interval width. LaTeX: \operatorname{Var}(C)=\frac{(10-0)^2}{12}=\frac{25}{3} LaTeX: \operatorname{SD}(C)=\frac{10}{\sqrt{12}}=2.8867513459\ldots ### Model For the eleven equally likely integers, use the first two discrete moments. LaTeX: E[D]=\frac{0+1+\cdots+10}{11}=5 LaTeX: E[D^2]=\frac{0^2+1^2+\cdots+10^2}{11}=35 ### Compute Convert the discrete moments to a standard deviation and compare the two spreads. LaTeX: \operatorname{Var}(D)=35-5^2=10 LaTeX: \operatorname{SD}(D)=\sqrt{10}=3.1622776602\ldots LaTeX: \left|\operatorname{SD}(D)-\operatorname{SD}(C)\right|=0.2755263142\ldots ### Answer The absolute difference rounds to 0.28. LaTeX: \boxed{0.28\quad\text{(C)}} ### Distractor analysis - A: This assumes equal endpoints and equal means force the continuous and discrete uniform distributions to have the same spread. - B: This effectively uses only ten discrete values by dropping one endpoint; the resulting standard-deviation difference is about 0.02. - D: This applies a half-unit continuity adjustment directly to a standard deviation instead of recomputing the discrete variance. - E: This subtracts the variances, 10-25/3=1.6667, rather than subtracting the standard deviations. ### Verification record - Main value: 0.2755263142202504 - Main method: Compute the continuous uniform variance from interval width, compute the discrete uniform variance from its moments, and subtract their standard deviations. - Independent check: The closed form variance for a discrete uniform distribution on 11 consecutive integers is (11^2-1)/12=10; comparing sqrt(10) with 10/sqrt(12) gives 0.2755263143. - Independent value: 0.2755263142202504 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: hybrid calibration offset A calibration offset X is generated in one of two ways. With probability 0.60 it is continuous uniform on [-2,2]. Otherwise it is selected uniformly from the three values -3, 0, and 3. Calculate the unconditional standard deviation of X. - A: 1.155 - B: 1.673 - C: 1.789 - D: 2.449 - E: 3.200 Both mixture components have mean zero. Their second moments are 4/3 and 6, so the unconditional variance is 0.60(4/3)+0.40(6)=3.2. Its square root is 1.788854, which rounds to choice C. #### Variant solution ### Setup Both component distributions are symmetric about zero, so the mixture mean is zero. LaTeX: E[X\mid M=1]=E[X\mid M=2]=0 LaTeX: E[X]=0 ### Model Calculate the conditional second moments for the continuous and discrete components. LaTeX: E[X^2\mid M=1]=\frac{(2-(-2))^2}{12}=\frac43 LaTeX: E[X^2\mid M=2]=\frac{(-3)^2+0^2+3^2}{3}=6 ### Compute Average the component second moments, which equal their variances because both conditional means are zero. LaTeX: \operatorname{Var}(X)=0.60\left(\frac43\right)+0.40(6)=3.2 LaTeX: \operatorname{SD}(X)=\sqrt{3.2}=1.7888543820\ldots ### Answer The unconditional standard deviation rounds to 1.789. LaTeX: \boxed{1.789\quad\text{(C)}} ### Verification record - Main value: 1.7888543819998317 - Main method: Mix the two conditional second moments, use the zero mixture mean, and take the square root of the resulting variance. - Independent check: The law of total variance gives E[Var(X|M)]=0.60(4/3)+0.40(6)=3.2 and Var(E[X|M])=0, so the standard deviation is sqrt(3.2). - Independent value: 1.7888543819998317 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #617 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-617/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Conditional Distributions; Variance Normalize the three joint masses in the specified column to obtain conditional probabilities 0.56, 0.32, and 0.12. The conditional second moment is 0.80 and the conditional mean is 0.56, so the variance is 0.4864 and choice D is correct. ### Independent solution ### Setup The conditioning column has total probability 0.25. Divide each mass in that column by this total before calculating any moment. LaTeX: \Pr(F=1)=0.14+0.08+0.03=0.25 LaTeX: \Pr(T=0,1,2\mid F=1)=(0.56,0.32,0.12) ### Model Use the first two raw moments of the resulting conditional distribution. LaTeX: \operatorname{E}[T\mid F=1]=0(0.56)+1(0.32)+2(0.12)=0.56 LaTeX: \operatorname{E}[T^2\mid F=1]=0^2(0.56)+1^2(0.32)+2^2(0.12)=0.80 ### Compute Subtract the square of the conditional mean from the conditional second moment. LaTeX: \operatorname{Var}(T\mid F=1)=0.80-(0.56)^2 LaTeX: \operatorname{Var}(T\mid F=1)=0.4864 ### Answer The variance rounds to 0.49. LaTeX: \boxed{0.49\quad\text{(D)}} ### Distractor analysis - A: Using the unnormalized column masses as though they formed a probability distribution gives 0.20-(0.14)^2=0.1804, which rounds to 0.18. - B: Subtracting the mean itself rather than its square from the second moment gives 0.80-0.56=0.24. - C: Ignoring the conditioning column and using the marginal distribution of T gives an unconditional variance of 0.45. - E: This is the conditional second moment 0.80; it omits subtraction of the squared conditional mean. ### Verification record - Main value: 0.4864 - Main method: Normalize the F=1 column, calculate the conditional first and second raw moments, and apply Var(T|F=1)=E[T^2|F=1]-E[T|F=1]^2. - Independent check: Calculate the centered moment directly about the conditional mean: 0.56(0-0.56)^2+0.32(1-0.56)^2+0.12(2-0.56)^2. - Independent value: 0.4864 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: satellite diagnostic update A satellite records zero, one, or two anomaly pings in a cycle with prior probabilities 0.50, 0.30, and 0.20. A diagnostic lamp activates with probability 0.10 after zero pings, 0.40 after one ping, and 0.80 after two pings. Given that the lamp activated, find the variance of the ping count. - A: 0.330 - B: 0.525 - C: 0.610 - D: 1.333 - E: 2.303 Bayes' rule turns the three prior-by-likelihood products into posterior masses 5/33, 12/33, and 16/33. Their variance is 52/99, or 0.525253, so choice B is correct. #### Variant solution ### Setup Let N be the ping count and A the event that the diagnostic lamp activates. Record the prior probabilities and the count-specific activation likelihoods. LaTeX: \bigl(\Pr(N=0),\Pr(N=1),\Pr(N=2)\bigr)=(0.50,0.30,0.20) LaTeX: \bigl(\Pr(A\mid N=0),\Pr(A\mid N=1),\Pr(A\mid N=2)\bigr)=(0.10,0.40,0.80) ### Model Multiply each prior by its activation likelihood. Their sum is the evidence used to normalize the posterior distribution. LaTeX: (0.50)(0.10),(0.30)(0.40),(0.20)(0.80)=(0.05,0.12,0.16) LaTeX: \Pr(A)=0.05+0.12+0.16=0.33 LaTeX: \bigl(\Pr(N=0\mid A),\Pr(N=1\mid A),\Pr(N=2\mid A)\bigr)=\left(\frac5{33},\frac{12}{33},\frac{16}{33}\right) ### Compute Calculate the first two posterior moments and apply the variance identity. LaTeX: \operatorname{E}[N\mid A]=\frac{12+2(16)}{33}=\frac43 LaTeX: \operatorname{E}[N^2\mid A]=\frac{12+4(16)}{33}=\frac{76}{33} LaTeX: \operatorname{Var}(N\mid A)=\frac{76}{33}-\left(\frac43\right)^2=\frac{52}{99}=0.525252\ldots ### Answer The posterior variance rounds to 0.525. LaTeX: \boxed{0.525\quad\text{(B)}} ### Verification record - Main value: 0.5252525252525253 - Main method: Apply Bayes' rule to the three activation likelihoods, then subtract the squared posterior mean from the posterior second moment. - Independent check: Using posterior mean 4/3, the centered sum (5/33)(0-4/3)^2+(12/33)(1-4/3)^2+(16/33)(2-4/3)^2 simplifies directly to 52/99. - Independent value: 0.5252525252525253 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #618 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-618/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Uniform Distribution; Conditional Probability The stated moments identify a uniform interval from 7 to 10. Conditioning on survival past 7.5 restricts the possible interval to length 2.5, while continuing past 9 leaves length 1, so the conditional probability is 1/2.5=0.40 and choice B is correct. ### Independent solution ### Setup Use the uniform mean and variance to recover the two endpoints. LaTeX: \frac{a+b}{2}=8.5 LaTeX: \frac{(b-a)^2}{12}=0.75 ### Model The interval width is positive, so the moment equations give a width of 3 and an endpoint sum of 17. LaTeX: b-a=\sqrt{12(0.75)}=3 LaTeX: a=7,\qquad b=10 ### Compute Given that the duration exceeds 7.5, the favorable portion above 9 has length 1 out of a remaining length of 2.5. LaTeX: \Pr(X>9\mid X>7.5)=\frac{10-9}{10-7.5} LaTeX: \Pr(X>9\mid X>7.5)=0.40 ### Answer The requested conditional probability is 0.40. LaTeX: \boxed{0.40\quad\text{(B)}} ### Distractor analysis - A: This is the unconditional probability of exceeding 9, namely the one-unit upper segment divided by the full width 3. - C: This uses the symmetry probability above the mean, 0.50, even though the cutoff is 9 and the sample is conditioned at 7.5. - D: This is the complement of the requested conditional tail, representing an endpoint at or below 9 after the condition is imposed. - E: This is the probability of the conditioning event itself, P(X>7.5)=2.5/3, rather than the conditional probability. ### Verification record - Main value: 0.4 - Main method: Recover the uniform endpoints from the given mean and variance, then divide the favorable interval length by the conditioning interval length. - Independent check: Evaluate the conditional probability as P(X>9)/P(X>7.5)=(1/3)/(2.5/3). - Independent value: 0.4 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: dual-courier arrival order Two couriers arrive independently at a loading bay, with each arrival time uniform over the same ten-minute receiving window. Given that at least one courier arrives during the first four minutes, calculate the probability that both couriers arrive during the first six minutes. - A: 0.320 - B: 0.360 - C: 0.500 - D: 0.563 - E: 0.640 The conditioning event occupies probability 0.64. Both arrivals lie before minute six and at least one lies before minute four with probability 0.36-0.04=0.32, so the conditional probability is 0.32/0.64=0.50 and choice C is correct. #### Variant solution ### Setup Let U and V be the two independent arrival times, each uniform from zero to ten. Express the condition and target using their minimum and maximum. LaTeX: C=\{\min(U,V)<4\} LaTeX: B=\{\max(U,V)<6\} ### Model Use the complement in which both couriers arrive at or after minute four to find the probability of the conditioning event. LaTeX: \Pr(C)=1-\Pr(U\ge4,V\ge4) LaTeX: \Pr(C)=1-\left(\frac6{10}\right)^2=0.64 ### Compute Within the square where both arrivals are before minute six, remove the smaller square where both fall between minutes four and six. LaTeX: \Pr(B\cap C)=\left(\frac6{10}\right)^2-\left(\frac2{10}\right)^2=0.32 LaTeX: \Pr(B\mid C)=\frac{0.32}{0.64}=0.50 ### Answer The conditional probability is one half. LaTeX: \boxed{0.500\quad\text{(C)}} ### Verification record - Main value: 0.5 - Main method: Use independence and uniform interval lengths to calculate the conditioning probability and the intersection involving the two order statistics. - Independent check: In the 10-by-10 arrival-time square, the conditioning region has area 100-36=64 and the favorable portion has area 36-4=32, giving the ratio 32/64. - Independent value: 0.5 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #619 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-619/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Expected Value Taking the ratio of the Poisson probabilities cancels their common exponential factor and gives λ^4/4!=54. Thus λ=6 per minute, and linearity of expectation gives 60(6)=360 over one hour, selecting choice E. ### Independent solution ### Setup Let the common mean count in one minute be the positive parameter lambda. LaTeX: \Pr(N=k)=e^{-\lambda}\frac{\lambda^k}{k!} ### Model Form the stated probability ratio. The exponential terms cancel. LaTeX: \frac{\Pr(N=4)}{\Pr(N=0)}=\frac{\lambda^4}{4!}=54 LaTeX: \lambda^4=54(24)=1296 ### Compute Use the positive fourth root for a Poisson mean, then scale the expectation to sixty minutes. LaTeX: \lambda=1296^{1/4}=6 LaTeX: \operatorname{E}[N_{60}]=60\lambda=360 ### Answer The expected count in the requested period is 360. LaTeX: \boxed{360\quad\text{(E)}} ### Distractor analysis - A: This corresponds to guessing a rate of 2 per minute instead of solving the fourth-power probability-ratio equation. - B: Omitting the factor 4! gives the incorrect equation lambda^4=54; rounding its fourth root to 3 produces 180 over sixty minutes. - C: Using the observed count label 4 as the mean rather than solving for lambda gives 60(4)=240. - D: This corresponds to using 5 as the fourth root of 1296; checking powers shows that 5^4 is only 625 while 6^4 is 1296. ### Verification record - Main value: 360 - Main method: Divide the Poisson mass at four by the mass at zero, solve lambda^4/4!=54 for the positive mean, and multiply the one-minute mean by 60. - Independent check: Substitute lambda=6 directly: [e^-6(6^4/24)]/[e^-6]=1296/24=54, and summing sixty one-minute expectations gives 360. - Independent value: 360 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: bright-meteor thinning Meteors reach an automated detector according to a Poisson process with mean rate 12 per hour. Each detected meteor is independently classified as bright with probability one-third and dim otherwise. During a thirty-minute interval, calculate the probability of exactly two bright meteors and at least one dim meteor. - A: 0.045 - B: 0.266 - C: 0.271 - D: 0.982 - E: 2.000 Poisson thinning produces independent bright and dim counts with means 2 and 4. Multiplying the probability of two bright meteors by the probability of at least one dim meteor gives 0.265713, so choice B is correct. #### Variant solution ### Setup Convert the hourly rate to the mean count over one half-hour. LaTeX: N\sim\operatorname{Poisson}\!\left(12\cdot\frac12\right)=\operatorname{Poisson}(6) ### Model Thin the process by classification. The bright and dim counts are independent Poisson variables with their respective split means. LaTeX: B\sim\operatorname{Poisson}\!\left(6\cdot\frac13\right)=\operatorname{Poisson}(2) LaTeX: D\sim\operatorname{Poisson}\!\left(6\cdot\frac23\right)=\operatorname{Poisson}(4) LaTeX: B\ \text{and}\ D\ \text{are independent} ### Compute Multiply the exact bright-count probability by the dim-count complement. LaTeX: \Pr(B=2,D\ge1)=\left(e^{-2}\frac{2^2}{2!}\right)\left(1-e^{-4}\right) LaTeX: \Pr(B=2,D\ge1)=0.2657130621\ldots ### Answer The requested joint probability rounds to 0.266. LaTeX: \boxed{0.266\quad\text{(B)}} ### Verification record - Main value: 0.26571306211989265 - Main method: Apply Poisson thinning to obtain independent means two and four, then multiply P(B=2) by P(D at least 1). - Independent check: Condition on the total count N: summing P(N=n) times C(n,2)(1/3)^2(2/3)^(n-2) over n at least 3 gives 0.2657130621. - Independent value: 0.26571306211989265 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #620 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-620/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Independence; Sums of Independent Random Variables More than two successes among four independent opportunities means exactly three or exactly four. Enumerating the four possible single-failure cases gives 0.27525, while all four successes has probability 0.07425. Their sum is 0.3495, which rounds to choice B. ### Independent solution ### Setup Represent each outcome by an independent Bernoulli indicator with the four supplied success probabilities. LaTeX: (p_1,p_2,p_3,p_4)=(0.55,0.45,0.50,0.60) LaTeX: \Pr(S>2)=\Pr(S=3)+\Pr(S=4) ### Model For exactly three successes, sum the four disjoint cases according to which opportunity fails. LaTeX: \Pr(S=3)=0.45(0.45)(0.50)(0.60)+0.55(0.55)(0.50)(0.60) LaTeX: \qquad\quad+0.55(0.45)(0.50)(0.60)+0.55(0.45)(0.50)(0.40)=0.27525 ### Compute Add the all-success case to the exactly-three probability. LaTeX: \Pr(S=4)=0.55(0.45)(0.50)(0.60)=0.07425 LaTeX: \Pr(S>2)=0.27525+0.07425=0.34950 ### Answer The probability rounds to 0.35. LaTeX: \boxed{0.35\quad\text{(B)}} ### Distractor analysis - A: This is the probability of exactly three successes, rounded from 0.27525, and it omits the all-four-success case. - C: Replacing all four unequal probabilities by 0.55 gives a binomial upper tail of about 0.391 instead of the required heterogeneous calculation. - D: Summing the four three-way success intersections without requiring the remaining outcome to fail gives 0.57225 and counts the all-success outcome four times. - E: This is the probability of at least two successes, 0.72575, which uses a threshold one success too low. ### Verification record - Main value: 0.3495 - Main method: Enumerate the four disjoint exactly-three-success cases, add the all-four-success case, and use independence to multiply within each case. - Independent check: Expand the generating polynomial (0.45+0.55z)(0.55+0.45z)(0.50+0.50z)(0.40+0.60z); its z^3 and z^4 coefficients are 0.27525 and 0.07425. - Independent value: 0.3495 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: conditional vial audit Eight sterile culture vials are tested independently, and each vial passes with probability 0.75. An audit report states that at least six of the eight vials passed. Conditional on that report, calculate the probability that exactly seven vials passed. - A: 0.148 - B: 0.267 - C: 0.393 - D: 0.459 - E: 0.679 Within the conditioning set, the binomial masses for six, seven, and eight passes have proportional weights 28, 24, and 9. The required conditional probability is therefore 24/61=0.393443, so choice C is correct. #### Variant solution ### Setup Let X be the number of passing vials. The common success probability makes X binomial. LaTeX: X\sim\operatorname{Binomial}(8,0.75) ### Model The audit restricts the possible counts to six, seven, or eight. Factor the same power of 0.75 and 0.25 from their three binomial masses. LaTeX: \Pr(X=6):\Pr(X=7):\Pr(X=8)=28:24:9 ### Compute Normalize the weight for seven passes over the three counts allowed by the report. LaTeX: \Pr(X=7\mid X\ge6)=\frac{24}{28+24+9} LaTeX: \Pr(X=7\mid X\ge6)=\frac{24}{61}=0.3934426229\ldots ### Answer The conditional probability rounds to 0.393. LaTeX: \boxed{0.393\quad\text{(C)}} ### Verification record - Main value: 0.39344262295081966 - Main method: Use binomial mass ratios for the three counts permitted by the condition and normalize the exactly-seven weight. - Independent check: Direct binomial evaluation gives P(X=7)=0.2669677734 and P(X at least 6)=0.6785430908; their quotient is 0.3934426229. - Independent value: 0.39344262295081966 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #621 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-621/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Memoryless Property The exponential memoryless property makes the elapsed event-free period irrelevant to the residual wait. Subtracting the survival probability beyond 25 days from the survival probability beyond 5 days gives 0.527656, which rounds to 0.53 and selects choice D. ### Independent solution ### Setup Let W denote the additional waiting time from the present. The original exponential mean determines its rate. LaTeX: \lambda=\frac{1}{15} ### Model By memorylessness, the residual wait is exponential with the same mean despite the elapsed quiet period. LaTeX: \Pr(W>t)=e^{-t/15} ### Compute The probability of landing between the two residual-time bounds is the difference of their survival probabilities. LaTeX: \Pr(510)=[e^-1-e^(-7/3)]/e^(-2/3). - Independent value: 0.5276557077362274 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: delayed calibration race At a remote observatory, the time until a calibration routine completes is exponential with mean 8 minutes. An independent safety watchdog stops the routine after an exponential time with mean 12 minutes. Both clocks start together. Calculate the probability that calibration finishes first and finishes more than 3 minutes after the start. - A: 0.214 - B: 0.321 - C: 0.535 - D: 0.600 - E: 0.687 The two clocks race at combined rate 5/24. Integrating the calibration completion density after minute three while the watchdog survives gives (3/5)e^(-5/8)=0.321157, so choice B is correct. #### Variant solution ### Setup Let C be the calibration completion time and W the independent watchdog time. Convert their means to rates. LaTeX: C\sim\operatorname{Exp}(1/8),\qquad W\sim\operatorname{Exp}(1/12) ### Model For calibration to finish first at time t, its density contributes while the watchdog must survive through t. LaTeX: \Pr(C3)=\int_3^\infty \frac18e^{-t/8}e^{-t/12}\,dt ### Compute Combine the exponential rates and evaluate the tail integral. LaTeX: \frac18+\frac1{12}=\frac5{24} LaTeX: \Pr(C3)=\frac{1/8}{5/24}e^{-(5/24)3} LaTeX: \Pr(C3)=\frac35e^{-5/8}=0.3211568571\ldots ### Answer The delayed calibration-win probability rounds to 0.321. LaTeX: \boxed{0.321\quad\text{(B)}} ### Verification record - Main value: 0.32115685711139413 - Main method: Integrate the calibration cause-specific density after minute three against the independent watchdog survival function. - Independent check: The first-clock time is exponential with rate 5/24 and calibration wins with probability 3/5, independently of that time; e^(-5/8)(3/5)=0.3211568571. - Independent value: 0.32115685711139413 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #622 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-622/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Uniform Distribution; Percentiles and Quantiles; Variance and Standard Deviation This problem recovers the endpoints of a uniform distribution from its midpoint and spread. The endpoints are approximately 3.14445 and 29.57555, whose ratio is 9.40563, so choice E is correct. ### Independent solution ### Setup Write the lower and upper endpoints as a and b. For a continuous uniform variable, the median is the midpoint of the support. LaTeX: \frac{a+b}{2}=16.36 LaTeX: a+b=32.72 ### Model The standard deviation of a uniform distribution equals its support width divided by the square root of twelve. LaTeX: \frac{b-a}{\sqrt{12}}=7.63 LaTeX: b-a=7.63\sqrt{12}=26.4310953235\ldots ### Compute Solve the sum-and-difference equations for the endpoints, then form the requested quotient. LaTeX: a=16.36-\frac{26.4310953235}{2}=3.1444523382\ldots LaTeX: b=16.36+\frac{26.4310953235}{2}=29.5755476618\ldots LaTeX: \frac{b}{a}=9.4056275880\ldots ### Answer The endpoint ratio rounds to 9.41. LaTeX: \boxed{9.41\quad\text{(E)}} ### Distractor analysis - A: This results from substituting 7.63 itself for the variance in (b-a)^2/12=7.63 instead of first squaring the stated standard deviation. That gives a width of sqrt(12 times 7.63) and a ratio near 1.83. - B: This value understates the support width substantially. It is tempting if the midpoint equation is used correctly but the standard deviation is treated as an endpoint displacement without the uniform-distribution scale factor. - C: This also comes from an interval that is too narrow, typically after mixing a half-width formula with the full-width formula. The standard deviation must be converted to the entire width before the endpoints are recovered. - D: Using (b-a)^2/10 as the variance formula gives a width of 7.63 sqrt(10) and a ratio of about 6.62. The denominator for a uniform distribution is twelve, not ten. ### Verification record - Main value: 9.405627587987356 - Main method: Use the uniform midpoint and standard-deviation formulas to solve the two endpoint equations and divide the resulting upper endpoint by the lower endpoint. - Independent check: A numerical root solve using the uniform CDF condition at the median and a direct integral of the squared centered density returns endpoints 3.1444523382 and 29.5755476618, hence the same ratio. - Independent value: 9.405627587987356 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: calibration-lag endpoint A robotics team models calibration lag as uniformly distributed across an unknown interval. Maintenance records give an expected lag of 18 minutes and a variance of 12 square minutes. What is the upper endpoint of the interval? - A: 6 minutes - B: 12 minutes - C: 18 minutes - D: 24 minutes - E: 30 minutes A uniform variance of 12 implies a support width of 12 minutes. Centering that interval at the mean of 18 puts its upper endpoint at 24 minutes, so choice D is correct. #### Variant solution ### Setup Let l and u denote the unknown endpoints. The mean fixes the center of a uniform interval. LaTeX: \frac{l+u}{2}=18 ### Model Relate the given variance to the full interval width. LaTeX: \operatorname{Var}(X)=\frac{(u-l)^2}{12}=12 ### Compute The width is twelve, so each endpoint lies six minutes from the center. LaTeX: u-l=\sqrt{144}=12 LaTeX: u=18+\frac{12}{2}=24 ### Answer The upper endpoint is 24 minutes. LaTeX: \boxed{24\text{ minutes}\quad\text{(D)}} ### Verification record - Main value: 24 - Main method: Solve the uniform variance formula for the twelve-minute width, then add half that width to the mean. - Independent check: Direct integration for a uniform density on [12,24] gives mean 18 and variance (24-12)^2/12=12, confirming the proposed endpoint. - Independent value: 24 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #623 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-623/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Normal Distribution; Percentiles and Quantiles; Standardization This is a normal-quantile calibration problem. The supplied center and lower quantile imply a standard deviation of about 15, making the target z-score 0.63337 and its cumulative probability 0.73675, so choice B is correct. ### Independent solution ### Setup A normal distribution is symmetric and unimodal, so its mode identifies its mean. Let sigma denote its unknown standard deviation. LaTeX: \mu=56 LaTeX: \Phi^{-1}(0.40)=-0.2533471031\ldots ### Model Standardize the known lower quantile to determine the scale. LaTeX: \frac{52.20-56}{\sigma}=-0.2533471031\ldots LaTeX: \sigma=14.9991847271\ldots ### Compute Convert the target score to a z-score and evaluate the standard-normal cumulative distribution. LaTeX: z=\frac{65.50-56}{14.9991847271}=0.6333677578\ldots LaTeX: \Phi(z)=0.7367532425\ldots ### Answer A cumulative probability of about 0.737 corresponds to the 74th percentile. LaTeX: \boxed{\text{74th percentile}\quad\text{(B)}} ### Distractor analysis - A: The 71st percentile corresponds to a z-score near 0.55. It results from inflating the inferred standard deviation instead of using the standard-normal 40th-percentile quantile accurately. - C: The 78th percentile requires a z-score near 0.77, so it reflects a standard deviation that is too small. The known quantile must first be converted with Phi inverse, not read as a raw distance. - D: The 81st percentile similarly overstates the target z-score by compressing the scale too much. Normal percentiles are not linearly spaced in the original score units. - E: Treating 0.40 itself as the magnitude of the known z-score gives sigma=9.5 and a target z-score of 1, which points near the 84th or 85th percentile. ### Verification record - Main value: 0.7367532425426623 - Main method: Set the normal mean equal to the mode, infer sigma from the 40th-percentile z-score, and evaluate the target score with the standard-normal CDF. - Independent check: Cancel sigma by taking the ratio of centered score gaps: the target z-score is (9.5/3.8) times the magnitude of Phi inverse of 0.40, or 0.6333678; a separate normal-table lookup gives 0.7368. - Independent value: 0.7368 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: paired calibration gap A calibration facility records independent readings A and B. Reading A is normal with mean 80 and standard deviation 6, while B is normal with mean 72 and standard deviation 8. A gap of 18 is observed for A-B. Approximately what percentile of the distribution of A-B does this gap represent? - A: 16th - B: 50th - C: 76th - D: 84th - E: 97th The difference A-B is normal with mean 8 and standard deviation 10. The observed gap has z=(18-8)/10=1, whose cumulative probability is 0.84134, so it is approximately the 84th percentile. #### Variant solution ### Setup Let D be the difference between the two independent calibration readings. LaTeX: D=A-B LaTeX: E[D]=80-72=8 ### Model A linear combination of independent normal variables is normal, and the variances add even though one variable is subtracted. LaTeX: \operatorname{Var}(D)=6^2+8^2=100 LaTeX: D\sim N(8,10^2) ### Compute Standardize the observed gap and evaluate the normal cumulative probability. LaTeX: z=\frac{18-8}{10}=1 LaTeX: \Pr(D\le18)=\Phi(1)=0.8413447461\ldots ### Answer A cumulative probability near 0.84 corresponds to the 84th percentile. LaTeX: \boxed{\text{84th percentile}\quad\text{(D)}} ### Verification record - Main value: 0.8413447460685429 - Main method: Form the distribution of the independent normal difference, standardize the observed gap, and evaluate its CDF. - Independent check: Conditioning on B and integrating P(A<=18+B | B) over the N(72,8^2) density gives 0.8413447461, confirming the distribution-of-a-difference calculation. - Independent value: 0.8413447460685429 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #624 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-624/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Normal Distribution; Linear Combinations of Independent Random Variables; Independence This is a sum of independent normal variables. Their aggregate has mean 300 and standard deviation sqrt(290000), so its positive-tail probability is 0.71127 and choice E is correct. ### Independent solution ### Setup Let X and Y denote the two independent normal contributions and define the combined result S. LaTeX: S=X+Y LaTeX: \operatorname{E}[S]=400+(-100)=300 ### Model An independent sum of normal variables is normal. Independence also makes the component variances additive. LaTeX: \operatorname{Var}(S)=200^2+500^2=290000 LaTeX: S\sim N\!\left(300,290000\right) ### Compute Standardize zero relative to the aggregate distribution and take the probability above that boundary. LaTeX: \operatorname{SD}(S)=\sqrt{290000}=538.5164807135\ldots LaTeX: \Pr(S>0)=\Phi\!\left(\frac{300}{538.5164807135}\right) LaTeX: \Pr(S>0)=\Phi(0.5570860145\ldots)=0.7112656688\ldots ### Answer The requested probability rounds to 0.71. LaTeX: \boxed{0.71\quad\text{(E)}} ### Distractor analysis - A: A value below one half reverses the sign or tail despite the aggregate mean being positive. The probability of a positive result must exceed one half here. - B: This understates the signal-to-noise ratio, often by retaining only the magnitude 100 when combining the two means. Signed means must be added before standardization. - C: This is consistent with using an overly large scale after blending means and standard deviations. Only variances, not unlike quantities, can be combined in the variance calculation. - D: Adding the standard deviations directly gives 700 and Phi(300/700)=0.6666, which rounds to 0.67. Independent standard deviations combine through their squared values. ### Verification record - Main value: 0.7112656687863502 - Main method: Add the independent normal means and variances, standardize the zero boundary, and evaluate the standard-normal CDF. - Independent check: Numerically integrate P(X>-y) against the second component's normal density, using the first component's normal survival function inside the integral; quadrature gives 0.7112656688. - Independent value: 0.7112656687863502 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: paired sensor correction A materials lab applies two independent normally distributed sensor corrections. Correction A has mean 12 units and standard deviation 3 units; correction B has mean -5 units and standard deviation 4 units. The final adjustment is D=A+2B. Find the probability that D is negative. - A: 0.081 - B: 0.407 - C: 0.428 - D: 0.500 - E: 0.593 The linear combination has mean 2 and variance 73. Standardizing zero gives Phi(-2/sqrt(73))=0.40746, so choice B is correct. #### Variant solution ### Setup Represent the final adjustment as the stated linear combination of the two independent inputs. LaTeX: D=A+2B ### Model A linear combination of independent normals is normal. Apply the coefficient to both the mean and the variance contribution. LaTeX: \operatorname{E}[D]=12+2(-5)=2 LaTeX: \operatorname{Var}(D)=3^2+2^2(4^2)=73 ### Compute Standardize the zero threshold under the resulting normal distribution. LaTeX: \Pr(D<0)=\Phi\!\left(\frac{0-2}{\sqrt{73}}\right) LaTeX: \Pr(D<0)=\Phi(-0.2340822944\ldots)=0.4074605456\ldots ### Answer The probability is approximately 0.407. LaTeX: \boxed{0.407\quad\text{(B)}} ### Verification record - Main value: 0.4074605455732343 - Main method: Compute the mean and variance of A+2B under independence and evaluate its normal CDF at zero. - Independent check: Direct numerical convolution integrates P(A<-2b) over the N(-5,16) density for B and returns 0.4074605456. - Independent value: 0.4074605455732343 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #625 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-625/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Conditional Probability; Independence Independent Poisson increments make the observed early count irrelevant to the remaining interval except for fixing the number still needed. The final three years must contain exactly two events, whose probability is 0.22404, so choice E is correct. ### Independent solution ### Setup The conditioning information fixes eight events in the completed seven-year portion. Reaching ten by the end therefore requires two more events. LaTeX: 10-8=2 ### Model Counts from disjoint years are independent. The count in the remaining three-year interval is Poisson with mean equal to three times the annual mean. LaTeX: R\sim\operatorname{Poisson}(3) ### Compute Evaluate the Poisson probability of exactly two events in that remaining interval. LaTeX: \Pr(R=2)=e^{-3}\frac{3^2}{2!} LaTeX: \Pr(R=2)=0.2240418077\ldots ### Answer The conditional probability rounds to 0.224. LaTeX: \boxed{0.224\quad\text{(E)}} ### Distractor analysis - A: Using e^(-3) calculates no events in the remaining three years. Two additional events, not zero, are required to reach the specified total. - B: This is approximately the unconditional probability of ten events over all ten years, using a Poisson mean of ten. It discards the observed seven-year count instead of conditioning on it. - C: This is approximately the probability of eight events in the first seven years, the event already supplied as information. The task is to evaluate what happens afterward. - D: This is close to a two-event probability under a one-year Poisson mean. The remaining exposure lasts three years, so its mean is three rather than one. ### Verification record - Main value: 0.22404180765538775 - Main method: Use independent Poisson increments: the final three-year count has mean three and must equal two. - Independent check: Apply Bayes with N(7) given N(10)=10 distributed as Binomial(10,0.7), together with the Poisson(10) and Poisson(7) probabilities; the resulting conditional probability is 0.2240418077. - Independent value: 0.22404180765538775 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: downlink timing within a pass Telemetry downlinks arrive according to a homogeneous Poisson process. During a four-hour satellite pass, exactly six downlinks are recorded. Conditional on that total, calculate the probability that exactly two of the downlinks occurred during the first hour. - A: 0.234 - B: 0.251 - C: 0.297 - D: 0.356 - E: 0.466 Given six arrivals in four hours, each arrival independently falls in the first hour with probability 1/4. The resulting binomial probability is 0.29663, so choice C is correct. #### Variant solution ### Setup Condition on the observed total of six arrivals over the complete four-hour pass. LaTeX: N(4)=6 ### Model Conditional on the total, homogeneous Poisson arrival times are uniformly distributed across the pass. Each arrival therefore lands in the first one-quarter of the interval with probability one-quarter. LaTeX: K\mid N(4)=6\sim\operatorname{Binomial}\!\left(6,\frac14\right) ### Compute Evaluate the conditional binomial mass at two. LaTeX: \Pr(K=2\mid N(4)=6)=\binom{6}{2}\left(\frac14\right)^2\left(\frac34\right)^4 LaTeX: \Pr(K=2\mid N(4)=6)=0.2966308594\ldots ### Answer The conditional probability rounds to 0.297. LaTeX: \boxed{0.297\quad\text{(C)}} ### Verification record - Main value: 0.296630859375 - Main method: Condition on the total Poisson count and use the binomial allocation of arrivals across a one-quarter subinterval. - Independent check: For any rate lambda, divide P(N(1)=2 and N(4)-N(1)=4) by P(N(4)=6); the exponential and rate terms cancel, leaving 15(1/4)^2(3/4)^4=0.2966308594. - Independent value: 0.296630859375 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #626 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-626/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Variance; Expectation and Variance For a Poisson variable, the variance equals its mean and the zero-count probability is exponential in that mean. Taking the logarithm of the stated probability ratio gives a variance difference of ln(2)=0.69315, so choice A is correct. ### Independent solution ### Setup Let the two Poisson means be lambda_1 and lambda_2. Their variances equal those same parameters. LaTeX: V_1=\lambda_1,\qquad V_2=\lambda_2 ### Model A Poisson count is zero with probability e raised to the negative mean. Express the given zero-count ratio with that identity. LaTeX: e^{-\lambda_1}=\frac{1}{2}e^{-\lambda_2} ### Compute Divide the exponentials and take natural logarithms to isolate the parameter difference. LaTeX: e^{-(\lambda_1-\lambda_2)}=\frac{1}{2} LaTeX: \lambda_1-\lambda_2=\ln 2=0.6931471806\ldots ### Answer The same difference applies to the two Poisson variances. LaTeX: \boxed{V_1-V_2=0.693\quad\text{(A)}} ### Distractor analysis - B: This exceeds ln(2) because it effectively replaces the exact one-half probability ratio by a smaller rounded ratio before taking the logarithm. The ratio is given exactly and should not be perturbed. - C: This treats the zero-count relation too much like a linear probability difference. Poisson zero probabilities are exponential in the means, so a logarithm is required. - D: This can arise from taking a logarithm after incorrectly adding a probability adjustment to the one-half ratio. The algebra requires division of the two zero-count probabilities. - E: This is another inflated parameter difference produced by altering the exact ratio. Once the exponential terms are divided, the result is fixed uniquely at the natural logarithm of two. ### Verification record - Main value: 0.6931471805599453 - Main method: Use P(N=0)=exp(-lambda) and Var(N)=lambda for Poisson counts, then take a natural logarithm of the zero-count probability ratio. - Independent check: Choose lambda_2=0.4 and solve the probability equation numerically; lambda_1=1.0931471806 makes exp(-lambda_1) exactly half of exp(-0.4), and their variance difference is 0.6931471806. - Independent value: 0.6931471805599453 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: triaged sensor alerts A monitoring station receives a Poisson number of sensor alerts with mean 6 per shift. Each alert is independently classified as urgent with probability 0.40; all other alerts are routine. Let U and R be the urgent and routine counts. Calculate Var(2U-R). - A: 2.4 - B: 6.0 - C: 8.4 - D: 13.2 - E: 36.0 Poisson thinning makes U and R independent Poisson counts with means 2.4 and 3.6. Therefore Var(2U-R)=4(2.4)+3.6=13.2, which is choice D. #### Variant solution ### Setup Split the mean total count between the urgent and routine categories. LaTeX: E[U]=6(0.40)=2.4 LaTeX: E[R]=6(0.60)=3.6 ### Model Independent thinning of a Poisson count produces independent Poisson category counts, each with variance equal to its mean. LaTeX: U\sim\operatorname{Poisson}(2.4),\qquad R\sim\operatorname{Poisson}(3.6) LaTeX: \operatorname{Cov}(U,R)=0 ### Compute Square the coefficients when taking the variance of the linear combination. LaTeX: \operatorname{Var}(2U-R)=2^2\operatorname{Var}(U)+(-1)^2\operatorname{Var}(R) LaTeX: \operatorname{Var}(2U-R)=4(2.4)+3.6=13.2 ### Answer The variance of the weighted alert score is 13.2. LaTeX: \boxed{13.2\quad\text{(D)}} ### Verification record - Main value: 13.2 - Main method: Use Poisson thinning and independence, then add the coefficient-squared category variances. - Independent check: Conditioning on total N gives 2U-R=3U-N, with Var(3U-N | N)=9N(0.4)(0.6) and E(3U-N | N)=0.2N; total variance is 9(0.4)(0.6)(6)+(0.2)^2(6)=13.2. - Independent value: 13.2 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #627 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-627/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Conditional Distributions; Conditional Expectation This is a conditional-expectation calculation from a joint probability table. Restricting the table to the qualifying rows gives probability 0.70 and a weighted numerator of 0.99, so the conditional mean is 99/70, or 1.414286, and choice C is correct. ### Independent solution ### Setup Retain the two rows allowed by the conditioning event. Their combined probability is the denominator of the conditional distribution. LaTeX: \Pr(X<2)=0.30+0.40=0.70 ### Model For each possible value of Y, add the retained joint masses and divide by 0.70. The conditional mean can be computed by dividing the corresponding weighted joint sum by the same denominator. LaTeX: \operatorname{E}[Y\mid X<2]=\frac{\sum_y y\,\Pr(Y=y,\,X<2)}{\Pr(X<2)} ### Compute The retained column masses for Y equal 0.14, 0.21, 0.27, and 0.08. Weighting them by 0, 1, 2, and 3 produces the numerator. LaTeX: \sum_y y\,\Pr(Y=y,\,X<2)=1(0.21)+2(0.27)+3(0.08)=0.99 LaTeX: \operatorname{E}[Y\mid X<2]=\frac{0.99}{0.70}=\frac{99}{70}=1.414285714\ldots ### Answer The conditional expected count rounds to 1.41. LaTeX: \boxed{1.41\quad\text{(C)}} ### Distractor analysis - A: Using a multiplier of two rather than three for the Y=3 column gives numerator 0.91 and 0.91/0.70=1.30. - B: Misadding the retained Y=3 column as 0.07 instead of 0.08 gives numerator 0.96 and 0.96/0.70=1.3714. - D: This is the unconditional mean of Y from all four rows. It ignores the condition on X and therefore uses the wrong distribution. - E: Using a multiplier of four rather than three for the Y=3 column gives numerator 1.07 and 1.07/0.70=1.5286. ### Verification record - Main value: 1.4142857142857144 - Main method: Sum the two qualifying rows, form the retained Y-column masses, and divide their weighted sum 0.99 by conditioning probability 0.70. - Independent check: Compute the conditional row means separately: the first qualifying row has mean 4/3 and the second has mean 59/40. Mixing them with conditional row weights 3/7 and 4/7 gives 99/70. - Independent value: 1.4142857142857144 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: microfluidic completion time Within a normalized one-minute microfluidic cycle, U records when a sample enters a channel and V records when processing finishes. Their joint density equals 2 on the triangular region 01/2 has probability 1/4, while the corresponding first-moment integral for V is 5/24. Their ratio is 5/6, or 0.8333, so choice D is correct. #### Variant solution ### Setup The conditioning event restricts U to the interval from one-half to one. For each such entry time, V ranges from U to one. LaTeX: A=\{U>1/2\} LaTeX: f_{U,V}(u,v)=2,\qquad \frac121/2 to obtain probability 1/4 and integrate v times the density over the same triangle to obtain 5/24. - Independent check: Given U=u, V is uniform from u to 1 and has mean (u+1)/2. The conditional mean of U above one-half is 2/3, so the tower property gives (1+2/3)/2=5/6. - Independent value: 0.8333333333333334 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #628 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-628/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Continuous Random Variables; Expectation; Distribution Functions This is a parameter-identification problem for a power density on the unit interval. The stated mean fixes the exponent at two, after which the upper-tail probability above one-half is 1-(1/2)^3=0.875, so choice E is correct. ### Independent solution ### Setup Write the mean as an integral under the stated one-parameter density. LaTeX: \operatorname{E}[R]=\int_0^1 r(\alpha+1)r^\alpha\,dr=\frac{\alpha+1}{\alpha+2} ### Model Equate the theoretical mean to the supplied value and solve for the positive parameter. LaTeX: \frac{\alpha+1}{\alpha+2}=\frac34 LaTeX: 4(\alpha+1)=3(\alpha+2) ### Compute The parameter is two, so the cumulative distribution function on the unit interval is the third power. Take the complement at one-half. LaTeX: \alpha=2,\qquad F_R(r)=r^3 LaTeX: \Pr(R>1/2)=1-F_R(1/2)=1-\left(\frac12\right)^3=\frac78=0.875 ### Answer The probability rounds to 0.88. LaTeX: \boxed{0.88\quad\text{(E)}} ### Distractor analysis - A: This treats the variable as uniform and copies the one-half cutoff into the answer, ignoring the increasing power density and the supplied mean. - B: This is approximately the probability above 0.75, since 1-0.75^3=0.578125; it uses the mean as the cutoff instead of the required one-half. - C: Substituting the numerical mean 0.75 directly for the parameter gives 1-0.5^1.75, approximately 0.70; the parameter must first be solved from the mean equation. - D: This simply reports the given expected value. A mean is not the same as the probability that the variable exceeds one-half. ### Verification record - Main value: 0.875 - Main method: Solve (alpha+1)/(alpha+2)=3/4 to obtain alpha=2, derive F(r)=r^3, and evaluate the upper tail at one-half. - Independent check: Insert alpha=2 directly into the density and integrate 3r^2 from one-half to one; the integral equals 7/8. - Independent value: 0.875 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: transformed optical-clarity score A scanner assigns raw optical clarity U a density f(u)=2u for 00.80 is equivalent to U>0.64. Integrating the raw density above 0.64 gives 1-0.64^2=0.5904, which selects choice D. #### Variant solution ### Setup The raw clarity distribution is already normalized on the unit interval, and the dashboard transformation is increasing. LaTeX: f_U(u)=2u,\qquad 00.80\}=\{\sqrt{U}>0.80\}=\{U>0.64\} ### Compute Integrate the raw density over the converted upper-tail region. LaTeX: \Pr(C>0.80)=\int_{0.64}^{1}2u\,du LaTeX: \Pr(C>0.80)=1-0.64^2=1-0.4096=0.5904 ### Answer The probability that the displayed score exceeds 0.80 is 0.5904. LaTeX: \boxed{0.5904\quad\text{(D)}} ### Verification record - Main value: 0.5904 - Main method: Translate C>0.80 into U>0.64 and integrate the density 2u from 0.64 to 1. - Independent check: Derive the transformed CDF directly: F_C(c)=P(U at most c squared)=c to the fourth power. Thus the upper tail at 0.80 is 1-0.80^4=0.5904. - Independent value: 0.5904 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #629 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-629/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Discrete Random Variables; Moments; Variance This is a discrete-moment problem in which the supplied second moment determines an unknown negative support point. That point is -5, the mean is 4, and the variance is 47.8-4^2=31.8, selecting choice D. ### Independent solution ### Setup Separate the unknown support point's contribution to the second moment from the three known contributions. LaTeX: 47.8=0.2a^2+0.1(1^2)+0.3(3^2)+0.4(10^2) ### Model Solve the second-moment equation and then use the stated sign restriction to choose the appropriate root. LaTeX: 0.2a^2=47.8-42.8=5 LaTeX: a^2=25,\qquad a=-5 ### Compute Calculate the first moment with the negative support value and apply the raw-moment variance identity. LaTeX: \operatorname{E}[Y]=0.2(-5)+0.1(1)+0.3(3)+0.4(10)=4 LaTeX: \operatorname{Var}(Y)=\operatorname{E}[Y^2]-\operatorname{E}[Y]^2=47.8-4^2=31.8 ### Answer The variance is 31.8. LaTeX: \boxed{31.8\quad\text{(D)}} ### Distractor analysis - A: The value 5.6 is the variance's square root rounded to one decimal place, so it is the standard deviation rather than the requested variance. - B: Choosing the positive root a=5 violates the sign restriction and changes the mean to 6; then 47.8-6^2=11.8. - C: Subtracting a^2=25 directly from the second moment gives 22.8, but the variance formula requires the square of the full mean, not a support value. - E: Adding the squared mean to the second moment instead of subtracting it gives 47.8+16=63.8. ### Verification record - Main value: 31.8 - Main method: Use the given second moment to solve a^2=25, apply a<0 to select -5, calculate E[Y]=4, and evaluate 47.8-16. - Independent check: Sum squared deviations about mean 4 directly: 0.2(−9)^2+0.1(−3)^2+0.3(−1)^2+0.4(6)^2=31.8. - Independent value: 31.8 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: courier-route time mixture An autonomous courier uses a direct route with probability 0.70 and an inspection route with probability 0.30. Conditional on the direct route, travel time has mean 8 minutes and variance 4 square minutes. Conditional on the inspection route, it has mean 14 minutes and variance 9 square minutes. Calculate the unconditional variance of travel time. - A: 5.29 - B: 5.50 - C: 7.56 - D: 9.80 - E: 13.06 The expected conditional variance is 5.50, while variation between the two route means contributes 7.56. Their sum is 13.06 square minutes, so choice E is correct. #### Variant solution ### Setup Let R denote the selected route and T the travel time. First combine the two conditional means. LaTeX: \operatorname{E}[T]=0.70(8)+0.30(14)=9.8 ### Model The law of total variance separates average within-route variability from variability of the route-specific means. LaTeX: \operatorname{Var}(T)=\operatorname{E}[\operatorname{Var}(T\mid R)]+\operatorname{Var}(\operatorname{E}[T\mid R]) ### Compute Evaluate the within-route and between-route components around the overall mean 9.8. LaTeX: \operatorname{E}[\operatorname{Var}(T\mid R)]=0.70(4)+0.30(9)=5.50 LaTeX: \operatorname{Var}(\operatorname{E}[T\mid R])=0.70(8-9.8)^2+0.30(14-9.8)^2=7.56 LaTeX: \operatorname{Var}(T)=5.50+7.56=13.06 ### Answer The unconditional travel-time variance is 13.06 square minutes. LaTeX: \boxed{13.06\quad\text{(E)}} ### Verification record - Main value: 13.06 - Main method: Apply the law of total variance, adding the weighted conditional variances 5.50 to the variance 7.56 of the two conditional means. - Independent check: Conditional second moments are 4+8^2=68 and 9+14^2=205. Their mixture is 109.1; subtracting the squared overall mean 9.8^2 gives 13.06. - Independent value: 13.06 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ## SOA Exam P Sample Question #630 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-630/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Expected Value; Deductible Loss Models This is an expected excess-loss calculation for an exponential severity above an ordinary deductible. The payment expectation equals the survival probability at two times the mean residual loss, giving e^{-2}, so choice A is correct. ### Independent solution ### Setup Let L be the loss amount and express the insurer's payment under an ordinary deductible of two. LaTeX: L\sim\operatorname{Exp}(1),\qquad P=(L-2)_+ ### Model A payment occurs only when the loss crosses the deductible. Conditional on that event, exponential memorylessness makes the excess have the original mean of one. LaTeX: \Pr(L>2)=e^{-2} LaTeX: \operatorname{E}[L-2\mid L>2]=1 ### Compute Multiply the chance of a positive payment by its conditional mean amount. LaTeX: \operatorname{E}[P]=\Pr(L>2)\operatorname{E}[L-2\mid L>2] LaTeX: \operatorname{E}[P]=e^{-2}(1)=e^{-2}=0.1353352832\ldots ### Answer The expected payment per loss is e to the power negative two. LaTeX: \boxed{e^{-2}\quad\text{(A)}} ### Distractor analysis - B: Multiplying the deductible amount two by the probability of exceeding it gives 2e^{-2}; the payment is the excess above the deductible, not the deductible itself. - C: This uses survival beyond one rather than beyond two, effectively replacing the stated deductible by the distribution's mean. - D: The quantity 3e^{-2} is E[L times the indicator that L exceeds 2], which includes the retained first two units and therefore does not subtract the deductible. - E: This is the mean loss before applying any deductible and consequently overstates the insurer's expected payment. ### Verification record - Main value: 0.1353352832366127 - Main method: Use exponential memorylessness: the payment probability is exp(-2) and the conditional excess mean is one. - Independent check: Integrate the payment directly: the integral from 2 to infinity of (x-2)exp(-x) dx equals exp(-2). - Independent value: 0.1353352832366127 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: capped equipment-repair layer Equipment damage severity L is exponentially distributed with mean 2 repair units. A service contract pays the portion of a loss lying between 1 and 4 units: it pays nothing below 1, L-1 between 1 and 4, and a maximum of 3 units above 4. Calculate the expected contract payment per incident. - A: 0.1353 - B: 0.4712 - C: 0.6065 - D: 0.9424 - E: 1.2131 The capped layer can be valued by integrating its payment tail from zero to three. This gives 2(e^{-1/2}-e^{-2})=0.942391, so choice D is correct. #### Variant solution ### Setup Represent the contract as a three-unit layer attaching when the severity reaches one. LaTeX: L\sim\operatorname{Exp}(1/2) LaTeX: W=\min\!\left((L-1)_+,3\right) ### Model For a nonnegative capped payment, integrate the probability that the payment exceeds each level from zero up to its cap. LaTeX: \operatorname{E}[W]=\int_0^3\Pr(W>t)\,dt LaTeX: \Pr(W>t)=\Pr(L>1+t)=e^{-(1+t)/2} ### Compute Evaluate the payment-tail integral across the full three-unit layer. LaTeX: \operatorname{E}[W]=\int_0^3 e^{-(1+t)/2}\,dt LaTeX: \operatorname{E}[W]=2\left(e^{-1/2}-e^{-2}\right)=0.9423907530\ldots ### Answer The expected capped-layer payment is approximately 0.9424 repair units. LaTeX: \boxed{0.9424\quad\text{(D)}} ### Verification record - Main value: 0.9423907529520414 - Main method: Use the payment-tail identity and integrate exp(-(1+t)/2) over payment levels from zero through three. - Independent check: Express the layer as (L-1) positive minus (L-4) positive. The two stop-loss means are 2exp(-1/2) and 2exp(-2), whose difference is 0.9423907530. - Independent value: 0.9423907529520414 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #631 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-631/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Conditional Distributions; Bernoulli Distribution; Conditional Variance This is a conditional-variance calculation for a binary random variable. Within the specified row, the conditional probability of one is 4/5, so the Bernoulli variance is (4/5)(1/5)=4/25 and choice A is correct. ### Independent solution ### Setup Add the two joint masses in the row selected by the conditioning event. LaTeX: \Pr(X=1)=\frac18+\frac12=\frac58 ### Model After conditioning, Y is Bernoulli. Its success probability is the joint mass for Y equal to one divided by the selected row total. LaTeX: p=\Pr(Y=1\mid X=1)=\frac{1/2}{5/8}=\frac45 ### Compute Apply the Bernoulli variance formula with conditional success probability four-fifths. LaTeX: \operatorname{Var}(Y\mid X=1)=p(1-p) LaTeX: \operatorname{Var}(Y\mid X=1)=\frac45\left(1-\frac45\right)=\frac4{25}=0.16 ### Answer The conditional variance is four twenty-fifths. LaTeX: \boxed{\frac4{25}\quad\text{(A)}} ### Distractor analysis - B: The fraction 15/64 is (5/8)(3/8), the variance of the indicator that X equals one. It uses the conditioning-row probability as a Bernoulli parameter instead of forming Y's conditional distribution. - C: Using the unnormalized joint mass 1/2 as though it were the conditional success probability gives (1/2)(1/2)=1/4. - D: The value 3/4 is the complement of the separate joint cell with mass 1/4; taking that complement does not condition Y on X equal to one. - E: The fraction 4/5 is the conditional probability and conditional mean of Y, not its conditional variance. ### Verification record - Main value: 0.16 - Main method: Normalize the X=1 row to obtain Bernoulli success probability 4/5 and calculate p(1-p)=4/25. - Independent check: Because Y is binary, both conditional raw moments equal 4/5; subtracting the squared mean gives 4/5-(4/5)^2=4/25. - Independent value: 0.16 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: warning-conditioned calibration spread A sensor line operates in fast mode with probability 0.60 and precision mode with probability 0.40. A warning appears with probability 0.20 in fast mode and 0.50 in precision mode. Given the mode, the warning is independent of calibration adjustment Y. In fast mode, Y has mean 0 and variance 1; in precision mode, it has mean 3 and variance 2. Given that a warning appeared, calculate Var(Y). - A: 0.3200 - B: 1.6250 - C: 1.8750 - D: 2.1094 - E: 3.7344 Bayes' rule gives posterior mode weights 3/8 for fast and 5/8 for precision after a warning. The conditional variance is the 1.625 within-mode component plus the 2.109375 between-mode component, totaling 3.734375 and selecting choice E. #### Variant solution ### Setup Let W denote a warning. Compute its probability and update the operating-mode weights using Bayes' rule. LaTeX: \Pr(W)=0.60(0.20)+0.40(0.50)=0.32 LaTeX: \Pr(F\mid W)=\frac{0.60(0.20)}{0.32}=\frac38,\qquad \Pr(P\mid W)=\frac58 ### Model Apply the conditional law of total variance across the two posterior modes. LaTeX: \operatorname{Var}(Y\mid W)=\operatorname{E}[\operatorname{Var}(Y\mid M,W)\mid W]+\operatorname{Var}(\operatorname{E}[Y\mid M,W]\mid W) ### Compute Conditional independence allows the supplied mode-specific moments to be used after observing the warning. LaTeX: \operatorname{E}[\operatorname{Var}(Y\mid M,W)\mid W]=\frac38(1)+\frac58(2)=1.625 LaTeX: \operatorname{E}[Y\mid W]=\frac38(0)+\frac58(3)=1.875 LaTeX: \operatorname{Var}(\operatorname{E}[Y\mid M,W]\mid W)=\frac38(0-1.875)^2+\frac58(3-1.875)^2=2.109375 LaTeX: \operatorname{Var}(Y\mid W)=1.625+2.109375=3.734375 ### Answer The calibration adjustment has conditional variance approximately 3.7344 after a warning. LaTeX: \boxed{3.7344\quad\text{(E)}} ### Verification record - Main value: 3.734375 - Main method: Use Bayes' rule for posterior mode weights and then add the conditional expected variance 1.625 to the variance 2.109375 of conditional means. - Independent check: The mode-specific second moments are 1 and 11. Their posterior mixture is 7.25, while the posterior mean is 1.875; 7.25-1.875 squared equals 3.734375. - Independent value: 3.734375 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ## SOA Exam P Sample Question #632 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-632/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Binomial Distribution; Limited Payments The number of covered occurrences is binomial with five trials and probability 0.5. Capping the count at three gives an expected payment of 30(73/32) = 68.4375, so the nearest listed value is D. ### Independent solution ### Setup Let N count the covered occurrences across the five independent periods. LaTeX: N\sim\operatorname{Binomial}(5,0.5) LaTeX: B=30\min(N,3) ### Model Group all counts of three or more at the contractual payment limit. LaTeX: E[\min(N,3)]=\sum_{k=0}^{2}k\Pr(N=k)+3\Pr(N\ge3) ### Compute Use the symmetric binomial probabilities with denominator 32. LaTeX: E[\min(N,3)]=\frac{5+20+3(10+5+1)}{32}=\frac{73}{32} LaTeX: E[B]=30\left(\frac{73}{32}\right)=68.4375 ### Answer The expected payment rounds to the listed value 68. LaTeX: \boxed{68\quad\text{(D)}} ### Distractor analysis - A: This treats the three-payment limit as three ordinary expected trials, giving 3(30)(0.5), instead of applying the limit to the five-period count. - B: This undercounts the upper tail as though only two payments were available and then rounds the result. - C: This is obtained by omitting the five-occurrence case from the capped upper tail, producing 65.625 before rounding. - E: This is the uncapped expectation 30(5)(0.5), so it ignores that counts above three receive no additional payment. ### Verification record - Main value: 68.4375 - Main method: Evaluate the expectation of 30 times a binomial count truncated at three. - Independent check: The tail-sum identity gives E[min(N,3)] = P(N>=1)+P(N>=2)+P(N>=3) = 31/32 + 26/32 + 16/32 = 73/32. - Independent value: 68.4375 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: fabrication safety triggers A fabrication line completes seven independent production runs. Its safety relay is triggered on any run with probability 0.25. A reliability rebate ignores the first two triggers, then pays 50 for each additional trigger, subject to at most three paid triggers. Calculate the expected rebate. - A: 0.000 - B: 12.180 - C: 16.351 - D: 16.422 - E: 83.258 For a binomial trigger count N, the number of paid triggers is min((N-2)+,3). Its expectation is the sum of the three binomial tails beginning at 3, 4, and 5, which equals 0.3270263672. Multiplication by 50 gives 16.35131836, so choice C is correct. #### Variant solution ### Setup Let N count relay triggers over the seven independent runs and translate the two-trigger threshold and three-payment ceiling into a payment variable. LaTeX: N\sim\operatorname{Binomial}(7,0.25) LaTeX: R=50\min\!\left((N-2)_+,3\right) ### Model Apply the tail-sum identity to the paid-trigger count, whose possible values are zero through three. LaTeX: E\!\left[\min\!\left((N-2)_+,3\right)\right] LaTeX: =\Pr(N\ge3)+\Pr(N\ge4)+\Pr(N\ge5) ### Compute Evaluate the three required binomial tails and multiply their sum by the rebate per paid trigger. LaTeX: \Pr(N\ge3)+\Pr(N\ge4)+\Pr(N\ge5) LaTeX: =0.2435913086+0.0705566406+0.0128784180 LaTeX: =0.3270263672 LaTeX: E[R]=50(0.3270263672)=16.35131836 ### Answer The expected reliability rebate rounds to 16.351. LaTeX: \boxed{16.351\quad\text{(C)}} ### Verification record - Main value: 16.351318359375 - Main method: Use the tail-sum identity for min((N-2)+,3), evaluate the binomial tails at 3, 4, and 5, and multiply by 50. - Independent check: Directly enumerate N=3 through 7 using payment counts 1, 2, 3, 3, 3; the probability-weighted payment is 16.351318359375. - Independent value: 16.351318359375 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #633 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-633/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Poisson Distribution; Conditional Distributions Conditioning two independent Poisson counts on their sum produces a binomial allocation with success probability 2/(2+3). The probability of six allocations to the first source out of eight is 0.04128768, so the answer is C. ### Independent solution ### Setup Let A and B be the independent counts with Poisson means 2 and 3. LaTeX: A\sim\operatorname{Poisson}(2),\qquad B\sim\operatorname{Poisson}(3) ### Model Given the combined count, each event is assigned to the first source with probability proportional to its Poisson rate. LaTeX: A\mid(A+B=8)\sim\operatorname{Binomial}\left(8,\frac25\right) ### Compute Evaluate the conditional binomial mass at six. LaTeX: \Pr(A=6\mid A+B=8)=\binom86\left(\frac25\right)^6\left(\frac35\right)^2 LaTeX: \Pr(A=6\mid A+B=8)=0.04128768 ### Answer The probability rounds to 0.041. LaTeX: \boxed{0.041\quad\text{(C)}} ### Distractor analysis - A: This is approximately the unconditioned joint probability of six first-source events and two second-source events; it omits division by the probability of a total of eight. - B: This is approximately the unconditional Poisson probability that the first source alone has count six. - D: This comes from perturbing the conditional source share instead of using the rate ratio 2/(2+3). - E: This substantially overweights allocations to the lower-rate source and does not follow the conditional binomial mass. ### Verification record - Main value: 0.04128768 - Main method: Use the conditional binomial distribution of one independent Poisson component given the total. - Independent check: Divide P(A=6,B=2) by P(A+B=8); cancellation of the common exp(-5) factor gives 8!/(6!2!) times 2^6 3^2 / 5^8. - Independent value: 0.04128768 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: wafer defect classification A scanner records three independent classes of microscopic wafer defects. Their respective Poisson means per wafer are 1, 2, and 3. Given that the scanner finds six defects in total on a wafer, calculate the probability that the three class counts are exactly 1, 2, and 3, respectively. - A: 0.002315 - B: 0.022309 - C: 0.082305 - D: 0.138889 - E: 0.312500 Conditioned on six total defects, the three class labels have multinomial probabilities 1/6, 2/6, and 3/6. The allocation probability for counts 1, 2, and 3 is 5/36=0.1388889, so choice D is correct. #### Variant solution ### Setup Let the three independent class counts have the stated Poisson means and condition on their combined count. LaTeX: (\lambda_1,\lambda_2,\lambda_3)=(1,2,3) LaTeX: \lambda_1+\lambda_2+\lambda_3=6 ### Model Independent Poisson categories conditioned on their total follow a multinomial allocation with rate-proportional cell probabilities. LaTeX: (K_1,K_2,K_3)\mid(K_1+K_2+K_3=6) LaTeX: \sim\operatorname{Multinomial}\!\left(6;\frac16,\frac26,\frac36\right) ### Compute Evaluate the multinomial mass at the ordered class-count vector. LaTeX: \Pr((K_1,K_2,K_3)=(1,2,3)\mid K_1+K_2+K_3=6) LaTeX: =\frac{6!}{1!2!3!}\left(\frac16\right)\left(\frac26\right)^2\left(\frac36\right)^3 LaTeX: =\frac5{36}=0.1388888889\ldots ### Answer The conditional allocation probability rounds to 0.138889. LaTeX: \boxed{0.138889\quad\text{(D)}} ### Verification record - Main value: 0.1388888888888889 - Main method: Condition the three independent Poisson counts on their total and evaluate the resulting multinomial mass at (1,2,3). - Independent check: Divide the unconditioned joint mass P(K1=1,K2=2,K3=3) by the Poisson(6) mass at 6; cancellation of exp(-6) gives 5/36. - Independent value: 0.1388888888888889 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #634 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-634/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Sums of Independent Random Variables; Coefficient of Variation Independence lets the second component inherit mean 0.8 and variance 1.35 from the stated totals. Its coefficient of variation is sqrt(1.35)/0.8 = 1.45237, so the answer is E. ### Independent solution ### Setup Recover the mean of the unobserved component by subtracting component means. LaTeX: E[Z]=E[X]-E[Y]=2.00-1.20=0.80 ### Model For independent components, the variance of their sum is the sum of their variances. LaTeX: \operatorname{Var}(Z)=\operatorname{Var}(X)-\operatorname{Var}(Y) LaTeX: \operatorname{Var}(Z)=2.25-0.90=1.35 ### Compute Divide the component standard deviation by its mean. LaTeX: \operatorname{CV}(Z)=\frac{\sqrt{1.35}}{0.80}=1.452368754\ldots ### Answer The coefficient of variation rounds to 1.45. LaTeX: \boxed{1.45\quad\text{(E)}} ### Distractor analysis - A: This results from mixing the first component's mean with the second component's variance rather than using a consistent standard-deviation-to-mean ratio. - B: This uses an incorrect variance subtraction or divides by a total-scale quantity instead of the mean of the requested component. - C: This is the standard deviation sqrt(1.35) by itself, before division by the component mean. - D: This is the variance 1.35, not the coefficient of variation. ### Verification record - Main value: 1.4523687548277815 - Main method: Subtract independent component means and variances, then divide the recovered standard deviation by the recovered mean. - Independent check: Recombining mean 0.8 and variance 1.35 with the known component reproduces total mean 2.0 and variance 2.25; sqrt(1.35)/0.8 then gives 1.452368755. - Independent value: 1.4523687548277815 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: weighted fabrication cost A fabrication shop models two independent random resource loads, X and Y. Their means are 3 and 5 units, and their standard deviations are 1 and 2 units, respectively. A batch cost index is defined by C=2X+Y. Calculate the coefficient of variation of C. - A: 0.223 - B: 0.257 - C: 0.354 - D: 0.364 - E: 0.727 The weighted cost has mean 2(3)+5=11. Independence gives variance 2 squared times 1 squared plus 2 squared, or 8. Therefore its coefficient of variation is sqrt(8)/11=0.2571297, so choice B is correct. #### Variant solution ### Setup Apply linearity of expectation to the weighted cost index. LaTeX: E[C]=2E[X]+E[Y]=2(3)+5=11 ### Model Independence removes the covariance term, while the coefficient on X is squared in the variance. LaTeX: \operatorname{Var}(C)=2^2\operatorname{Var}(X)+\operatorname{Var}(Y) LaTeX: \operatorname{Var}(C)=4(1^2)+2^2=8 ### Compute Divide the standard deviation of the cost index by its mean. LaTeX: \operatorname{CV}(C)=\frac{\sqrt{\operatorname{Var}(C)}}{E[C]} LaTeX: \operatorname{CV}(C)=\frac{\sqrt8}{11}=0.2571297386\ldots ### Answer The cost-index coefficient of variation rounds to 0.257. LaTeX: \boxed{0.257\quad\text{(B)}} ### Verification record - Main value: 0.25712973861329 - Main method: Use linearity for the mean, square the linear coefficient in the independent variance calculation, and divide the resulting standard deviation by the mean. - Independent check: Expand the second moment: E[X^2]=10, E[Y^2]=29, and independence gives E[XY]=15, so E[C^2]=4(10)+29+4(15)=129 and Var(C)=129-11^2=8. - Independent value: 0.25712973861329 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #635 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-635/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Exponential Distribution; Means and Variances The second moment condition first forces the mean of Y to equal 2. Substitution into the remaining mean-variance identity gives alpha squared minus alpha minus 2 equal to zero, whose positive root is 2, so the answer is D. ### Independent solution ### Setup Use the fact that an exponential variable with mean m has variance m squared. LaTeX: E[U]=\alpha+\beta+4 LaTeX: \operatorname{Var}(U)=\alpha^2+\beta^2+16 LaTeX: E[V]=\alpha-\beta,\qquad \operatorname{Var}(V)=\alpha^2+\beta^2 ### Model The relation involving the differences of the two means and variances isolates beta. LaTeX: E[U]-E[V]=2\beta+4 LaTeX: \frac{\operatorname{Var}(U)-\operatorname{Var}(V)}{2}=8 LaTeX: 2\beta+4=8 ### Compute Solve for beta, then apply the remaining equality and retain the positive exponential mean. LaTeX: \beta=2 LaTeX: \alpha+6=\alpha^2+4 LaTeX: \alpha^2-\alpha-2=0 LaTeX: \alpha\in\{2,-1\} ### Answer An exponential mean must be positive, so alpha equals 2. LaTeX: \boxed{\alpha=2.0\quad\text{(D)}} ### Distractor analysis - A: This can result from treating the fixed exponential mean 4 as a variance of 4 rather than 16. - B: This is obtained by keeping an incorrect linearized version of the exponential variance relation. - C: This can arise from averaging the positive and inadmissible algebraic roots instead of enforcing positivity. - E: This results from a sign or constant error when rearranging alpha squared minus alpha minus 2. ### Verification record - Main value: 2 - Main method: Translate exponential means to variances, use the difference condition to solve beta, and solve the resulting quadratic for alpha. - Independent check: At alpha=beta=2, E(U)=8 equals Var(V)=8, while E(U)-E(V)=8 and [Var(U)-Var(V)]/2=(24-8)/2=8. - Independent value: 2 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: routed maintenance time A maintenance request is routed to a quick workflow with probability 0.40 and to an extensive workflow otherwise. Conditional on the route, completion time is exponential with mean 2 hours for the quick workflow and 5 hours for the extensive workflow. Calculate the unconditional variance of completion time. - A: 3.80 - B: 14.44 - C: 16.60 - D: 18.76 - E: 33.20 The unconditional first moment is 3.8 and the second moment is 33.2 because an exponential variable with mean m has second moment 2m squared. Subtracting 3.8 squared gives variance 18.76, so choice D is correct. #### Variant solution ### Setup Let X be completion time and M identify the selected workflow. Average the two conditional means using the routing probabilities. LaTeX: E[X]=0.40(2)+0.60(5)=3.8 ### Model For an exponential variable with mean m, the second moment is twice the square of the mean. LaTeX: E[X^2\mid M=\text{quick}]=2(2^2)=8 LaTeX: E[X^2\mid M=\text{extensive}]=2(5^2)=50 ### Compute Mix the conditional second moments and subtract the square of the unconditional mean. LaTeX: E[X^2]=0.40(8)+0.60(50)=33.2 LaTeX: \operatorname{Var}(X)=33.2-(3.8)^2=18.76 ### Answer The unconditional completion-time variance is 18.76 square hours. LaTeX: \boxed{18.76\quad\text{(D)}} ### Verification record - Main value: 18.76 - Main method: Mix the exponential conditional first and second moments, then apply Var(X)=E[X^2]-E[X]^2. - Independent check: The law of total variance gives E[Var(X|M)]=16.60 and Var(E[X|M])=0.40(2-3.8)^2+0.60(5-3.8)^2=2.16; their sum is 18.76. - Independent value: 18.76 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #636 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-636/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Central Limit Theorem; Sums of Independent Random Variables The aggregate has mean 3,000,000 and standard deviation 50,000 times the square root of 60. A central-limit approximation gives an upper-tail probability of 0.09835, so the nearest listed value is choice A. ### Independent solution ### Setup Let S be the aggregate of the 60 independent exponential loss variables. LaTeX: S=\sum_{i=1}^{60}X_i LaTeX: E[S]=60(50000)=3000000 ### Model An exponential variable has standard deviation equal to its mean. Independence makes the component variances additive, and the central limit theorem supplies a normal approximation. LaTeX: \operatorname{SD}(S)=50000\sqrt{60}=387298.3346\ldots LaTeX: S\ \dot\sim\ N\!\left(3000000,\,60(50000)^2\right) ### Compute Standardize the aggregate threshold and evaluate the upper standard-normal tail. LaTeX: z=\frac{3500000-3000000}{50000\sqrt{60}}=\frac{10}{\sqrt{60}}=1.2909944487\ldots LaTeX: \Pr(S>3500000)\approx1-\Phi(1.2909944487)=0.0983528012\ldots ### Answer The approximation is closest to 0.10. LaTeX: \boxed{0.10\quad\text{(A)}} ### Distractor analysis - B: This can result from rounding the standardized threshold too aggressively before taking the normal tail. - C: This understates the standardized distance by using an inflated aggregate standard deviation. - D: This divides the threshold excess by the aggregate mean rather than by its standard deviation, producing a small z-score and an upper tail near 0.43. - E: This treats the threshold as though it were equal to the aggregate mean. ### Verification record - Main value: 0.0983528012 - Main method: Add the independent exponential means and variances, standardize the threshold, and evaluate the normal upper tail. - Independent check: The exact aggregate is Erlang with shape 60 and scale 50000; its survival probability at 3500000 is 0.1023971494, independently confirming choice A. - Independent value: 0.1023971494 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: capsule shipment weight A shipment contains 50 independent sensor capsules and 30 independent shielding clips. Capsule weights have mean 12 grams and standard deviation 2 grams; clip weights have mean 20 grams and standard deviation 3 grams. All item weights are mutually independent. Using a normal approximation, estimate the probability that the shipment weighs between 1170 and 1230 grams. - A: 0.083 - B: 0.417 - C: 0.834 - D: 0.917 - E: 0.970 The two item groups give total mean 1200 grams and variance 470 square grams. The interval endpoints are 1.383797 standard deviations below and above the mean, giving central normal probability 0.8335793 and selecting choice C. #### Variant solution ### Setup Add the expected weights from the two item groups. LaTeX: T=\sum_{i=1}^{50}X_i+\sum_{j=1}^{30}Y_j LaTeX: E[T]=50(12)+30(20)=1200 ### Model Mutual independence makes every item variance additive. Approximate the heterogeneous sum by a normal distribution. LaTeX: \operatorname{Var}(T)=50(2^2)+30(3^2)=470 LaTeX: T\ \dot\sim\ N(1200,470) ### Compute Standardize both interval endpoints and evaluate the central normal probability. LaTeX: z=\frac{30}{\sqrt{470}}=1.3837968120\ldots LaTeX: \Pr(11701000000)\approx 1-\Phi(0.0441911530)=0.4823760170\ldots ### Answer The approximate probability rounds to 0.48. LaTeX: \boxed{0.48\quad\text{(B)}} ### Distractor analysis - A: Using 130 as the standard deviation of the entire sum gives z=100/130 and an upper tail near 0.22; the aggregate standard deviation must grow by the square root of 303. - C: Rounding the aggregate mean of 999900 to the cutoff before standardizing makes the z-score zero and produces 0.50. - D: This is the lower-tail probability Phi(0.044191), whereas the requested aggregate lies above the cutoff. - E: This is approximately the lower-tail complement of choice A, combining the unscaled one-day standard deviation with the wrong tail direction. ### Verification record - Main value: 0.48237601700553145 - Main method: Add the 303 independent means and variances, then evaluate the normal upper tail at z=100/(130 sqrt(303)). - Independent check: Re-express the event using the sample mean: its standard error is 130/sqrt(303), and the threshold mean is 1000000/303; a two-decimal normal-table lookup at z=0.04 gives 0.4840, which independently confirms choice B. - Independent value: 0.484 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: culture-vial usable volume A laboratory prepares 144 culture vials. The usable liquid volume in each vial is independent, with mean 50 milliliters and standard deviation 12 milliliters. Use a normal approximation to find the probability that the combined usable volume exceeds 7300 milliliters. - A: 0.000 - B: 0.244 - C: 0.477 - D: 0.756 - E: 1.000 The combined mean is 7200 milliliters and the combined standard deviation is 144 milliliters. The upper-tail probability at z=100/144 is 0.243702, so choice B is correct. #### Variant solution ### Setup Let V_i denote usable volume from vial i and let W be the combined volume. LaTeX: W=\sum_{i=1}^{144}V_i LaTeX: \operatorname{E}[W]=144(50)=7200 ### Model The independent vial contributions allow variance addition, and the large count supports a central-limit approximation. LaTeX: \operatorname{SD}(W)=12\sqrt{144}=144 LaTeX: W\ \dot\sim\ N(7200,144^2) ### Compute Convert 7300 milliliters to a standard-normal score and take the upper tail. LaTeX: z=\frac{7300-7200}{144}=0.6944444444\ldots LaTeX: \Pr(W>7300)\approx 1-\Phi(0.6944444444)=0.2437017648\ldots ### Answer The requested probability is approximately 0.244. LaTeX: \boxed{0.244\quad\text{(B)}} ### Verification record - Main value: 0.24370176483801265 - Main method: Apply the normal approximation to a sum with mean 144(50) and standard deviation 12 sqrt(144), then evaluate the upper tail at 7300. - Independent check: Using the sample mean instead gives a threshold of 7300/144 and standard error 12/sqrt(144); this yields the same z-score 25/36 and probability. - Independent value: 0.24370176483801265 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #638 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-638/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Conditional Distributions; Variance This is a conditional-moment calculation from one row of a joint probability table. Normalizing that row gives a conditional mean of 86/33 and a second moment of 274/33, so the variance is 1646/1089, or 1.511478, and choice B is correct. ### Independent solution ### Setup Condition on the stated value of the first variable. The corresponding five joint masses have total probability 0.33. LaTeX: \Pr(S=2)=0.10+0.03+0.07+0.05+0.08=0.33 LaTeX: \Pr(B=b\mid S=2)=\frac{p_{2,b}}{0.33},\qquad b=1,\ldots,5 ### Model Compute the first two conditional raw moments and then use the variance identity. LaTeX: \operatorname{Var}(B\mid S=2)=\operatorname{E}[B^2\mid S=2]-\operatorname{E}[B\mid S=2]^2 ### Compute Weight the five possible values by the normalized row probabilities. LaTeX: \operatorname{E}[B\mid S=2]=\frac{1(0.07)+2(0.10)+3(0.08)+4(0.05)+5(0.03)}{0.33}=\frac{86}{33} LaTeX: \operatorname{E}[B^2\mid S=2]=\frac{1^2(0.07)+2^2(0.10)+3^2(0.08)+4^2(0.05)+5^2(0.03)}{0.33}=\frac{274}{33} LaTeX: \operatorname{Var}(B\mid S=2)=\frac{274}{33}-\left(\frac{86}{33}\right)^2=\frac{1646}{1089}=1.5114784206\ldots ### Answer The conditional variance rounds to 1.51. LaTeX: \boxed{1.51\quad\text{(B)}} ### Distractor analysis - A: The value 1.23 is approximately the conditional standard deviation, obtained by taking the square root of the required variance. - C: Using the joint row entries without dividing by their row total gives 2.74-0.86 squared, approximately 2.00; those entries do not form a conditional distribution until normalized. - D: Dividing the first-moment numerator 0.86 by an incorrectly added row total of 0.35 gives about 2.46, and that quantity would be a mean rather than a variance. - E: A value near 6.06 is a squared location estimate, not a variance computed as the second moment minus the square of the mean; it also exceeds the correct moment calculation by a wide margin. ### Verification record - Main value: 1.5114784205693297 - Main method: Normalize the S=2 row and evaluate E[B^2|S=2]-E[B|S=2]^2 exactly as 1646/1089. - Independent check: Sum the normalized squared deviations (b-86/33)^2 directly over b=1 through 5; this gives the same fraction 1646/1089. - Independent value: 1.5114784205693297 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: night-mode inspection flags An automated panel inspector records its operating mode M and the number D of flags raised. In the night-mode row of the joint probability table, the entries for D=0, 1, 2, and 3 are 0.03, 0.09, 0.12, and 0.06; the remaining joint mass belongs to other modes. Given that the inspector is in night mode, calculate the variance of D. - A: 0.300 - B: 0.850 - C: 0.810 - D: 1.700 - E: 3.700 The night-mode row has mass 0.30. After normalization, the conditional mean is 1.70 and the conditional second moment is 3.70, so the variance is 3.70 minus 1.70 squared, or 0.81; this selects choice C. #### Variant solution ### Setup First find the probability of the conditioning mode from its joint-table row. LaTeX: \Pr(M=\text{night})=0.03+0.09+0.12+0.06=0.30 ### Model Divide each row entry by 0.30 to obtain the conditional distribution of the flag count. LaTeX: \Pr(D=d\mid M=\text{night})=\frac{(0.03,0.09,0.12,0.06)_d}{0.30},\qquad d=0,1,2,3 ### Compute Calculate both conditional moments before applying the variance identity. LaTeX: \operatorname{E}[D\mid M=\text{night}]=\frac{0(0.03)+1(0.09)+2(0.12)+3(0.06)}{0.30}=1.70 LaTeX: \operatorname{E}[D^2\mid M=\text{night}]=\frac{0^2(0.03)+1^2(0.09)+2^2(0.12)+3^2(0.06)}{0.30}=3.70 LaTeX: \operatorname{Var}(D\mid M=\text{night})=3.70-1.70^2=0.81 ### Answer The conditional variance of the number of flags is 0.81. LaTeX: \boxed{0.810\quad\text{(C)}} ### Verification record - Main value: 0.81 - Main method: Normalize the night-mode row, compute conditional moments 1.70 and 3.70, and subtract the squared mean. - Independent check: Evaluate the conditional squared-deviation sum with probabilities 0.10, 0.30, 0.40, and 0.20 around mean 1.70. - Independent value: 0.81 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #639 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-639/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Independence The two independent Poisson counts must fall at or below their respective coverage limits. Their cumulative probabilities are 2.5e to the negative one and 1.3e to the negative 0.3, whose product is 0.885728; this is choice B. ### Independent solution ### Setup Let X and Y be the two annual event counts. The required outcome occurs when each count stays within its own limit. LaTeX: X\sim\operatorname{Poisson}(1),\qquad Y\sim\operatorname{Poisson}(0.3) LaTeX: \{\text{both limits met}\}=\{X\le2\}\cap\{Y\le1\} ### Model Independence allows the joint event probability to factor into two Poisson cumulative probabilities. LaTeX: \Pr(X\le2,Y\le1)=\Pr(X\le2)\Pr(Y\le1) ### Compute Sum each Poisson mass from zero through the applicable upper bound. LaTeX: \Pr(X\le2)=e^{-1}\left(1+1+\frac{1}{2}\right)=2.5e^{-1} LaTeX: \Pr(Y\le1)=e^{-0.3}(1+0.3)=1.3e^{-0.3} LaTeX: \Pr(X\le2,Y\le1)=(2.5e^{-1})(1.3e^{-0.3})=0.8857283274\ldots ### Answer The product matches the expression in choice B. LaTeX: \boxed{(2.5e^{-1})(1.3e^{-0.3})\quad\text{(B)}} ### Distractor analysis - A: This multiplies the probabilities of landing exactly at both limits. Counts below either limit are also fully accommodated and must be included. - C: This is the complement of the probability that both limits are met, so it describes at least one count exceeding its allowance. - D: This multiplies the two exceedance probabilities and therefore requires both counts to be over their limits. - E: This is the complement of both counts exceeding their limits; it allows one count to exceed its limit and is therefore too broad. ### Verification record - Main value: 0.8857283273605411 - Main method: Multiply the independent Poisson cumulative probabilities P(X<=2)=2.5e^-1 and P(Y<=1)=1.3e^-0.3. - Independent check: Compute the two tail probabilities, 0.0803013971 and 0.0369363131, then use inclusion-exclusion for the chance that neither tail occurs. - Independent value: 0.885728327360541 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: greenhouse service thresholds During a greenhouse season, the number of pump resets is Poisson with mean 0.6, and the number of sensor recalibrations is Poisson with mean 1.4. The counts are independent. A basic service plan covers at most one reset and at most two recalibrations. Calculate the probability that every service event is covered. - A: 0.020 - B: 0.080 - C: 0.268 - D: 0.732 - E: 0.980 The two relevant Poisson cumulative probabilities are 1.6e to the negative 0.6 and 3.38e to the negative 1.4. Their independent product is 0.731893, which rounds to choice D. #### Variant solution ### Setup Let R count pump resets and let C count sensor recalibrations during the season. LaTeX: R\sim\operatorname{Poisson}(0.6),\qquad C\sim\operatorname{Poisson}(1.4) LaTeX: \{\text{all events covered}\}=\{R\le1,C\le2\} ### Model Because the two counts are independent, multiply their within-threshold probabilities. LaTeX: \Pr(R\le1,C\le2)=\Pr(R\le1)\Pr(C\le2) ### Compute Evaluate each short Poisson sum and then multiply. LaTeX: \Pr(R\le1)=e^{-0.6}(1+0.6)=1.6e^{-0.6} LaTeX: \Pr(C\le2)=e^{-1.4}\left(1+1.4+\frac{1.4^2}{2}\right)=3.38e^{-1.4} LaTeX: \Pr(R\le1,C\le2)=5.408e^{-2}=0.7318932117\ldots ### Answer The service plan covers every event with probability about 0.732. LaTeX: \boxed{0.732\quad\text{(D)}} ### Verification record - Main value: 0.7318932117436014 - Main method: Evaluate and multiply the Poisson CDFs through one at mean 0.6 and through two at mean 1.4. - Independent check: Compute upper-tail probabilities 0.1219013822 and 0.1665022619, then take the complement of their union using independence. - Independent value: 0.7318932117436014 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #640 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-640/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Conditional Probability; Binomial Distribution Let N be the number of adverse outcomes among the three independent trials. Dividing the expanded probability for N=1 by the expanded positive-count probability and cancelling p gives 3(1-p) squared over p squared minus 3p plus 3. This is algebraically equivalent to choice E. ### Independent solution ### Setup Represent the number of adverse outcomes among the three independent trials by N. LaTeX: N\sim\operatorname{Binomial}(3,p) ### Model Because exactly one adverse outcome implies at least one, the conditional numerator reduces to the probability that N equals one. LaTeX: \Pr(N=1\mid N\ge1)=\frac{\Pr(N=1)}{\Pr(N\ge1)} ### Compute Expand both probabilities as polynomials, then cancel their common factor p. LaTeX: \Pr(N=1)=3p-6p^2+3p^3 LaTeX: \Pr(N\ge1)=3p-3p^2+p^3 LaTeX: \Pr(N=1\mid N\ge1)=\frac{3-6p+3p^2}{3-3p+p^2}=\frac{3(1-p)^2}{p^2-3p+3} ### Answer The simplified expression is equivalent to the expression in choice E. LaTeX: \boxed{\frac{3(1-p)^2}{p^2-3p+3}\quad\text{(E)}} ### Distractor analysis - A: This is the chance that the other two trials avoid the adverse outcome after a particular trial is fixed as adverse; it omits both the three possible positions and the conditioning denominator. - B: This is the unconditional probability of exactly one adverse outcome and does not divide by the probability that at least one occurs. - C: This results from dividing the exactly-one probability by (1-p) cubed, the zero-outcome probability, instead of by its complement. - D: This conditions correctly but counts only one specified trial as the sole adverse outcome, omitting the factor of three. ### Verification record - Main value: 3(1-p)^2/(p^2-3p+3) - Main method: Apply the conditional-probability ratio for a binomial count, expand numerator and denominator, and cancel their common factor. - Independent check: At p=0.2, direct enumeration gives 0.384 mass across the three exactly-one outcomes and 0.488 across all seven nonzero outcomes, whose ratio 0.786885246 agrees with the formula. - Independent value: 0.7868852459016398 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: acoustic sensor self-test Four acoustic sensors run independent self-tests. Each sensor fails its test with probability q, where q is strictly between zero and one. Given that the array reports at least one failure, determine the probability that exactly two sensors fail. - A: q^2(1-q)^2 - B: 6q^2(1-q)^2 - C: 6q^2(1-q)^2 / [1-(1-q)^4] - D: [1-(1-q)^4] / [6q^2(1-q)^2] - E: 1 - 6q^2(1-q)^2 / [1-(1-q)^4] The unconditional probability of two failures is 6q squared times (1-q) squared, while at least one failure has probability 1-(1-q) to the fourth power. Their ratio is choice C. #### Variant solution ### Setup Let F be the number of failed self-tests in the four-sensor array. LaTeX: F\sim\operatorname{Binomial}(4,q) ### Model The exactly-two event is contained in the positive-failure event, so use a simple conditional ratio. LaTeX: \Pr(F=2\mid F\ge1)=\frac{\Pr(F=2)}{\Pr(F\ge1)} ### Compute The binomial coefficient counts the six possible failed pairs, and a complement gives the denominator. LaTeX: \Pr(F=2)=\binom42q^2(1-q)^2=6q^2(1-q)^2 LaTeX: \Pr(F\ge1)=1-(1-q)^4 LaTeX: \Pr(F=2\mid F\ge1)=\frac{6q^2(1-q)^2}{1-(1-q)^4} ### Answer The conditional probability is the expression in choice C. LaTeX: \boxed{\frac{6q^2(1-q)^2}{1-(1-q)^4}\quad\text{(C)}} ### Verification record - Main value: 6q^2(1-q)^2/[1-(1-q)^4] - Main method: Divide the binomial probability of two failures among four sensors by the complement probability of no failures. - Independent check: For q=0.2, enumerate the 16 labeled pass-fail patterns: the six two-failure patterns total 0.1536, while the 15 nonzero patterns total 0.5904; their ratio is 0.260162602. - Independent value: 0.2601626016260164 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #641 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-641/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Conditional Probability; Independence List the three mutually exclusive ways that exactly one loss category can occur. The target category alone has probability 0.072 and all exactly-one cases total 0.306, so the conditional probability is 0.235294 and choice B is correct. ### Independent solution ### Setup Write F, W, and T for occurrence of a loss in the first, second, and third categories. Their loss probabilities are the complements of the given no-loss probabilities. LaTeX: \Pr(F)=0.10,\qquad \Pr(W)=0.10,\qquad \Pr(T)=0.20 ### Model Under independence, multiply across categories for each of the three disjoint exactly-one cases. LaTeX: \Pr(F\cap W^c\cap T^c)=0.10(0.90)(0.80)=0.072 LaTeX: \Pr(F^c\cap W\cap T^c)=0.90(0.10)(0.80)=0.072 LaTeX: \Pr(F^c\cap W^c\cap T)=0.90(0.90)(0.20)=0.162 ### Compute Add the exactly-one cases for the conditioning probability, then divide the target case by that total. LaTeX: \Pr(\text{exactly one category})=0.072+0.072+0.162=0.306 LaTeX: \Pr(F\mid\text{exactly one category})=\frac{0.072}{0.306}=0.2352941176\ldots ### Answer The conditional probability rounds to 0.235. LaTeX: \boxed{0.235\quad\text{(B)}} ### Distractor analysis - A: This is the joint probability that the target category is the sole loss category, before division by the exactly-one conditioning probability. - C: Normalizing the three marginal loss probabilities 0.10, 0.10, and 0.20 gives 0.10/0.40=0.25, but the exactly-one cases require the accompanying no-loss factors. - D: This is the probability of exactly one loss category, which belongs in the denominator of the conditional probability. - E: Normalizing the no-loss probabilities gives 0.90/(0.90+0.90+0.80)=0.346; the conditioning event concerns loss occurrences, not no-loss weights. ### Verification record - Main value: 0.23529411764705882 - Main method: Divide the target-only independent product 0.1(0.9)(0.8) by the sum of all three exactly-one products. - Independent check: Factor out the probability of no losses from all three exactly-one cells; their relative weights become loss odds 1/9, 1/9, and 1/4, so the target share is (1/9)/(1/9+1/9+1/4)=4/17. - Independent value: 0.23529411764705882 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: greenhouse alert diagnosis A greenhouse controller monitors four independent alert channels: moisture, frost, pests, and backup power. Their probabilities of triggering during a shift are 0.20, 0.10, 0.05, and 0.02, respectively. The shift log shows that exactly one channel triggered. Calculate the conditional probability that it was the moisture channel. - A: 0.1676 - B: 0.2000 - C: 0.2910 - D: 0.5405 - E: 0.5758 The moisture-only probability is 0.16758, and the four mutually exclusive one-alert probabilities total 0.29102. Their ratio is 0.575837, so the correct choice is E. #### Variant solution ### Setup Denote the four trigger probabilities by 0.20, 0.10, 0.05, and 0.02 in the order listed. LaTeX: (p_M,p_F,p_P,p_B)=(0.20,0.10,0.05,0.02) ### Model For a named channel to be the sole trigger, that channel must trigger while all other independent channels remain quiet. LaTeX: w_i=p_i\prod_{j\ne i}(1-p_j) ### Compute Evaluate each one-trigger cell and normalize the moisture cell by their sum. LaTeX: (w_M,w_F,w_P,w_B)=(0.16758,0.07448,0.03528,0.01368) LaTeX: \Pr(\text{exactly one trigger})=0.29102 LaTeX: \Pr(M\mid\text{exactly one trigger})=\frac{0.16758}{0.29102}=0.5758367123\ldots ### Answer Given a single alert, the moisture channel accounts for about 57.58% of cases. LaTeX: \boxed{0.5758\quad\text{(E)}} ### Verification record - Main value: 0.5758367122534532 - Main method: Compute all four independent one-trigger products and divide the moisture-only product by their total. - Independent check: Cancel the common all-quiet product from the one-trigger cells, leaving odds weights 0.25, 1/9, 1/19, and 1/49; the moisture share of those weights is 0.5758367123. - Independent value: 0.5758367122534533 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ## SOA Exam P Sample Question #642 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-642/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Conditional Distributions; Memoryless Property This is an exponential memorylessness problem. Conditioning on survival to the stated age shifts the lifetime but leaves the residual variance equal to the square of the nine-year mean, so choice C is correct. ### Independent solution ### Setup Let T denote the lifetime and condition on the information that T has reached age five. An exponential variable with mean nine has rate 1/9. LaTeX: T\sim\operatorname{Exp}(1/9),\qquad \operatorname{Var}(T)=9^2 ### Model The memoryless property says that the additional lifetime beyond age five has the original exponential distribution. LaTeX: T\mid(T\ge 5)\overset{d}{=}5+Y,\qquad Y\sim\operatorname{Exp}(1/9) ### Compute Adding a constant changes the conditional mean but not the variance. The variance therefore comes entirely from the residual variable Y. LaTeX: \operatorname{Var}(T\mid T\ge5)=\operatorname{Var}(5+Y) LaTeX: \operatorname{Var}(5+Y)=\operatorname{Var}(Y)=\frac{1}{(1/9)^2}=81 ### Answer The conditional lifetime variance is 81 square years. LaTeX: \boxed{81\quad\text{(C)}} ### Distractor analysis - A: The value 14 is the conditional mean, obtained by adding the five elapsed years to the nine-year expected residual lifetime; it is not a variance. - B: The value 16 comes from subtracting the elapsed age from the original mean and then squaring. Memorylessness does not shorten the residual mean from nine to four. - D: The value 106 adds the square of the five-year shift to 81. A deterministic shift contributes zero variance, so that extra 25 must not be included. - E: The value 196 is the square of the conditional mean, 14 squared. Variance is the second central moment, not the squared mean. ### Verification record - Main value: 81 - Main method: Apply exponential memorylessness to write the conditional lifetime as five plus an independent-form residual exponential variable with mean nine. - Independent check: Direct conditional moments give E[T given T at least 5]=14 and E[T squared given T at least 5]=277, hence 277 minus 14 squared equals 81. - Independent value: 81 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: archive-drive inspection A cold-storage archive drive has an exponentially distributed operating lifetime with mean six months. A technician inspects a drive that is still running after four months. Calculate the conditional variance, in square months, of its total operating lifetime. - A: 4 - B: 10 - C: 16 - D: 36 - E: 100 The four completed months are a fixed shift, while the remaining lifetime is exponential with mean six months. Its variance is therefore 6 squared, or 36, selecting choice D. #### Variant solution ### Setup Let L be the total operating lifetime. The exponential mean of six corresponds to rate 1/6. LaTeX: L\sim\operatorname{Exp}(1/6) ### Model After conditioning on operation through month four, memorylessness supplies a fresh residual variable R with the same exponential law. LaTeX: L\mid(L\ge4)\overset{d}{=}4+R,\qquad R\sim\operatorname{Exp}(1/6) ### Compute The constant four has no variance. An exponential variable's variance is the square of its mean. LaTeX: \operatorname{Var}(L\mid L\ge4)=\operatorname{Var}(R)=6^2=36 ### Answer The requested conditional variance is 36 square months. LaTeX: \boxed{36\quad\text{(D)}} ### Verification record - Main value: 36 - Main method: Use memorylessness and the exponential variance formula to obtain the variance of the six-month residual lifetime. - Independent check: Conditional raw moments are E[L given L at least 4]=10 and E[L squared given L at least 4]=136, so 136 minus 10 squared equals 36. - Independent value: 36 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #643 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-643/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Discrete Random Variables; Poisson Distribution This is a Poisson cumulative-probability calculation. Summing the masses from zero through four for a Poisson mean of two gives 7 times exp(-2), approximately 0.947347, which matches choice E. ### Independent solution ### Setup Let N be the count in the stated one-year interval. Its Poisson mean is two. LaTeX: N\sim\operatorname{Poisson}(2),\qquad \Pr(N=k)=e^{-2}\frac{2^k}{k!} ### Model Not exceeding four means that the possible counts are zero, one, two, three, and four. LaTeX: \Pr(N\le4)=\sum_{k=0}^{4}e^{-2}\frac{2^k}{k!} ### Compute Factor out the common exponential term and add the five finite-series coefficients. LaTeX: \Pr(N\le4)=e^{-2}\left(1+2+2+\frac{4}{3}+\frac{2}{3}\right) LaTeX: \Pr(N\le4)=7e^{-2}=0.9473469827 ### Answer The cumulative probability through count four is 7 exp(-2). LaTeX: \boxed{7e^{-2}\quad\text{(E)}} ### Distractor analysis - A: This expression is only the single Poisson mass at count four. A cumulative probability must also include counts zero through three. - B: This takes the complement of the cumulative probability through three, producing the probability of four or more rather than no more than four. - C: This is the cumulative probability through count three. It is missing the nonzero mass at count four. - D: This is the probability of at least one occurrence, found by excluding only count zero; it does not impose the upper cutoff at four. ### Verification record - Main value: 0.9473469826562889 - Main method: Evaluate the Poisson cumulative sum from k=0 through k=4 at mean two and simplify the coefficient to seven. - Independent check: Generate successive masses from p(0)=exp(-2) using p(k)=2p(k-1)/k and add the five values numerically. - Independent value: 0.9473469826562889 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: meteor calibration window During a twenty-minute calibration window, an observatory camera records a Poisson number of meteor streaks with mean 1.5. Calculate the probability that the camera records no more than three streaks during the window. - A: 0.1255 - B: 0.2231 - C: 0.5578 - D: 0.9344 - E: 0.9814 Adding the Poisson masses for zero through three streaks gives 0.9343575. Rounded to four decimals, this is 0.9344 and selects choice D. #### Variant solution ### Setup Let M denote the number of recorded streaks in the calibration window. LaTeX: M\sim\operatorname{Poisson}(1.5) ### Model The requested event contains the four counts from zero through three. LaTeX: \Pr(M\le3)=e^{-1.5}\sum_{k=0}^{3}\frac{1.5^k}{k!} ### Compute Evaluate the short cumulative sum and then apply the exponential factor. LaTeX: \Pr(M\le3)=e^{-1.5}(1+1.5+1.125+0.5625) LaTeX: \Pr(M\le3)=4.1875e^{-1.5}=0.9343575456 ### Answer The four-decimal probability is 0.9344. LaTeX: \boxed{0.9344\quad\text{(D)}} ### Verification record - Main value: 0.9343575456215498 - Main method: Sum the four Poisson masses from zero through three at mean 1.5. - Independent check: Starting at exp(-1.5), use the Poisson mass recurrence with multiplier 1.5/k and total the first four terms. - Independent value: 0.9343575456215498 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #644 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-644/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Conditional Probability; Combinatorial Probability; Exchangeability This is a conditional exchangeability problem. Once exactly three of six equally exposed units are known to be damaged, each three-unit subset is equally likely; four of the twenty subsets avoid both units in the distinguished pair, giving 0.20 and choice D. ### Independent solution ### Setup Condition on exactly three damaged units among the six. Because every unit has the same independent damage probability, the identity of the damaged three is uniformly distributed over all three-unit subsets. LaTeX: \#\{\text{possible damaged subsets}\}=\binom{6}{3}=20 ### Model For neither member of the two-unit distinguished group to be damaged, all three damaged units must come from the other four. LaTeX: \#\{\text{favorable damaged subsets}\}=\binom{4}{3}=4 ### Compute Take the ratio of favorable conditional subsets to all possible conditional subsets. LaTeX: \Pr(\text{distinguished pair undamaged}\mid\text{three damaged}) LaTeX: =\frac{\binom{4}{3}}{\binom{6}{3}}=\frac{4}{20}=0.20 ### Answer The required conditional probability is 0.2000. LaTeX: \boxed{0.2000\quad\text{(D)}} ### Distractor analysis - A: This is the unconditional joint probability that exactly three of the four nondistinguished units are damaged and every other unit is undamaged. It is the numerator before conditioning. - B: This result can be produced by retaining only one undamaged factor after selecting three damaged units from the group of four. It fails to account for both members of the distinguished pair. - C: This is the unconditional probability that exactly three of all six units are damaged. It is the denominator event, not the requested conditional ratio. - E: This is the unconditional probability that the two distinguished units are both undamaged. Knowing that three of the six are damaged changes that probability. ### Verification record - Main value: 0.2 - Main method: Under the condition of three damages, divide the four favorable three-subsets drawn from the other four units by all 20 three-subsets of six units. - Independent check: Bayes calculation gives joint numerator 4(0.1)^3(0.9)^3=0.002916 and conditioning probability 20(0.1)^3(0.9)^3=0.01458; their ratio is 0.2. - Independent value: 0.2 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: navigation-beacon shock test Eight sealed navigation beacons ride through the same shock test. Five use a standard casing and three use a ceramic casing. Each beacon independently fails with probability 0.20. After learning that exactly two beacons failed, calculate the probability that both failed beacons used the standard casing. - A: 0.1049 - B: 0.2936 - C: 0.3571 - D: 0.3906 - E: 0.6250 Given two failures among equally exposed beacons, every pair is equally likely. Ten of the 28 pairs come from the five standard-casing beacons, so the conditional probability is 5/14, approximately 0.3571, and choice C is correct. #### Variant solution ### Setup Condition on the observed total of two failed beacons. Equal independent failure probabilities make the failed pair uniform among all pairs of the eight labeled beacons. LaTeX: \#\{\text{possible failed pairs}\}=\binom{8}{2}=28 ### Model A favorable pair contains two of the five standard-casing beacons. LaTeX: \#\{\text{favorable pairs}\}=\binom{5}{2}=10 ### Compute The conditional probability is the favorable-pair count divided by the total pair count. LaTeX: \Pr(\text{both standard}\mid\text{two failures})=\frac{10}{28} LaTeX: \frac{10}{28}=\frac{5}{14}=0.3571428571 ### Answer Rounded to four decimal places, the result is 0.3571. LaTeX: \boxed{0.3571\quad\text{(C)}} ### Verification record - Main value: 0.35714285714285715 - Main method: Count ten standard-casing pairs among the 28 possible failed pairs after conditioning on exactly two failures. - Independent check: The joint probability is 10(0.2)^2(0.8)^6=0.1048576 and the probability of exactly two failures is 28(0.2)^2(0.8)^6=0.29360128; their ratio is 5/14. - Independent value: 0.35714285714285715 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #645 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-645/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Conditional Probability; Independence; Combinatorial Probability This is a conditional placement problem for two event-years among five independent years. Of the ten equally likely pairs of event-years, three have year two as the first, so the conditional probability is 3/10 and choice C is correct. ### Independent solution ### Setup Record only whether at least one event occurs in each year. Conditioning on exactly two event-years makes every pair of positions among the five years equally likely. LaTeX: \#\{\text{two-year position sets}\}=\binom{5}{2}=10 ### Model For year two to be the first event-year, year one must be clear, year two must be selected, and the other selected year can be any of years three through five. LaTeX: \#\{\text{favorable position sets}\}=\binom{3}{1}=3 ### Compute Divide the three favorable placements by all ten placements allowed by the conditioning event. LaTeX: \Pr(\text{first event-year is 2}\mid\text{two event-years}) LaTeX: =\frac{\binom{3}{1}}{\binom{5}{2}}=\frac{3}{10}=0.30 ### Answer The conditional probability is 0.300. LaTeX: \boxed{0.300\quad\text{(C)}} ### Distractor analysis - A: The value 0.160 is the unconditional product of a clear first year and an event in year two. It ignores both the exact-two-years condition and the remaining three years. - B: The value 0.250 can result from treating the possible first event positions as four equally likely choices. Their probabilities are not uniform once two event-years must fit into five positions. - D: The value 0.384 is the probability of exactly one event-year among the last three years. It omits the required statuses of years one and two and does not divide by the conditioning event. - E: The value 0.400 is the conditional chance that year two is one of the two selected years. It also counts pairs containing year one, for which year two is not the first event-year. ### Verification record - Main value: 0.3 - Main method: Count the three pairs consisting of year two and one later year among all ten two-year subsets of five positions. - Independent check: Using event-year probability 0.2, the joint numerator is (0.8)(0.2) times 3(0.2)(0.8)^2=0.06144, while the exact-two denominator is 10(0.2)^2(0.8)^3=0.2048; the ratio is 0.3. - Independent value: 0.3 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: aquaculture alarm nights An aquaculture sensor independently raises an alarm on each of seven nightly checks with probability 0.35. After the monitoring period, the log shows alarms on exactly two nights. Calculate the conditional probability that the first alarm was on the third night. - A: 0.0569 - B: 0.1905 - C: 0.2857 - D: 0.2985 - E: 0.3500 There are 21 equally likely pairs of alarm nights. Four pairs have night three first—night three paired with one of nights four through seven—so the probability is 4/21, approximately 0.1905, and choice B is correct. #### Variant solution ### Setup Given exactly two alarm nights, the selected pair is uniform among all pairs of the seven night positions. LaTeX: \#\{\text{possible alarm-night pairs}\}=\binom{7}{2}=21 ### Model If the first alarm is on night three, nights one and two are not selected, night three is selected, and the other alarm can occur on any later night. LaTeX: \#\{\text{favorable pairs}\}=\binom{4}{1}=4 ### Compute Form the conditional ratio from the favorable and total pair counts. LaTeX: \Pr(\text{first alarm on night 3}\mid\text{two alarm nights})=\frac{4}{21} LaTeX: \frac{4}{21}=0.1904761905 ### Answer The rounded conditional probability is 0.1905. LaTeX: \boxed{0.1905\quad\text{(B)}} ### Verification record - Main value: 0.19047619047619047 - Main method: Count the four pairs containing night three and one later night among the 21 possible two-night subsets. - Independent check: The joint numerator is (0.65)^2(0.35) times 4(0.35)(0.65)^3=0.056854240625, while the exact-two denominator is 21(0.35)^2(0.65)^5=0.29848476328125; their ratio is 4/21. - Independent value: 0.19047619047619047 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #646 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-646/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Complement Rule; Independence This is a complement-rule calculation for independent days. The chance of a claim-free seven-day period is r to the seventh power, so the chance of at least one claim is 1 minus r to the seventh power and choice D is correct. ### Independent solution ### Setup For each of the seven days, the no-claim probability is r, and the daily claim indicators are mutually independent. LaTeX: \Pr(\text{no claim on one day})=r ### Model The complement of at least one claim is a period with no claim on every day. LaTeX: \Pr(\text{at least one claim})=1-\Pr(\text{no claims in seven days}) ### Compute Independence allows the seven daily no-claim probabilities to be multiplied. LaTeX: \Pr(\text{no claims in seven days})=\prod_{j=1}^{7}r=r^7 LaTeX: \Pr(\text{at least one claim})=1-r^7 ### Answer The required probability is the complement of seven claim-free days. LaTeX: \boxed{1-r^7\quad\text{(D)}} ### Distractor analysis - A: The expression r to the seventh power is the probability of no claims during the entire period, which is the complement of the requested event. - B: This is the probability of a claim on one specified day and no claims on the other six days. It does not include every way claims can occur. - C: This is the probability of exactly one claim-day anywhere in the period. The requested event also includes outcomes with claims on multiple days. - E: The expression (1-r) to the seventh power requires a claim on every day. That is only one narrow part of the at-least-one event. ### Verification record - Main value: 1-r^7 - Main method: Multiply the seven independent daily no-claim probabilities and take the complement. - Independent check: Summing the binomial probabilities for one through seven claim-days gives the binomial expansion sum 1-r^7. - Independent value: 1-r^7 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: soil-probe self-test Twelve remote soil probes run independent overnight self-tests. Each probe passes its test with probability 0.96. Calculate the probability that at least one probe requires recalibration after the tests. - A: 0.0400 - B: 0.3064 - C: 0.3873 - D: 0.4800 - E: 0.6127 All twelve probes pass with probability 0.96 to the twelfth power. Taking the complement gives 0.3872902, which rounds to 0.3873 and selects choice C. #### Variant solution ### Setup A recalibration is required when a probe does not pass. The tests are independent across the twelve probes. LaTeX: \Pr(\text{one probe passes})=0.96 ### Model Use the complement event in which every probe passes its self-test. LaTeX: \Pr(\text{at least one recalibration})=1-\Pr(\text{all pass}) ### Compute Multiply the twelve pass probabilities, then subtract the result from one. LaTeX: \Pr(\text{all pass})=0.96^{12}=0.6127097573 LaTeX: 1-0.96^{12}=0.3872902427 ### Answer The probability that at least one probe needs recalibration is 0.3873 to four decimals. LaTeX: \boxed{0.3873\quad\text{(C)}} ### Verification record - Main value: 0.38729024267023293 - Main method: Take one minus the product 0.96 raised to the twelfth power for all probes passing. - Independent check: Sum the binomial probabilities for one through twelve recalibrations with per-probe probability 0.04. - Independent value: 0.38729024267023293 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #647 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-647/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Law of Total Probability; Mixture Distributions This is a two-branch mixture whose unknown branch probability is determined by a cumulative probability. Solving the mixture equation gives a weighted-die branch probability of 0.666, after which the probability of rolling a four is 0.203667 and choice C. ### Independent solution ### Setup Let h be the probability of selecting the weighted die. Normalizing its face weights gives probability 1/9 on each odd face and 2/9 on each even face. LaTeX: h=\Pr(\text{weighted-die branch}) LaTeX: \Pr(X=k\mid\text{weighted})=\begin{cases}1/9,&k\text{ odd},\\2/9,&k\text{ even}.\end{cases} ### Model Apply the law of total probability to the event that the displayed value is at most three. LaTeX: \Pr(X\le3\mid\text{weighted})=\frac{1+2+1}{9}=\frac49 LaTeX: \Pr(X\le3\mid\text{fair})=\frac12 LaTeX: 0.463=h\left(\frac49\right)+(1-h)\left(\frac12\right) ### Compute Solve for the mixing probability, then mix the two conditional probabilities for face four. LaTeX: h=18(0.500-0.463)=0.666 LaTeX: \Pr(X=4)=0.666\left(\frac29\right)+0.334\left(\frac16\right) LaTeX: \Pr(X=4)=0.203666\ldots ### Answer The requested probability rounds to 0.204 at the precision of the choices. LaTeX: \boxed{\Pr(X=4)\approx0.204\quad\text{(C)}} ### Distractor analysis - A: The value 0.167 is the fair-die probability of a four. It ignores the branch using the weighted die. - B: The value 0.194 is the simple average of 1/6 and 2/9, which incorrectly assumes that the two branches are equally likely. - D: The value 0.537 is 1-0.463, the probability of a value above three; it includes faces five and six as well as face four. - E: The value 0.667 is the inferred probability of selecting the weighted branch, not the unconditional probability of the requested face. ### Verification record - Main value: 0.2036666666666666 - Main method: Solve 0.463=h(4/9)+(1-h)(1/2), then evaluate h(2/9)+(1-h)(1/6). - Independent check: The resulting marginal mass is 1/6-h/18 on every odd face and 1/6+h/18 on every even face; at h=0.666 these six masses sum to one, their first three sum to 0.463, and the fourth equals 0.2036666667. - Independent value: 0.2036666666666666 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: adaptive severity labels A diagnostic platform routes a record to an adaptive classifier with probability r and to a legacy classifier otherwise. The adaptive classifier assigns severity labels 1 through 5 with probabilities 0.10, 0.10, 0.20, 0.20, and 0.40. The legacy classifier assigns the five labels uniformly. Historical output has P(S at most 3)=0.47. Calculate P(S=5). - A: 0.200 - B: 0.260 - C: 0.300 - D: 0.330 - E: 0.650 The adaptive and legacy cumulative probabilities through label three are 0.40 and 0.60. Their mixture equals 0.47 only when r=0.65, and the resulting probability of label five is 0.33, selecting choice D. #### Variant solution ### Setup Compute the conditional cumulative probabilities through label three for the two classifiers. LaTeX: \Pr(S\le3\mid\text{adaptive})=0.10+0.10+0.20=0.40 LaTeX: \Pr(S\le3\mid\text{legacy})=3(0.20)=0.60 ### Model Use the observed cumulative probability to identify the routing proportion. LaTeX: 0.47=0.40r+0.60(1-r) ### Compute Solve the mixture equation and then combine the two label-five probabilities. LaTeX: r=\frac{0.60-0.47}{0.20}=0.65 LaTeX: \Pr(S=5)=0.65(0.40)+0.35(0.20)=0.33 ### Answer The platform assigns severity label five with probability 0.330. LaTeX: \boxed{\Pr(S=5)=0.330\quad\text{(D)}} ### Verification record - Main value: 0.33 - Main method: Recover r from the two-component cumulative mixture and apply total probability to label five. - Independent check: The two joint label-five cells have masses 0.65(0.40)=0.26 and 0.35(0.20)=0.07; their sum is 0.33, while the corresponding through-three cells sum to the supplied 0.47. - Independent value: 0.33 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #648 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-648/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Continuous Random Variables; Raw Moments Differentiate the cumulative distribution to obtain density 2x/25 on the interval from zero to five. Integrating x squared against this density gives a second raw moment of 12.50, so choice D is correct. ### Independent solution ### Setup Differentiate the nonconstant part of the cumulative distribution to recover the density on its support. LaTeX: f_X(x)=F_X'(x)=\frac{2x}{25},\qquad 00 ### Model A continuous median m places one-half of the probability at or below m. LaTeX: F_X(m)=\frac12 LaTeX: 1-e^{-4m^2}=\frac12 ### Compute Isolate the exponential term, take logarithms, and retain the positive root because the support is positive. LaTeX: 4m^2=\log 2 LaTeX: m=\sqrt{\frac{\log 2}{4}}=0.416277\ldots ### Answer The median rounds to 0.416. LaTeX: \boxed{m\approx0.416\quad\text{(C)}} ### Distractor analysis - A: The value 0.173 is log(2)/4, which is the computed value of m squared; taking the positive square root is still required. - B: The value 0.268 is approximately the first quartile, obtained by setting the cumulative probability to 0.25 instead of 0.50. - D: The value 0.693 reports log(2) itself, omitting both division by four and the square root. - E: The value 0.833 is sqrt(log(2)); it takes the square root but fails to divide by two afterward. ### Verification record - Main value: 0.41627730557884884 - Main method: Integrate the density, set 1-exp(-4m^2)=1/2, and solve for the positive root. - Independent check: Substitution of m=0.4162773056 into the cumulative distribution gives 1-exp(-4m^2)=0.5000000000. - Independent value: 0.5 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: acoustic detection radius The distance R, in kilometers, from an ocean sensor to a detected acoustic event has density 18r exp(-9r squared) for positive r and zero otherwise. Determine the 75th percentile of the detection distance. - A: 0.278 - B: 0.333 - C: 0.392 - D: 0.462 - E: 0.680 The cumulative distribution is 1 minus exp(-9r squared). Equating it to 0.75 gives a percentile of 0.392470 kilometer, selecting choice C. #### Variant solution ### Setup Integrate the density over its positive support. LaTeX: F_R(r)=\int_0^r 18t e^{-9t^2}\,dt=1-e^{-9r^2} ### Model Let q be the 75th percentile and set its cumulative probability to 0.75. LaTeX: 1-e^{-9q^2}=0.75 ### Compute Solve the exponential equation for the positive distance. LaTeX: e^{-9q^2}=0.25 LaTeX: q=\sqrt{\frac{\log 4}{9}}=0.392470\ldots ### Answer The 75th-percentile detection radius is approximately 0.392 kilometer. LaTeX: \boxed{q\approx0.392\quad\text{(C)}} ### Verification record - Main value: 0.3924700075051582 - Main method: Integrate the density and solve 1-exp(-9q^2)=0.75 for the positive quantile. - Independent check: At q=0.3924700075, the survival probability exp(-9q^2) is 0.25, so the cumulative probability is exactly 0.75. - Independent value: 0.75 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #650 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-650/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Multinomial Distribution; Linearity of Expectation The two category counts have expectations 0.4 and 0.2. Applying the two loss amounts linearly gives expected total loss 2(0.4)+20(0.2)=4.80, which matches choice E. ### Independent solution ### Setup Let I and J count the two nonzero outcome categories among the two trials. Their joint distribution is multinomial, so each count has expectation equal to the trial count times its category probability. LaTeX: \operatorname{E}[I]=2(0.20)=0.40 LaTeX: \operatorname{E}[J]=2(0.10)=0.20 ### Model Express the total loss as a linear combination of the category counts. LaTeX: L=2I+20J LaTeX: \operatorname{E}[L]=2\operatorname{E}[I]+20\operatorname{E}[J] ### Compute Substitute the two expected counts. No independence between I and J is needed for this expectation. LaTeX: \operatorname{E}[L]=2(0.40)+20(0.20) LaTeX: \operatorname{E}[L]=0.80+4.00=4.80 ### Answer The expected total loss is 4.80. LaTeX: \boxed{\operatorname{E}[L]=4.80\quad\text{(E)}} ### Distractor analysis - A: The value 0.60 is the expected number of nonzero-category outcomes, 2(0.20+0.10), rather than the expected monetary loss. - B: The value 1.00 is the sum of the probabilities in the multinomial distribution. Normalization alone does not weight outcomes by their losses. - C: The value 1.44 can arise by treating every nonzero outcome as a loss of two, obtaining the incorrect proxy 1.20, and then squaring it; expectation is linear and the second category has loss 20. - D: The value 3.92 omits the mixed outcome having one count of each type. Its probability-weighted contribution is 0.04(22)=0.88, and 4.80-0.88=3.92. ### Verification record - Main value: 4.8 - Main method: Use multinomial count means and linearity: E[2I+20J]=2(2·0.2)+20(2·0.1). - Independent check: Enumerating the six count pairs gives weighted contributions 0, 0.56, 0.16, 2.80, 0.88, and 0.40; these sum to 4.80. - Independent value: 4.8 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: instrument repair tickets A laboratory receives four instrument-repair tickets. Each ticket independently requires no replacement part with probability 0.60, a sensor costing 5 credits with probability 0.30, or a rush controller costing 20 credits with probability 0.10. Calculate the expected total replacement-part cost for the four tickets. - A: 1.60 - B: 3.50 - C: 6.00 - D: 8.00 - E: 14.00 The expected sensor and rush-controller counts are 1.2 and 0.4. Their expected cost contributions are 6 and 8 credits, respectively, for a total of 14 credits and choice E. #### Variant solution ### Setup Let S and R be the numbers of sensor and rush-controller replacements in the four tickets. LaTeX: \operatorname{E}[S]=4(0.30)=1.20 LaTeX: \operatorname{E}[R]=4(0.10)=0.40 ### Model Write total parts cost as the corresponding linear combination of these counts. LaTeX: C=5S+20R LaTeX: \operatorname{E}[C]=5\operatorname{E}[S]+20\operatorname{E}[R] ### Compute Evaluate the contribution from each replacement category. LaTeX: \operatorname{E}[C]=5(1.20)+20(0.40) LaTeX: \operatorname{E}[C]=6.00+8.00=14.00 ### Answer The four tickets generate an expected replacement-part cost of 14 credits. LaTeX: \boxed{\operatorname{E}[C]=14.00\quad\text{(E)}} ### Verification record - Main value: 14 - Main method: Apply multinomial count means to C=5S+20R for four tickets. - Independent check: One ticket has expected cost 0.60(0)+0.30(5)+0.10(20)=3.50; summing four identically distributed ticket costs gives 14.00. - Independent value: 14 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ## SOA Exam P Sample Question #651 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-651/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Marginal Distributions; Expected Loss Only 40% of the second-period loss remains unreimbursed, while the first-period loss is fully covered. Marginalizing the joint table gives a second-period mean of 0.90, so the expected unreimbursed amount is 0.36 and choice A. ### Independent solution ### Setup Let Y be the loss in the partially covered period. Sum each row of the joint table to obtain the marginal distribution of Y. LaTeX: \Pr(Y=0)=0.80,\quad \Pr(Y=2)=0.10 LaTeX: \Pr(Y=5)=0.06,\quad \Pr(Y=10)=0.04 ### Model The fully covered period contributes nothing to the unreimbursed amount. The other period leaves 40% of its loss with the patient. LaTeX: U=0.40Y LaTeX: \operatorname{E}[U]=0.40\operatorname{E}[Y] ### Compute Calculate the marginal mean and apply the uncovered percentage. LaTeX: \operatorname{E}[Y]=0(0.80)+2(0.10)+5(0.06)+10(0.04)=0.90 LaTeX: \operatorname{E}[U]=0.40(0.90)=0.36 ### Answer The expected unreimbursed loss over the two periods is 0.36. LaTeX: \boxed{\operatorname{E}[U]=0.36\quad\text{(A)}} ### Distractor analysis - B: The value 0.76 is 40% of the first-period mean 1.90. That period is fully covered, so it should not create any unreimbursed loss. - C: The value 0.90 is the second-period expected loss before applying its 40% uncovered share. - D: This amount exceeds the entire second-period mean of 0.90, so it cannot be the expectation of only 40% of that nonnegative loss; it reflects a misread table margin or coverage rate. - E: The value 2.80 is the expected total loss from both periods before either reimbursement provision is applied. ### Verification record - Main value: 0.36 - Main method: Marginalize the second-period loss from the joint table and multiply its mean 0.90 by the uncovered share 0.40. - Independent check: The expected reimbursement on the second-period loss is 0.60(0.90)=0.54, and 0.90-0.54=0.36; the first-period unreimbursed amount is zero. - Independent value: 0.36 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: festival vendor credit A theatre festival vendor may incur a booth charge B and an electrical surcharge E. A sponsor pays all of B and 70% of E. Joint probabilities for columns B=0, 4, 10 are: row E=0 has 0.40, 0.15, 0.05; row E=5 has 0.10, 0.08, 0.02; and row E=12 has 0.08, 0.07, 0.05. Calculate the vendor's expected amount not paid by the sponsor. - A: 0.30 - B: 0.40 - C: 1.02 - D: 1.74 - E: 3.40 The surcharge marginal probabilities are 0.60, 0.20, and 0.20 at 0, 5, and 12. Its mean is 3.40, of which the vendor pays 30%, giving 1.02 and choice C. #### Variant solution ### Setup Sum across booth-charge columns to obtain the marginal distribution of the electrical surcharge. LaTeX: \Pr(E=0)=0.40+0.15+0.05=0.60 LaTeX: \Pr(E=5)=0.10+0.08+0.02=0.20 LaTeX: \Pr(E=12)=0.08+0.07+0.05=0.20 ### Model The booth charge is fully sponsored. The vendor's unpaid amount is therefore 30% of the electrical surcharge. LaTeX: V=0.30E LaTeX: \operatorname{E}[V]=0.30\operatorname{E}[E] ### Compute Evaluate the surcharge mean and then its uncovered portion. LaTeX: \operatorname{E}[E]=0(0.60)+5(0.20)+12(0.20)=3.40 LaTeX: \operatorname{E}[V]=0.30(3.40)=1.02 ### Answer The vendor expects to pay 1.02 credits beyond the sponsor's contribution. LaTeX: \boxed{\operatorname{E}[V]=1.02\quad\text{(C)}} ### Verification record - Main value: 1.02 - Main method: Marginalize the electrical surcharge, compute its mean, and apply the 30% vendor share. - Independent check: The vendor pays 0 with probability 0.60, 1.50 with probability 0.20, and 3.60 with probability 0.20; the direct weighted sum is 1.02. - Independent value: 1.02 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #652 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-652/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Marginal Distributions; Moments Summing the joint mass over the second coordinate gives marginal probabilities 0.1, 0.3, and 0.6 at the positive values of X. The resulting first and second moments are 2.5 and 6.7, so the variance is 0.45 and choice A. ### Independent solution ### Setup Marginalize the supplied joint mass function over every allowed value of the second coordinate. LaTeX: p_X(x)=\sum_{y=0}^{x}\frac{x+2y}{40},\qquad x=0,1,2,3 ### Model The finite arithmetic sum reduces to a simple marginal mass function. LaTeX: \sum_{y=0}^{x}(x+2y)=x(x+1)+x(x+1)=2x(x+1) LaTeX: p_X(x)=\frac{x(x+1)}{20} ### Compute Calculate both raw moments from the marginal probabilities. LaTeX: \Pr(X=1)=0.1,\qquad \Pr(X=2)=0.3,\qquad \Pr(X=3)=0.6 LaTeX: \operatorname{E}[X]=1(0.1)+2(0.3)+3(0.6)=2.5 LaTeX: \operatorname{E}[X^2]=1^2(0.1)+2^2(0.3)+3^2(0.6)=6.7 LaTeX: \operatorname{Var}(X)=6.7-(2.5)^2=0.45 ### Answer The marginal variance is 0.45. LaTeX: \boxed{\operatorname{Var}(X)=0.45\quad\text{(A)}} ### Distractor analysis - B: This is the raw second moment 6.70 with the decimal point shifted; a variance must subtract the squared mean. - C: This candidate fails the direct mean-centered sum and is too large once the mass concentrated near X = 3 is centered at 2.5. - D: This remains on the scale of the raw second moment because it subtracts only a small correction instead of the full squared mean 6.25. - E: This is E[X²] = 6.70, not Var(X). ### Verification record - Main value: 0.45 - Main method: Marginalize the joint mass function, then evaluate E[X^2] - E[X]^2. - Independent check: The mean-centered sum 0.1(1-2.5)^2 + 0.3(2-2.5)^2 + 0.6(3-2.5)^2 equals 0.45. - Independent value: 0.45 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: inspection tags and manual reviews A quality team records the number A of priority tags on a production lot and whether a manual review B is requested. Their joint probabilities are P(A=0,B=0)=0.10, P(A=0,B=1)=0.05, P(A=1,B=0)=0.20, P(A=1,B=1)=0.25, P(A=2,B=0)=0.15, and P(A=2,B=1)=0.25. Calculate the variance of A. - A: 0.4000 - B: 0.4500 - C: 0.4875 - D: 1.2500 - E: 2.0500 Marginalizing over the review indicator gives probabilities 0.15, 0.45, and 0.40 for A = 0, 1, and 2. The first two raw moments are 1.25 and 2.05, so the variance is 0.4875 and choice C. #### Variant solution ### Setup Add the two joint cells in each row to obtain the distribution of the tag count. LaTeX: \Pr(A=0)=0.10+0.05=0.15 LaTeX: \Pr(A=1)=0.20+0.25=0.45 LaTeX: \Pr(A=2)=0.15+0.25=0.40 ### Model Use the first and second raw moments of this three-point marginal distribution. LaTeX: \operatorname{Var}(A)=\operatorname{E}[A^2]-\operatorname{E}[A]^2 ### Compute Evaluate the moments and subtract the squared mean. LaTeX: \operatorname{E}[A]=0(0.15)+1(0.45)+2(0.40)=1.25 LaTeX: \operatorname{E}[A^2]=0^2(0.15)+1^2(0.45)+2^2(0.40)=2.05 LaTeX: \operatorname{Var}(A)=2.05-1.25^2=0.4875 ### Answer The priority-tag count has variance 0.4875. LaTeX: \boxed{0.4875\quad\text{(C)}} ### Verification record - Main value: 0.4875 - Main method: Marginalize the six-cell joint table and apply the raw-moment variance identity. - Independent check: The centered sum 0.15(0-1.25)^2 + 0.45(1-1.25)^2 + 0.40(2-1.25)^2 equals 0.4875. - Independent value: 0.4875 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #653 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-653/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Independence; Complement Rule On one trial the three possible totals have probabilities 1/9, 4/9, and 4/9. The two independent totals agree with probability 33/81, so they differ with probability 48/81 = 16/27 and choice C. ### Independent solution ### Setup Determine the distribution of the total on a single trial from the two face-value probabilities. LaTeX: \Pr(D=1)=\frac13,\qquad \Pr(D=2)=\frac23 LaTeX: \Pr(S=2)=\frac19,\qquad \Pr(S=3)=\frac49,\qquad \Pr(S=4)=\frac49 ### Model It is shorter to calculate the probability that the two independent trial totals match, then take its complement. LaTeX: \Pr(S_1=S_2)=\sum_{s=2}^{4}\Pr(S=s)^2 ### Compute Square and add the three single-trial probabilities. LaTeX: \Pr(S_1=S_2)=\left(\frac19\right)^2+\left(\frac49\right)^2+\left(\frac49\right)^2=\frac{33}{81} LaTeX: \Pr(S_1\ne S_2)=1-\frac{33}{81}=\frac{48}{81}=\frac{16}{27} ### Answer The probability of different totals is sixteen twenty-sevenths. LaTeX: \boxed{\frac{16}{27}\quad\text{(C)}} ### Distractor analysis - A: This undercounts the ordered pairs of unequal totals; all six off-diagonal pairs must be included. - B: This also omits some off-diagonal probability mass and does not complement the exact match probability 33/81. - D: This is 2/3, the answer obtained by incorrectly treating the three possible totals as equally likely. - E: This overstates the result by assigning too little probability to a repeated total, especially the two totals that each occur with probability 4/9. ### Verification record - Main value: 0.5925925925925926 - Main method: Find the one-trial total distribution and complement the sum of its squared probabilities. - Independent check: Directly summing the six ordered unequal-total pairs gives 2[(1/9)(4/9) + (1/9)(4/9) + (4/9)(4/9)] = 48/81. - Independent value: 0.5925925925925926 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: diagnostic status changes A diagnostic run independently reports Low, Medium, or High status with probabilities 0.20, 0.30, and 0.50. Two runs are made under unchanged conditions. Calculate the probability that the reported status differs between the two runs. - A: 0.38 - B: 0.50 - C: 0.58 - D: 0.62 - E: 0.70 The two statuses match with probability 0.20² + 0.30² + 0.50² = 0.38. The complement is 0.62, so choice D. #### Variant solution ### Setup List the three single-run probabilities. LaTeX: p_L=0.20,\qquad p_M=0.30,\qquad p_H=0.50 ### Model Independence makes the probability of matching in a category the square of that category probability. LaTeX: \Pr(\text{same})=p_L^2+p_M^2+p_H^2 ### Compute Calculate the matching probability and take its complement. LaTeX: \Pr(\text{same})=0.20^2+0.30^2+0.50^2=0.38 LaTeX: \Pr(\text{different})=1-0.38=0.62 ### Answer The two diagnostic statuses differ with probability 0.62. LaTeX: \boxed{0.62\quad\text{(D)}} ### Verification record - Main value: 0.62 - Main method: Complement the sum of the three independent matching-category probabilities. - Independent check: The six directed changes total 2[(0.20)(0.30) + (0.20)(0.50) + (0.30)(0.50)] = 0.62. - Independent value: 0.62 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #654 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-654/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Order Statistics; Uniform Distribution For three independent Uniform(0,10) observations, the maximum has distribution function (x/10)³. Integrating its density gives expected maximum 3(10)/4 = 7.5, which is choice B. ### Independent solution ### Setup Let M be the largest of the three independent uniform observations. LaTeX: M=\max(X_1,X_2,X_3),\qquad X_i\sim\operatorname{Uniform}(0,10) ### Model The maximum does not exceed x exactly when all three observations do not exceed x. LaTeX: F_M(x)=\Pr(X_1\le x,X_2\le x,X_3\le x)=\left(\frac{x}{10}\right)^3 LaTeX: f_M(x)=\frac{3x^2}{10^3},\qquad 0\le x\le10 ### Compute Integrate x against the maximum density. LaTeX: \operatorname{E}[M]=\int_0^{10}x\frac{3x^2}{10^3}\,dx LaTeX: \operatorname{E}[M]=\frac{3}{1000}\left[\frac{x^4}{4}\right]_0^{10}=\frac{30}{4}=7.5 ### Answer The expected largest observation is 7.5. LaTeX: \boxed{7.5\quad\text{(B)}} ### Distractor analysis - A: This is approximately the expected maximum of only two Uniform(0,10) observations, 20/3. - C: This is the expected maximum for four observations, not three. - D: This is approximately the five-observation value 50/6 and therefore assumes too many observations. - E: This places still more mass near the upper endpoint than a three-observation maximum can have. ### Verification record - Main value: 7.5 - Main method: Derive the maximum density from the cube of the uniform CDF and integrate its first moment. - Independent check: The tail-integral identity gives E[M] = integral_0^10 [1 - (x/10)^3] dx = 10 - 2.5 = 7.5. - Independent value: 7.5 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: strongest calibration reading Four independent calibration readings are uniformly distributed between 60 and 90. A technician records only the largest reading. Calculate its expected value. - A: 75.0 - B: 80.0 - C: 82.5 - D: 84.0 - E: 85.0 After scaling to Uniform(0,1), the maximum of four observations has mean 4/5. Mapping back to [60,90] gives 60 + 30(4/5) = 84, so choice D. #### Variant solution ### Setup Standardize each reading to a unit-uniform variable. LaTeX: U_i=\frac{X_i-60}{30}\sim\operatorname{Uniform}(0,1) LaTeX: X_{(4)}=60+30U_{(4)} ### Model The maximum of four unit-uniform observations has density four times u cubed. LaTeX: F_{U_{(4)}}(u)=u^4,\qquad f_{U_{(4)}}(u)=4u^3 ### Compute Find the standardized expected maximum and transform it back to the original scale. LaTeX: \operatorname{E}[U_{(4)}]=\int_0^1u(4u^3)\,du=\frac45 LaTeX: \operatorname{E}[X_{(4)}]=60+30\left(\frac45\right)=84 ### Answer The expected strongest calibration reading is 84. LaTeX: \boxed{84.0\quad\text{(D)}} ### Verification record - Main value: 84 - Main method: Scale to four unit-uniform observations, use E[U_(4)] = 4/5, and map back to [60,90]. - Independent check: The tail integral 60 + integral_60^90 [1 - ((x-60)/30)^4] dx equals 60 + 30(4/5) = 84. - Independent value: 84 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #655 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-655/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Variance and Standard Deviation Scaling the constant Poisson rate from thirty years to ten gives mean and variance 2/3. The standard deviation is the square root of 2/3, approximately 0.816, so the answer is D. ### Independent solution ### Setup Convert the stated long-window count rate into the requested time window. LaTeX: \lambda_{10}=2\left(\frac{10}{30}\right)=\frac23 ### Model A Poisson count has variance equal to its mean. LaTeX: N_{10}\sim\operatorname{Poisson}\left(\frac23\right) LaTeX: \operatorname{Var}(N_{10})=\frac23 ### Compute Take the square root of the variance to obtain the standard deviation. LaTeX: \operatorname{SD}(N_{10})=\sqrt{\frac23}=0.8164965809\ldots ### Answer The requested standard deviation rounds to 0.816. LaTeX: \boxed{0.816\quad\text{(D)}} ### Distractor analysis - A: This divides the correct standard deviation by the square root of ten, as though averaging ten observations rather than counting events over ten years. - B: This corresponds to using variance 2/9, which scales the thirty-year parameter by an extra factor of one third. - C: This is the ten-year Poisson mean and variance 2/3, not their square root. - E: This is the square root of the full thirty-year mean 2 and ignores the shorter time window. ### Verification record - Main value: 0.816496580927726 - Main method: Scale the constant Poisson mean to ten years and take the square root of the resulting variance. - Independent check: Ten independent one-year increments each have variance 1/15; adding them gives variance 10/15 = 2/3 and standard deviation sqrt(2/3). - Independent value: 0.816496580927726 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: priority server alerts Priority alerts arrive at a server as a Poisson process with an average of 3.6 alerts per twelve-hour monitoring window. Calculate the standard deviation of the number of alerts during the next five hours. - A: 0.417 - B: 1.000 - C: 1.225 - D: 1.500 - E: 1.897 The five-hour mean and variance are 3.6(5/12) = 1.5. The standard deviation is the square root of 1.5, approximately 1.225, so choice C. #### Variant solution ### Setup Scale the twelve-hour rate to five hours. LaTeX: \lambda_5=3.6\left(\frac{5}{12}\right)=1.5 ### Model The five-hour alert count is Poisson, so its variance is also 1.5. LaTeX: A_5\sim\operatorname{Poisson}(1.5) LaTeX: \operatorname{Var}(A_5)=1.5 ### Compute Take the square root of the five-hour variance. LaTeX: \operatorname{SD}(A_5)=\sqrt{1.5}=1.224744871\ldots ### Answer The standard deviation rounds to 1.225. LaTeX: \boxed{1.225\quad\text{(C)}} ### Verification record - Main value: 1.224744871391589 - Main method: Scale the Poisson parameter to five hours and take the square root of the resulting variance. - Independent check: Five independent hourly increments each have variance 3.6/12 = 0.3; their variance sum is 1.5, giving standard deviation 1.224744871. - Independent value: 1.224744871391589 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #656 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-656/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Binomial and Multinomial Distributions; Complement Rule Each of the two independent outcomes avoids the target category with probability 0.9. Therefore, the probability that the target occurs at least once is 1 - 0.9^2 = 0.19, so the answer is D. ### Independent solution ### Setup For one outcome, the target category has probability 0.1, so its complement has probability 0.9. LaTeX: P(\text{not target})=1-0.1=0.9 ### Model Use independence to calculate the probability that neither of the two outcomes falls in the target category. LaTeX: P(\text{no target in two outcomes})=(0.9)^2 ### Compute Take the complement of the event that the target never occurs. LaTeX: P(\text{at least one target})=1-(0.9)^2=0.19 ### Answer The required probability is 0.19. LaTeX: \boxed{0.19\quad\text{(D)}} ### Distractor analysis - A: This halves the single-outcome target probability even though two independent opportunities increase the chance of seeing the target. - B: This adds only part of the second opportunity and omits valid outcomes. - C: This starts from the exact-one probability 0.18 and incorrectly subtracts the both-target probability 0.01. - E: This applies the same complement calculation to a category with probability 0.2 rather than the requested category with probability 0.1. ### Verification record - Main value: 0.19 - Main method: Use the complement rule: one minus the probability that both independent outcomes avoid the target category. - Independent check: Sum the disjoint cases of exactly one target and two targets: 2(0.1)(0.9) + (0.1)^2 = 0.19. - Independent value: 0.19 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: component inspection outcomes Each of three independently inspected components is classified as pass with probability 0.80, manual review with probability 0.15, or quarantine with probability 0.05. Calculate the probability that at least one component is sent for manual review. - A: 0.150 - B: 0.325 - C: 0.386 - D: 0.450 - E: 0.614 A component avoids manual review with probability 0.85. All three avoid review with probability 0.85^3, so the requested probability is 1 - 0.85^3 = 0.385875, which rounds to choice C. #### Variant solution ### Setup Combine the two classifications that are not manual review. LaTeX: P(\text{not review})=0.80+0.05=0.85 ### Model Independence makes the probability that all three components avoid review the product of their individual probabilities. LaTeX: P(\text{no reviews})=(0.85)^3 ### Compute Take the complement to include one, two, or three manual-review outcomes. LaTeX: P(\text{at least one review})=1-(0.85)^3=0.385875 ### Answer The probability rounds to 0.386. LaTeX: \boxed{0.386\quad\text{(C)}} ### Verification record - Main value: 0.385875 - Main method: Use the complement of three independent non-review outcomes. - Independent check: Sum the binomial probabilities for exactly one, two, and three reviews: 3(0.15)(0.85)^2 + 3(0.15)^2(0.85) + (0.15)^3. - Independent value: 0.385875 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #657 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-657/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Inclusion–Exclusion; Probability Bounds This is an inclusion–exclusion bounds problem. Writing the unknown overlap as q makes the probability of neither event 0.1+q; the feasible overlap runs from 0 to 0.3, so the requested range is 0.1 to 0.4 and choice C is correct. ### Independent solution ### Setup Let the two indicated actions be events A and B, and denote their unknown overlap probability by q. LaTeX: \Pr(A)=0.30,\qquad \Pr(B)=0.60,\qquad q=\Pr(A\cap B) ### Model Use inclusion–exclusion for the union, then take its complement to express the probability p of neither action. LaTeX: \Pr(A\cup B)=0.30+0.60-q LaTeX: p=1-\Pr(A\cup B)=0.10+q ### Compute The Fréchet bounds restrict the intersection to the interval from zero through the smaller marginal probability. Substituting those endpoints gives the full feasible range for p. LaTeX: \max(0,0.30+0.60-1)\le q\le \min(0.30,0.60) LaTeX: 0\le q\le 0.30\quad\Longrightarrow\quad 0.10\le p\le 0.40 ### Answer Every intermediate overlap is feasible, so the probability of neither action can be any value in the closed interval shown. LaTeX: \boxed{0.10\le p\le 0.40\quad\text{(C)}} ### Distractor analysis - A: This interval can arise from giving the overlap the wrong sign in the complement, writing 0.10-q, and then truncating the resulting negative values at zero. - B: The value 0.10 is only the endpoint obtained when the two events are disjoint; the information does not force their intersection to be zero. - D: The value 0.30 corresponds to selecting q=0.20 in p=0.10+q, but no condition singles out that particular overlap. - E: The range 0.70 to 1.00 belongs to the complement of the intersection, 1-q, which means 'not both' rather than 'neither.' ### Verification record - Main value: 0.100000–0.400000 - Main method: Apply inclusion–exclusion and the Fréchet bounds to p=1-0.30-0.60+q. - Independent check: Construct endpoint probability tables: q=0 gives p=0.10, while q=0.30 gives p=0.40; convex mixtures realize every value between them. - Independent value: 0.100000–0.400000 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: overnight processing features For an overnight data-processing batch, event A means that edge compression is used and event B means that redundant storage is used. Historical records give P(A)=0.55 and P(B)=0.35, but they do not specify the dependence between the events. Determine the full possible range of the probability that exactly one of the two features is used. - A: 0.000 to 0.350 - B: 0.200 to 0.550 - C: 0.200 to 0.900 - D: 0.550 to 0.900 - E: 0.650 If q is the probability that both features are used, the exactly-one probability is 0.90-2q. Since q can range from 0 to 0.35, the answer ranges from 0.20 to 0.90, selecting choice C. #### Variant solution ### Setup Represent the unspecified dependence by the intersection probability q. LaTeX: q=\Pr(A\cap B) LaTeX: 0\le q\le \min(0.55,0.35)=0.35 ### Model Exactly one feature is used in either A without B or B without A. Each marginal contains the intersection once. LaTeX: \Pr(\text{exactly one})=\Pr(A)+\Pr(B)-2\Pr(A\cap B) LaTeX: \Pr(\text{exactly one})=0.90-2q ### Compute The expression decreases with q, so its endpoints occur at the two feasible endpoint overlaps. LaTeX: q=0\quad\Longrightarrow\quad \Pr(\text{exactly one})=0.90 LaTeX: q=0.35\quad\Longrightarrow\quad \Pr(\text{exactly one})=0.20 ### Answer The possible exactly-one probability fills the interval from 0.20 through 0.90. LaTeX: \boxed{0.20\le \Pr(\text{exactly one})\le 0.90\quad\text{(C)}} ### Verification record - Main value: 0.200000–0.900000 - Main method: Bound q=P(A intersection B) and evaluate 0.90-2q at both endpoints. - Independent check: Two explicit joint tables with q=0 and q=0.35 yield exactly-one masses 0.90 and 0.20, and linear interpolation preserves valid cell probabilities. - Independent value: 0.200000–0.900000 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #658 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-658/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Conditional Probability; Inclusion–Exclusion This problem combines conditional probability with a three-event union. Regrouping the union makes its probability 0.80-0.80(0.15r)+0.17r; setting this equal to one gives r=4, which is choice D. ### Independent solution ### Setup Group the full union as event B together with the union of A and C. The overlap between those two grouped events simplifies because B and C cannot occur together. LaTeX: B\cap(A\cup C)=(A\cap B)\cup(B\cap C)=A\cap B LaTeX: \Pr(A\cap B)=\Pr(B)\Pr(A\mid B) ### Model Apply the two-set addition rule to the grouped events and substitute the supplied expressions. LaTeX: \Pr(A\cup B\cup C)=\Pr(B)+\Pr(A\cup C)-\Pr(A\cap B) LaTeX: 1=0.80+0.17r-0.80(0.15r) ### Compute Combine the two coefficients of r and isolate the parameter. LaTeX: 1=0.80+(0.17-0.12)r=0.80+0.05r LaTeX: r=\frac{0.20}{0.05}=4 ### Answer The parameter value satisfying all of the event relationships is four. LaTeX: \boxed{r=4\quad\text{(D)}} ### Distractor analysis - A: Setting r to zero leaves the grouped union probability at 0.80, so it cannot satisfy the stated total probability of one. - B: The value 20/17 comes from solving 0.80+0.17r=1 and therefore treating A union C as disjoint from B; it omits the A intersection B overlap. - C: The value 100/29 comes from the incorrect equation 1=0.12r+0.17r, which drops the standalone 0.80 mass and adds the overlap rather than subtracting it. - E: Treating 0.15r as P(A intersection B), instead of as the conditional probability P(A given B), produces r=10 and falsely suggests that none of A through D applies. ### Verification record - Main value: 4 - Main method: Regroup the union as B union (A union C), convert the conditional probability to an intersection, and solve the resulting linear equation. - Independent check: At r=4, P(A intersection B)=0.80(0.60)=0.48 and P(A union C)=0.68, so 0.80+0.68-0.48=1. - Independent value: 1 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: diagnostic signals Every diagnostic batch produces at least one of three signals: drift D, latency L, or thermal T. Records show that P(D union T)=0.58 and P(D given L)=0.25. A latency signal and a thermal signal can never occur in the same batch. Calculate P(L). - A: 0.315 - B: 0.420 - C: 0.560 - D: 0.580 - E: 0.670 Because latency and thermal signals are mutually exclusive, the overlap of L with D union T is just D intersection L. The full-union equation becomes 1=b+0.58-0.25b, giving b=0.56 and choice C. #### Variant solution ### Setup Write b for the latency probability. The conditional drift rate among latency batches determines their intersection. LaTeX: b=\Pr(L) LaTeX: \Pr(D\cap L)=\Pr(L)\Pr(D\mid L)=0.25b ### Model Combine L with D union T. Since L and T are disjoint, their grouped overlap contains only D intersection L. LaTeX: \Pr(L\cap(D\cup T))=\Pr(D\cap L)=0.25b LaTeX: 1=\Pr(L)+\Pr(D\cup T)-\Pr(L\cap(D\cup T)) ### Compute Insert the known union probability and solve the one-variable equation. LaTeX: 1=b+0.58-0.25b=0.58+0.75b LaTeX: b=\frac{0.42}{0.75}=0.56 ### Answer The latency signal occurs in 56% of diagnostic batches. LaTeX: \boxed{\Pr(L)=0.560\quad\text{(C)}} ### Verification record - Main value: 0.56 - Main method: Use the grouped union identity 1=P(L)+0.58-P(D intersection L) with P(D intersection L)=0.25P(L). - Independent check: Substitution gives P(D intersection L)=0.14 and grouped union probability 0.56+0.58-0.14. - Independent value: 1 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #659 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-659/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Negative Binomial Distribution; Waiting-Time Probabilities Let X be the trial number of the second success. The negative-binomial probabilities for X=10, 11, and 12 are 0.046695, 0.036318, and 0.027965; their sum is 0.110978, which rounds to choice D. ### Independent solution ### Setup Model the outcomes as independent Bernoulli trials with success probability 0.30, and let X mark the trial on which success number two occurs. LaTeX: p=0.30,\qquad 1-p=0.70 ### Model For the second success to occur on trial x, exactly one of the first x-1 trials succeeds and trial x also succeeds. LaTeX: \Pr(X=x)=\binom{x-1}{1}(0.30)^2(0.70)^{x-2},\qquad x=2,3,\ldots ### Compute Evaluate the three mutually exclusive arrival times and add their probabilities. LaTeX: \Pr(X=10)=9(0.30)^2(0.70)^8=0.0466948881 LaTeX: \Pr(X=11)=10(0.30)^2(0.70)^9=0.0363182463 LaTeX: \Pr(X=12)=11(0.30)^2(0.70)^{10}=0.0279650497 LaTeX: \Pr(10\le X\le12)=0.1109781841 ### Answer The requested three-trial window has probability approximately 0.1110. LaTeX: \boxed{0.1110\quad\text{(D)}} ### Distractor analysis - A: The value 0.0280 is only P(X=12), so it retains the final endpoint but omits the valid arrival times 10 and 11. - B: The value 0.0363 is only P(X=11), the middle term of the required three-term sum. - C: The value 0.0856 results from the off-by-one sum of x(0.30)^2(0.70)^(x-1) for x=10 through 12, advancing both the eligible-position count and the failure count. - E: The value 0.1223 uses x rather than x-1 as the number of possible locations for the first success, even though the final trial is already fixed as success number two. ### Verification record - Main value: 0.11097818405099992 - Main method: Sum the negative-binomial mass (x-1)(0.30)^2(0.70)^(x-2) over x=10, 11, and 12. - Independent check: Use the binomial-count identity P(X≤n)=P(Binomial(n,0.30)≥2) and subtract the values at n=12 and n=9. - Independent value: 0.11097818405100002 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: corrupted telemetry frames An automated observatory independently labels each incoming telemetry frame as corrupted with probability 0.20. Its emergency channel opens when the third corrupted frame is observed. Calculate the probability that the channel opens on frame 8, 9, or 10. - A: 0.058720 - B: 0.139335 - C: 0.174168 - D: 0.224395 - E: 0.524288 For the third corrupted frame to arrive at position x, two of the first x-1 frames must be corrupted and frame x must also be corrupted. Summing x=8 through 10 gives 0.174168, so choice C is correct. #### Variant solution ### Setup Let T be the position of the third corrupted frame in an independent sequence with corruption probability 0.20. LaTeX: p=0.20,\qquad q=0.80 ### Model An arrival at position x requires two corruptions among the preceding x-1 frames, followed by a corruption at x. LaTeX: \Pr(T=x)=\binom{x-1}{2}p^3q^{x-3} ### Compute Add the three disjoint arrival-time probabilities. LaTeX: \Pr(T=8)=\binom{7}{2}(0.20)^3(0.80)^5=0.05505024 LaTeX: \Pr(T=9)=\binom{8}{2}(0.20)^3(0.80)^6=0.058720256 LaTeX: \Pr(T=10)=\binom{9}{2}(0.20)^3(0.80)^7=0.0603979776 LaTeX: \Pr(8\le T\le10)=0.1741684736 ### Answer The emergency channel opens in the stated frame window with probability about 0.174168. LaTeX: \boxed{0.174168\quad\text{(C)}} ### Verification record - Main value: 0.1741684736000001 - Main method: Sum C(x-1,2)(0.20)^3(0.80)^(x-3) for x=8, 9, and 10. - Independent check: Compute P(Binomial(10,0.20)≥3)-P(Binomial(7,0.20)≥3), which isolates third-arrival times 8 through 10. - Independent value: 0.17416847360000007 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #660 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-660/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Survival Functions An exponential variable with mean 2.5 has rate 0.4. Its probability of exceeding 2.5 is exp[-(0.4)(2.5)]=exp(-1), so the correct response is choice B. ### Independent solution ### Setup Convert the exponential mean into the corresponding rate parameter. LaTeX: \lambda=\frac{1}{2.50}=0.40 ### Model The question asks for an upper-tail probability, so use the exponential survival function rather than its complement. LaTeX: S(t)=\Pr(X\ge t)=e^{-\lambda t} ### Compute At a time equal to the distribution's mean, the rate-time product equals one. LaTeX: \Pr(X\ge2.50)=e^{-(0.40)(2.50)}=e^{-1} LaTeX: e^{-1}=0.3678794412\ldots ### Answer The required survival probability is the exponential constant e to the power negative one. LaTeX: \boxed{e^{-1}\quad\text{(B)}} ### Distractor analysis - A: The expression 1-exp(-0.4) both omits the time factor 2.5 from the exponent and takes the cumulative probability instead of the survival probability. - C: The expression 1-exp(-1) is the probability that the waiting time ends by 2.5, the complement of the requested at-least probability. - D: The expression exp(-0.4) has the correct survival form but uses the rate alone as the exponent, omitting multiplication by 2.5. - E: The expression 1-exp(-6.25) multiplies the time by the mean rather than by the reciprocal mean, and it also selects the lower tail. ### Verification record - Main value: 0.36787944117144233 - Main method: Convert mean 2.5 to rate 0.4 and evaluate the exponential survival function at time 2.5. - Independent check: Directly integrate the density 0.4 exp(-0.4t) from 2.5 to infinity. - Independent value: 0.36787944117144233 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: filtration cartridge life The operating life of a filtration cartridge is exponentially distributed with mean 8 weeks. A cartridge has already operated for 5 weeks without failing. Calculate the conditional probability that it continues operating for at least 6 additional weeks. - A: 0.252840 - B: 0.527633 - C: 0.472367 - D: 0.535261 - E: 0.747160 Exponential memorylessness makes the remaining lifetime exponential with the original mean. The six-week residual survival probability is exp(-6/8)=0.472367, which selects choice C. #### Variant solution ### Setup Let T be total cartridge life. Translate the wording into a conditional survival probability. LaTeX: \Pr(T\ge11\mid T\ge5) ### Model The exponential distribution is memoryless, so only the six additional weeks enter the residual survival calculation. LaTeX: \Pr(T\ge5+6\mid T\ge5)=\Pr(T\ge6) LaTeX: \lambda=\frac{1}{8} ### Compute Evaluate the exponential survival function for the required additional duration. LaTeX: \Pr(T\ge6)=e^{-(1/8)6}=e^{-3/4} LaTeX: e^{-3/4}=0.4723665527\ldots ### Answer The conditional probability of at least six more operating weeks is approximately 0.472367. LaTeX: \boxed{0.472367\quad\text{(C)}} ### Verification record - Main value: 0.4723665527410147 - Main method: Apply exponential memorylessness and evaluate exp[-6/8]. - Independent check: Divide the unconditional survivals S(11)/S(5)=exp(-11/8)/exp(-5/8). - Independent value: 0.4723665527410147 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #661 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-661/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Discrete Random Variables; Conditional Probability Let N be the given count. The target event within the condition is 1≤N≤3, whose probability is 0.045, while the conditioning event N≤3 has probability 0.965. Their ratio is 9/193=0.046632, which matches choice A. ### Independent solution ### Setup Translate the two verbal events into bounds on the count N. At least one affected unit means N is positive, while at least one unaffected unit rules out the maximum count of four. LaTeX: \{\text{at least one affected}\}=\{N\ge1\} LaTeX: \{\text{at least one unaffected}\}=\{N\le3\} ### Model Intersect the target with the conditioning event and apply the conditional-probability ratio. LaTeX: \Pr(N\ge1\mid N\le3)=\frac{\Pr(1\le N\le3)}{\Pr(N\le3)} ### Compute Add the appropriate entries from the supplied probability distribution. LaTeX: \Pr(1\le N\le3)=0.015+0.010+0.020=0.045 LaTeX: \Pr(N\le3)=0.920+0.015+0.010+0.020=0.965 LaTeX: \frac{0.045}{0.965}=\frac{9}{193}=0.0466321244\ldots ### Answer The conditional probability rounds to 0.0466. LaTeX: \boxed{0.0466\quad\text{(A)}} ### Distractor analysis - B: The value 0.0800 is the unconditional probability P(N≥1)=1-P(N=0), so it fails to impose the condition that N cannot equal four. - C: The value 0.0829 divides 0.080 by 0.965 but leaves the excluded N=4 mass in the numerator after removing it from the conditioning denominator. - D: The value 0.5625 equals 0.045/0.080 and reverses the conditioning, calculating the chance of at least one unaffected unit given at least one affected unit. - E: The value 0.7500 equals 0.015/0.020 and arises by replacing both 'at least one' statements with exactly-one point probabilities before taking an unrelated ratio. ### Verification record - Main value: 0.046632124352331605 - Main method: Divide the probability mass for N=1,2,3 by the probability mass for N=0,1,2,3. - Independent check: Use complements to obtain the denominator 1-P(N=4)=0.965, then subtract P(N=0)=0.920 from it for the numerator. - Independent value: 0.046632124352331605 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: delayed shipment count Let N be the number of delayed shipments in a five-shipment dispatch. Its distribution is P(N=0)=0.64, P(N=1)=0.16, P(N=2)=0.09, P(N=3)=0.06, P(N=4)=0.03, and P(N=5)=0.02. Given that at least one shipment arrives on time, calculate the probability that N is odd. - A: 0.220000 - B: 0.224490 - C: 0.240000 - D: 0.244898 - E: 0.775510 The condition removes N=5 and has probability 0.98. Within that event, an odd delayed count means N=1 or N=3, with total mass 0.22; dividing gives 0.224490 and choice B. #### Variant solution ### Setup At least one on-time shipment means not all five shipments are delayed. LaTeX: H=\{N\le4\} LaTeX: \Pr(H)=1-\Pr(N=5)=1-0.02=0.98 ### Model Inside the conditioning event, the possible odd values are one and three; the odd value five has been excluded. LaTeX: \{N\text{ odd}\}\cap H=\{N=1\}\cup\{N=3\} ### Compute Add the two eligible point probabilities and normalize by the conditioning probability. LaTeX: \Pr(N\text{ odd},H)=0.16+0.06=0.22 LaTeX: \Pr(N\text{ odd}\mid H)=\frac{0.22}{0.98}=0.2244897959\ldots ### Answer The conditional probability of an odd delayed count is approximately 0.224490. LaTeX: \boxed{0.224490\quad\text{(B)}} ### Verification record - Main value: 0.22448979591836735 - Main method: Condition the discrete distribution on N≤4 and sum the remaining odd masses at N=1 and N=3. - Independent check: Normalize all six masses after setting the excluded N=5 cell to zero; the conditional cells at N=1 and N=3 sum to 0.16/0.98+0.06/0.98. - Independent value: 0.22448979591836732 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #662 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-662/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Discrete Random Variables; Sums of Independent Random Variables; Convolution This is a discrete convolution for two independent copies of the hourly count. Adding the six qualifying ordered-pair probabilities gives 57/400, so the official answer is choice E. ### Independent solution ### Setup Let X and Y be the two technicians' counts. They are independent and share the stated probability mass function. LaTeX: \Pr(X=k)=\Pr(Y=k)=\frac{(1,2,4,8,3,2)_k}{20},\qquad k=0,1,\ldots,5 ### Model A total of at least eight is possible only for six ordered pairs. Independence turns each joint probability into a product of marginal masses. LaTeX: \{(x,y):x+y\ge 8\}=\{(3,5),(4,4),(5,3),(4,5),(5,4),(5,5)\} LaTeX: \Pr(X=x,Y=y)=\Pr(X=x)\Pr(Y=y) ### Compute Use the distribution weights directly and retain both orders whenever the counts differ. LaTeX: \Pr(X+Y\ge 8)=\frac{8\cdot2+3\cdot3+2\cdot8+3\cdot2+2\cdot3+2\cdot2}{20^2} LaTeX: \Pr(X+Y\ge 8)=\frac{57}{400}=0.1425 ### Answer The required probability is 57/400. LaTeX: \boxed{\frac{57}{400}\quad\text{(E)}} ### Distractor analysis - A: This counts only outcomes in which both technicians resolve at least four problems. It omits the qualifying ordered pairs (3,5) and (5,3). - B: This total counts each unequal qualifying pair in only one order. The technicians are distinguishable, so both (x,y) and (y,x) must be retained. - C: This is the probability of exactly eight resolved problems, obtained from (3,5), (4,4), and (5,3), rather than the probability of at least eight. - D: This counts qualifying outcomes having at least one count equal to five, but it leaves out the additional qualifying outcome (4,4). ### Verification record - Main value: 0.1425 - Main method: Enumerate the six ordered pairs whose independent discrete counts sum to at least eight and add their product probabilities. - Independent check: Convolve the integer weights and sum the complement coefficients for totals zero through seven; they total 343 out of 400, leaving 57 out of 400. - Independent value: 0.1425 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: two-drone parcel sortie Two rescue drones independently complete a parcel sortie. The number delivered by either drone has probabilities 0.10, 0.10, 0.20, 0.40, and 0.20 for counts 0, 1, 2, 3, and 4, respectively. Calculate the probability that the two drones deliver at least six parcels in total. - A: 24% - B: 28% - C: 32% - D: 36% - E: 44% Enumerating the ordered count pairs with total at least six gives probability 0.44. The result selects choice E. #### Variant solution ### Setup Let R and S be the parcel counts for the first and second drone. Their common probability weights for counts zero through four are 1, 1, 2, 4, and 2 out of 10. LaTeX: \Pr(R=k)=\Pr(S=k)=\frac{(1,1,2,4,2)_k}{10},\qquad k=0,1,\ldots,4 ### Model The qualifying ordered pairs are those with a sum of six, seven, or eight. LaTeX: \{(r,s):r+s\ge6\}=\{(2,4),(3,3),(4,2),(3,4),(4,3),(4,4)\} ### Compute Multiply the independent weights for each ordered pair and add. LaTeX: \Pr(R+S\ge6)=\frac{2\cdot2+4\cdot4+2\cdot2+4\cdot2+2\cdot4+2\cdot2}{10^2} LaTeX: \Pr(R+S\ge6)=\frac{44}{100}=0.44 ### Answer The chance of delivering at least six parcels is 0.44. LaTeX: \boxed{0.44\quad\text{(E)}} ### Verification record - Main value: 0.44 - Main method: Add the independent product probabilities for all six ordered pairs with total at least six. - Independent check: Multiply the probability-generating polynomial with coefficients (1,1,2,4,2)/10 by itself and sum its coefficients at powers six through eight. - Independent value: 0.44 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ## SOA Exam P Sample Question #663 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-663/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Conditional Probability; Normal Distribution; Standardization Condition on the completion time being above its normal mean, an event with probability one-half. Standardizing the upper endpoint gives a conditional probability of 0.261117, which rounds to choice B. ### Independent solution ### Setup Let T be the total completion time. Standardize T using the given mean and standard deviation. LaTeX: T\sim N(7.5,1.5^2),\qquad Z=\frac{T-7.5}{1.5} ### Model The requested event lies inside the conditioning event, so conditional probability is the interval probability divided by the upper-half probability. LaTeX: \Pr(T\le8\mid T>7.5)=\frac{\Pr(7.57.5)} LaTeX: \Pr(T>7.5)=\frac12 ### Compute The upper endpoint is one-third of a standard deviation above the mean. LaTeX: \frac{8-7.5}{1.5}=\frac13 LaTeX: \Pr(T\le8\mid T>7.5)=\frac{\Phi(1/3)-\Phi(0)}{1/2} LaTeX: =2\Phi(1/3)-1=0.2611173196\ldots ### Answer The conditional probability is approximately 0.26. LaTeX: \boxed{0.26\quad\text{(B)}} ### Distractor analysis - A: This is the unconditional interval probability between 7.5 and 8 hours. It omits division by the conditioning probability of one-half. - C: This is approximately the unconditional upper-tail probability beyond 8 hours, rather than the probability of finishing by 8 after the stated condition. - D: This is approximately the unconditional probability of completion by 8 hours, so it includes all times below 7.5 that the condition rules out. - E: This is the complement within the conditioned upper half: the probability of still working after 8 given a completion time above 7.5. ### Verification record - Main value: 0.26111731963647267 - Main method: Standardize the interval and evaluate [Phi(1/3)-Phi(0)]/[1-Phi(0)]. - Independent check: Numerically integrate the N(7.5,1.5^2) density from 7.5 to 8 and divide by its independently integrated upper-half mass. - Independent value: 0.26111731963647267 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: laboratory batch runtime The runtime of an automated laboratory batch is normally distributed with mean 42 minutes and standard deviation 6 minutes. A dashboard shows that a batch is still running at minute 42. Calculate the conditional probability that it completes by minute 48. - A: 0.1587 - B: 0.3173 - C: 0.5000 - D: 0.6827 - E: 0.8413 Given that runtime has passed its mean, the relevant normal mass is the interval from zero to one standard deviation divided by one-half. The result is 0.682689, so choice D is correct. #### Variant solution ### Setup Let R denote runtime and convert minutes to standard normal units. LaTeX: R\sim N(42,6^2),\qquad Z=\frac{R-42}{6} ### Model The dashboard condition restricts R to the upper half of its symmetric distribution. LaTeX: \Pr(R\le48\mid R>42)=\frac{\Pr(4242)} ### Compute Minute 48 is one standard deviation above the mean, and the conditioning event has probability one-half. LaTeX: \Pr(R\le48\mid R>42)=\frac{\Phi(1)-\Phi(0)}{1/2} LaTeX: =2\Phi(1)-1=0.6826894921\ldots ### Answer The conditioned batch has probability about 0.6827 of finishing by minute 48. LaTeX: \boxed{0.6827\quad\text{(D)}} ### Verification record - Main value: 0.6826894921370859 - Main method: Evaluate the conditional standard-normal interval [Phi(1)-Phi(0)]/0.5. - Independent check: Use the error-function identity 2Phi(1)-1=erf(1/sqrt(2)). - Independent value: 0.6826894921370859 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #664 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-664/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Binomial Distribution; Survival Functions An exponential observation exceeds its own mean with probability e^{-1}. The number of exceedances in ten independent observations is binomial, and its nine-or-ten tail equals 0.000825499, matching choice B. ### Independent solution ### Setup For an exponential variable, divide the threshold by the mean in the survival function. Here that ratio is one. LaTeX: \Pr(X>20000)=\exp\!\left(-\frac{20000}{20000}\right)=e^{-1} ### Model Let N count observations above the threshold in the independent sample. LaTeX: N\sim\operatorname{Bin}\!\left(10,e^{-1}\right) LaTeX: \Pr(N\ge9)=\Pr(N=9)+\Pr(N=10) ### Compute Evaluate the two possible binomial outcomes and combine them. LaTeX: \Pr(N\ge9)=\binom{10}{9}e^{-9}(1-e^{-1})+e^{-10} LaTeX: \Pr(N\ge9)=\frac{10e-9}{e^{10}}=0.0008254987\ldots ### Answer The required binomial tail is represented by choice B. LaTeX: \boxed{\Pr(N\ge9)=\frac{10e-9}{e^{10}}\quad\text{(B)}} ### Distractor analysis - A: This reverses the coefficients that arise after factoring the two binomial tail terms. The factor must be 10-9/e, not 9-10/e. - C: This uses the nonexceedance probability 1-e^{-1} as though it were the exceedance probability and also applies an incompatible extra scaling. - D: This is the probability of exactly nine exceedances. It omits the additional qualifying outcome in which all ten observations exceed the threshold. - E: This raises the nonexceedance probability to the tenth power and attaches a factor of nine; it is neither an exact binomial mass nor the requested tail. ### Verification record - Main value: 0.0008254986730044322 - Main method: Use the exponential survival probability e^-1 as the binomial success probability and add the masses at nine and ten. - Independent check: Enumerate the ten patterns with exactly one nonexceedance and the single all-exceedance pattern, then add their independent product probabilities. - Independent value: 0.0008254986730044322 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: long-duration sensor outages Outage durations for a remote sensor are independent and exponentially distributed with mean 4 hours. During six outages selected at random, calculate the probability that at least five last longer than 4 hours. - A: 0.002479 - B: 0.004259 - C: 0.006738 - D: 0.025555 - E: 0.028034 Each outage exceeds four hours with probability e^{-1}. Adding the binomial masses for five and six exceedances gives 0.0280339, so choice E is correct. #### Variant solution ### Setup The threshold equals the exponential mean, so the individual survival probability has a simple form. LaTeX: p=\Pr(T>4)=e^{-4/4}=e^{-1} ### Model Let K count long outages among the six independent observations. LaTeX: K\sim\operatorname{Bin}(6,e^{-1}) LaTeX: \Pr(K\ge5)=\Pr(K=5)+\Pr(K=6) ### Compute Add the exactly-five and all-six probabilities. LaTeX: \Pr(K\ge5)=6e^{-5}(1-e^{-1})+e^{-6} LaTeX: \Pr(K\ge5)=0.0280339211\ldots ### Answer The probability that at least five outages are long is approximately 0.028034. LaTeX: \boxed{0.028034\quad\text{(E)}} ### Verification record - Main value: 0.028033921111181012 - Main method: Evaluate the binomial tail at five and six with n=6 and p=e^-1. - Independent check: Sum six disjoint product probabilities for which one specified outage is short, then add the all-long product probability. - Independent value: 0.028033921111181012 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ## SOA Exam P Sample Question #665 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-665/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Sums of Independent Random Variables; Cumulative Probabilities Independent Poisson counts add, so the full-document count is Poisson with mean 10. Complementing its cumulative probability through four gives 0.970747, which rounds to choice C. ### Independent solution ### Setup Let X_j be the count on page j and let S be the count over the entire document. LaTeX: X_j\sim\operatorname{Poisson}(0.10),\qquad S=\sum_{j=1}^{100}X_j ### Model The independent Poisson components combine into another Poisson variable whose mean is the sum of the page means. LaTeX: S\sim\operatorname{Poisson}(100\cdot0.10)=\operatorname{Poisson}(10) ### Compute More than four is the complement of counts zero through four. LaTeX: \Pr(S>4)=1-e^{-10}\sum_{k=0}^{4}\frac{10^k}{k!} LaTeX: \Pr(S>4)=1-e^{-10}\left(1+10+50+\frac{1000}{6}+\frac{10000}{24}\right) LaTeX: \Pr(S>4)=0.9707473119\ldots ### Answer The upper-tail probability rounds to 0.97. LaTeX: \boxed{0.97\quad\text{(C)}} ### Distractor analysis - A: This is approximately the probability of four or fewer counts only through three, P(S<=3)=0.01034; it both uses the lower tail and drops the boundary value four. - B: This is the lower-tail probability P(S<=4)=0.02925. It is the complement of the requested result. - D: This is approximately 1-P(S=4)=0.98108, which excludes exactly four but incorrectly keeps counts zero through three. - E: This is approximately P(S>=4)=0.98966, an off-by-one tail that includes the four-count outcome. ### Verification record - Main value: 0.9707473119230389 - Main method: Aggregate the 100 independent Poisson means to 10 and subtract the CDF through four from one. - Independent check: Generate masses recursively from p0=e^-10 using p(k+1)=10p(k)/(k+1), add p0 through p4, and take the complement. - Independent value: 0.9707473119230389 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: satellite mosaic false detections A satellite image mosaic is divided into 40 independently processed zones. The number of false detections in each zone is Poisson with mean 0.075. Calculate the probability that the completed mosaic contains more than two false detections. - A: 0.0498 - B: 0.2240 - C: 0.4232 - D: 0.5768 - E: 0.8009 The 40 independent zone counts produce a Poisson total with mean 3. Its probability above two is 0.576810, selecting choice D. #### Variant solution ### Setup Let F be the total false-detection count across all zones. LaTeX: F=\sum_{j=1}^{40}F_j,\qquad F_j\sim\operatorname{Poisson}(0.075) ### Model Poisson superposition adds the 40 zone means. LaTeX: F\sim\operatorname{Poisson}(40\cdot0.075)=\operatorname{Poisson}(3) ### Compute Use the complement of the zero-, one-, and two-count masses. LaTeX: \Pr(F>2)=1-e^{-3}\left(1+3+\frac{3^2}{2}\right) LaTeX: \Pr(F>2)=0.5768099189\ldots ### Answer The probability of more than two false detections is approximately 0.5768. LaTeX: \boxed{0.5768\quad\text{(D)}} ### Verification record - Main value: 0.5768099188731565 - Main method: Aggregate the independent Poisson counts to mean 3 and complement the CDF through two. - Independent check: Use p0=e^-3, p1=3p0, and p2=(3/2)p1; subtract their recursively obtained sum from one. - Independent value: 0.5768099188731565 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #666 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-666/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Normal Distribution; Percentiles; Scale Parameters The permitted interval is symmetric about the normal mean, so 80% central coverage leaves 10% in each tail. Dividing the half-width by z_{0.90} gives a largest standard deviation of 0.390152, matching choice B. ### Independent solution ### Setup Write the tolerance interval as a symmetric distance of 0.5 from the mean. LaTeX: X\sim N(11,\sigma^2),\qquad \Pr(|X-11|\le0.5)\ge0.80 ### Model For a centered normal distribution, the interval probability decreases as sigma increases. The largest allowable sigma therefore makes the probability exactly 0.80. LaTeX: \Pr\!\left(-\frac{0.5}{\sigma}\le Z\le\frac{0.5}{\sigma}\right)=0.80 LaTeX: \Phi\!\left(\frac{0.5}{\sigma}\right)=0.90 ### Compute Use the 90th standard-normal percentile because the two outside tails each have probability 0.10. LaTeX: z_{0.90}=1.2815515655\ldots LaTeX: \sigma_{\max}=\frac{0.5}{z_{0.90}}=0.3901520730\ldots ### Answer The largest standard deviation rounds to 0.39. LaTeX: \boxed{\sigma_{\max}\approx0.39\quad\text{(B)}} ### Distractor analysis - A: This smaller spread would satisfy the coverage requirement, but it is unnecessarily conservative and therefore cannot be the largest permitted value. - C: This uses the 80th percentile as the upper standardized endpoint. Central 80% coverage instead requires the 90th percentile after splitting the remaining probability between two tails. - D: This results from an inverted or width-scaled standardization. Substitution gives only about 24.5% central coverage, far below the required level. - E: This treats a normal quantile as a standard deviation multiplier rather than dividing the half-width by it; its central coverage is only about 15.5%. ### Verification record - Main value: 0.39015207303618954 - Main method: Set the symmetric central coverage to 0.80 and divide the half-width 0.5 by the standard-normal 90th percentile. - Independent check: Substitute the computed standard deviation into Phi(0.5/sigma)-Phi(-0.5/sigma). - Independent value: 0.7999999999999998 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: precision pouch filling Fill weights for a precision powder pouch are normally distributed with mean 50 grams and standard deviation s. A quality rule requires at least 90% of pouches to weigh between 48.8 and 51.2 grams. Calculate the largest allowable value of s. - A: 0.466 - B: 0.730 - C: 0.936 - D: 1.974 - E: 3.948 A central probability of 90% places 5% in each tail. Dividing the 1.2-gram half-width by z_{0.95} gives 0.729548 grams, so choice B is correct. #### Variant solution ### Setup The permitted range extends 1.2 grams to either side of the mean. LaTeX: W\sim N(50,s^2),\qquad \Pr(|W-50|\le1.2)\ge0.90 ### Model At the largest scale, the central interval contains exactly 90%, leaving 5% in each tail. LaTeX: \Phi\!\left(\frac{1.2}{s}\right)=0.95 ### Compute Divide the half-width by the 95th standard-normal percentile. LaTeX: z_{0.95}=1.6448536270\ldots LaTeX: s_{\max}=\frac{1.2}{z_{0.95}}=0.7295481983\ldots ### Answer The maximum allowable standard deviation is approximately 0.730 grams. LaTeX: \boxed{s_{\max}\approx0.730\quad\text{(B)}} ### Verification record - Main value: 0.729548198294123 - Main method: Divide the symmetric half-width 1.2 by the standard-normal 95th percentile. - Independent check: Evaluate Phi(1.2/s)-Phi(-1.2/s) at the computed scale. - Independent value: 0.8999999999999999 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #667 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-667/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Normal Distribution; Exponential Distribution; Percentiles; Moments Match the exponential 80th percentile, 500 ln(5), to the normal percentile 500+z0.80 sigma. This gives sigma=362.062 and second moment 500 squared plus sigma squared, or 381,089, which rounds to choice C. ### Independent solution ### Setup Write the exponential distribution through its survival function and solve for its 80th percentile q. LaTeX: 0.20=e^{-q/500} LaTeX: q=500\log 5=804.7189562\ldots ### Model For the normal model, the same percentile is the mean plus the 80th standard-normal quantile times the unknown standard deviation. LaTeX: q=500+z_{0.80}\sigma LaTeX: z_{0.80}=0.8416212336\ldots ### Compute Solve for the normal standard deviation, then use the identity relating the raw second moment to variance and squared mean. LaTeX: \sigma=\frac{500\log 5-500}{0.8416212336}=362.0618683\ldots LaTeX: \operatorname{E}[X^2]=\sigma^2+\mu^2=(362.0618683)^2+500^2 LaTeX: \operatorname{E}[X^2]=381088.7965\ldots ### Answer Rounding the second moment to the nearest thousand gives 381,000. LaTeX: \boxed{381{,}000\quad\text{(C)}} ### Distractor analysis - A: This is the normal variance, approximately 131,089, and omits the squared mean 500 squared required for the raw second moment. - B: This treats the percentile probability 0.80 as the normal z-quantile and then reports the resulting variance, approximately 145,084. - D: This uses 0.80 in place of z0.80=0.841621 but does add the squared mean, producing approximately 395,084. - E: This is the exponential model's second moment, 2(500 squared), rather than the requested second moment under the normal model. ### Verification record - Main value: 381088.79646715557 - Main method: Equate the two 80th-percentile formulas, solve for the normal standard deviation, and evaluate variance plus squared mean. - Independent check: Substituting sigma=362.0618683 into the normal CDF at 500 ln(5) gives cumulative probability 0.800000. - Independent value: 0.8 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: parcel-handling duration models A logistics team models a parcel-handling duration by a normal distribution with mean 40 minutes. A separate benchmark uses an exponential distribution with mean 50 minutes. The 75th percentiles of the two models coincide. Calculate the second raw moment of the duration under the normal model. - A: 1600 - B: 1889 - C: 3489 - D: 4805 - E: 5000 The shared percentile is 50 ln(4). Matching it to 40+z0.75 sigma gives variance 1888.951, and adding 40 squared gives second moment 3488.951, so choice C. #### Variant solution ### Setup Find the benchmark's 75th percentile from its exponential survival probability. LaTeX: q=50\log 4=69.3147181\ldots ### Model Express the same percentile in the normal model using the 75th standard-normal quantile. LaTeX: q=40+z_{0.75}\sigma LaTeX: z_{0.75}=0.6744897502\ldots ### Compute Recover the variance and then the raw second moment. LaTeX: \sigma=\frac{50\log4-40}{0.6744897502}=43.4620660\ldots LaTeX: \operatorname{E}[Y^2]=40^2+\sigma^2=3488.9511831\ldots ### Answer The normal model's second raw moment rounds to 3489 square minutes. LaTeX: \boxed{3489\quad\text{(C)}} ### Verification record - Main value: 3488.951183133967 - Main method: Match the normal and exponential 75th percentiles, then add the inferred normal variance to the squared mean. - Independent check: The normal CDF with mean 40 and sigma=43.4620660 evaluates to 0.750000 at 50 ln(4). - Independent value: 0.75 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #668 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-668/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Uniform Distribution; Percentiles; Moments For a uniform interval [a,b], the median condition gives a+b=12 and the 90th-percentile condition gives a+0.9(b-a)=13.2. Thus a=-3 and b=15, yielding second moment 63 and choice E. ### Independent solution ### Setup Let a and b be the lower and upper endpoints. Uniform percentiles advance linearly across the interval. LaTeX: q_p=a+p(b-a) ### Model Translate the median and 90th-percentile information into two endpoint equations. LaTeX: \frac{a+b}{2}=6 LaTeX: a+0.9(b-a)=13.2 ### Compute Solve the two equations, then use the raw second-moment formula for a continuous uniform distribution. LaTeX: a=-3,\qquad b=15 LaTeX: \operatorname{E}[X^2]=\frac{a^2+ab+b^2}{3} LaTeX: \operatorname{E}[X^2]=\frac{9-45+225}{3}=63 ### Answer The second raw moment equals 63. LaTeX: \boxed{63\quad\text{(E)}} ### Distractor analysis - A: This reports the median, which is also the uniform mean, rather than the second raw moment. - B: This is the variance (15-(-3)) squared divided by 12, so it omits the squared mean needed for the second moment. - C: This adds the mean 6 to the variance 27; the second-moment identity requires the squared mean 36 instead. - D: This is the squared mean 6 squared and omits the variance 27. ### Verification record - Main value: 63 - Main method: Recover the uniform endpoints from the two quantile equations and evaluate the closed-form raw second moment. - Independent check: Using mean 6 and variance 18 squared divided by 12 gives E[X^2]=27+36=63. - Independent value: 63 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: calibration-error interval A calibration error U is modeled as continuous uniform on an unknown interval. Its 25th percentile is -2 units and its 75th percentile is 6 units. Calculate the second raw moment of U. - A: 2.000 - B: 4.000 - C: 16.000 - D: 21.333 - E: 25.333 The two quartile equations imply endpoints -6 and 10. The variance is 64/3 and the squared mean is 4, so the second raw moment is 76/3=25.333 and choice E. #### Variant solution ### Setup Let c and d denote the unknown endpoints and write the linear uniform quantile formula. LaTeX: q_p=c+p(d-c) ### Model Apply the formula to the two stated quartiles. LaTeX: c+0.25(d-c)=-2 LaTeX: c+0.75(d-c)=6 ### Compute Solve for the endpoints and substitute them into the uniform second-moment formula. LaTeX: c=-6,\qquad d=10 LaTeX: \operatorname{E}[U^2]=\frac{c^2+cd+d^2}{3} LaTeX: \operatorname{E}[U^2]=\frac{36-60+100}{3}=\frac{76}{3} ### Answer The second raw moment is approximately 25.333 square units. LaTeX: \boxed{25.333\quad\text{(E)}} ### Verification record - Main value: 25.333333333333332 - Main method: Solve the two quartile equations for the uniform endpoints and evaluate the second raw moment. - Independent check: The inferred interval has mean 2 and variance 64/3, so variance plus squared mean is 76/3. - Independent value: 25.333333333333332 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ## SOA Exam P Sample Question #669 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-669/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Normal Distribution; Independence; Complements Independence converts the two-year at-least-one probability into a one-year negative-profit probability of 0.20. Hence -100/sigma is the 20th standard-normal quantile, giving sigma=118.818 and choice C. ### Independent solution ### Setup Let p be the probability of a negative result in one year. Use the complement of seeing no negative year in two independent trials. LaTeX: 1-(1-p)^2=0.36 ### Model Solve for the annual probability, then standardize the zero threshold under the normal model. LaTeX: 1-p=\sqrt{0.64}=0.80\quad\Longrightarrow\quad p=0.20 LaTeX: \Phi\!\left(\frac{0-100}{\sigma}\right)=0.20 ### Compute Use the 20th standard-normal quantile and solve for the positive standard deviation. LaTeX: \Phi^{-1}(0.20)=-0.8416212336\ldots LaTeX: -\frac{100}{\sigma}=-0.8416212336 LaTeX: \sigma=118.8182950\ldots ### Answer The annual standard deviation rounds to 119. LaTeX: \boxed{119\quad\text{(C)}} ### Distractor analysis - A: This multiplies 100 by the 80th-percentile z-value 0.8416; standardizing zero requires dividing 100 by that magnitude. - B: This first approximates the annual probability as 0.36/2=0.18, ignoring the overlap between the two independent annual events. - D: This uses the one-year nonnegative probability 0.80 directly as a z-score magnitude, giving 100/0.80=125. - E: With standard deviation 127, the one-year negative probability is about 0.216 and the two-year at-least-one probability is about 0.385, not 0.36. ### Verification record - Main value: 118.818294989389 - Main method: Recover the one-year tail probability through the independent two-year complement, then invert the standard normal CDF. - Independent check: At sigma=118.818295, the annual negative probability is 0.20 and 1-(1-0.20)^2 reconstructs 0.36. - Independent value: 0.36 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: greenhouse energy shortfalls A greenhouse's daily net energy output is normally distributed with mean 60 kWh and an unknown standard deviation that is unchanged from day to day. Daily outputs are independent. The probability of a negative output on at least one of the next three days is 0.271. Calculate the daily standard deviation. - A: 44.8 kWh - B: 46.8 kWh - C: 76.9 kWh - D: 98.4 kWh - E: 600.0 kWh The three-day complement implies a one-day shortfall probability of 0.10. Standardizing zero then gives sigma=60/1.281552=46.818 kWh, so choice B. #### Variant solution ### Setup Let r be the chance of negative output on one day and complement the three-day event. LaTeX: 1-(1-r)^3=0.271 ### Model Solve the independence equation and connect the resulting tail probability to the unknown normal scale. LaTeX: 1-r=\sqrt[3]{0.729}=0.9\quad\Longrightarrow\quad r=0.1 LaTeX: \Phi\!\left(-\frac{60}{\sigma}\right)=0.1 ### Compute Invert the standard normal CDF at 0.10 and isolate sigma. LaTeX: \Phi^{-1}(0.1)=-1.2815515655\ldots LaTeX: \sigma=\frac{60}{1.2815515655}=46.8182488\ldots ### Answer The daily standard deviation is approximately 46.8 kWh. LaTeX: \boxed{46.8\text{ kWh}\quad\text{(B)}} ### Verification record - Main value: 46.81824876434275 - Main method: Use the three-day independent complement to obtain the daily lower-tail probability, then invert the normal CDF. - Independent check: At sigma=46.8182488, the daily negative-output probability is 0.10 and the three-day union probability is 1-0.9^3=0.271. - Independent value: 0.271 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #670 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-670/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Normal Distribution; Percentiles; Insurance Pricing A normal median equals its mean. The upper-tail condition places the premium at the 95th percentile, so its excess above the median is z0.95 times 500, or 822.427, which rounds to choice A. ### Independent solution ### Setup A normal distribution is symmetric, so its median is its mean. LaTeX: m=2500 ### Model The stated upper-tail probability makes the premium the 95th percentile of the loss distribution. LaTeX: \Pr(L>P)=0.05\quad\Longrightarrow\quad \Pr(L\le P)=0.95 LaTeX: P=2500+500z_{0.95} ### Compute Subtract the median before evaluating the standard-normal quantile. LaTeX: P-m=500z_{0.95} LaTeX: P-m=500(1.6448536270)=822.4268135\ldots ### Answer The premium exceeds the median loss by approximately 822. LaTeX: \boxed{822\quad\text{(A)}} ### Distractor analysis - B: This uses the two-sided 95% critical value 1.96, giving 1.96(500)=980, instead of the one-sided 95th percentile 1.64485. - C: This reports the median loss itself, not the amount by which the premium exceeds it. - D: This is approximately the premium 2500+822, while the requested quantity is only its excess above the median. - E: This both uses the two-sided critical value 1.96 and reports the resulting premium 2500+980 rather than the excess. ### Verification record - Main value: 822.426813475736 - Main method: Identify the premium as the normal 95th percentile and subtract the mean-median. - Independent check: Substitution of P=3322.4268135 into the normal survival function with mean 2500 and standard deviation 500 gives 0.050000. - Independent value: 0.05 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: wind-farm reserve threshold A wind farm's hourly output is normally distributed with mean 180 MW and standard deviation 24 MW. An operator chooses a reserve threshold R so that output exceeds R in 10% of hours. Calculate the amount by which R exceeds the median hourly output. - A: 24.000 MW - B: 30.757 MW - C: 39.476 MW - D: 180.000 MW - E: 210.757 MW The threshold is the 90th percentile and the median is 180 MW. Their difference is 24z0.90=30.757 MW, so choice B. #### Variant solution ### Setup Use normal symmetry to identify the median with the stated mean. LaTeX: m=180\text{ MW} ### Model An upper-tail probability of 0.10 places the threshold at cumulative probability 0.90. LaTeX: R=180+24z_{0.90} ### Compute Subtract the median and evaluate the 90th standard-normal quantile. LaTeX: R-m=24(1.2815515655) LaTeX: R-m=30.7572376\ldots\text{ MW} ### Answer The reserve threshold is 30.757 MW above the median. LaTeX: \boxed{30.757\text{ MW}\quad\text{(B)}} ### Verification record - Main value: 30.75723757307041 - Main method: Convert the upper-tail requirement to the 90th normal percentile and subtract the mean-median. - Independent check: The normal survival probability at 180+30.7572376 with standard deviation 24 is 0.100000. - Independent value: 0.1 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #671 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-671/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Linear Combinations of Independent Random Variables; Normal Distribution; Independence The average of the two independent normals is normal with mean 75 and variance 149. Standardizing 80 gives z=0.4096 and an upper tail of 0.341044, so choice C. ### Independent solution ### Setup Let A be the arithmetic average of the two independent normal lifetimes. LaTeX: A=\frac{X+Y}{2} ### Model A linear combination of independent normal variables is normal. Means combine linearly and variances use squared coefficients. LaTeX: \operatorname{E}[A]=\frac{70+80}{2}=75 LaTeX: \operatorname{Var}(A)=\frac{14^2+20^2}{4}=149 ### Compute Standardize the threshold using the standard deviation of the average and evaluate the upper tail. LaTeX: z=\frac{80-75}{\sqrt{149}}=0.40961596\ldots LaTeX: \Pr(A>80)=1-\Phi(0.40961596)=0.34104384\ldots ### Answer The probability that the average exceeds 80 is approximately 0.341. LaTeX: \boxed{0.341\quad\text{(C)}} ### Distractor analysis - A: This is approximately Pr(X>80)Pr(Y>80), which incorrectly replaces the event that the average exceeds 80 by the stricter event that both values exceed 80. - B: This is approximately Pr(X>80) for the first distribution alone and ignores the second value in the average. - D: This is close to the result of averaging the component standard deviations directly; independent variances must be combined with squared coefficients. - E: This is the lower-tail complement Pr(A less than or equal to 80), rather than the requested upper tail. ### Verification record - Main value: 0.3410438435934441 - Main method: Form the normal distribution of the independent average and evaluate its standardized upper tail. - Independent check: Equivalently, X+Y is normal with mean 150 and variance 596; its upper tail above 160 is 0.3410438436. - Independent value: 0.3410438435934441 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: blended laboratory score Two independent laboratory instruments produce normally distributed quality scores. Instrument A has mean 10 and standard deviation 6; instrument B has mean 14 and standard deviation 8. A blended score is W=0.25A+0.75B. Calculate the probability that W exceeds 15. - A: 0.091 - B: 0.373 - C: 0.395 - D: 0.450 - E: 0.627 The weighted normal score has mean 13 and variance 38.25. Its threshold z-score is 0.32338, giving upper-tail probability 0.373203 and choice B. #### Variant solution ### Setup Treat the reported score as a linear combination of the two independent readings. LaTeX: W=0.25A+0.75B ### Model Combine the normal means and independent variances using the linear weights. LaTeX: \operatorname{E}[W]=0.25(10)+0.75(14)=13 LaTeX: \operatorname{Var}(W)=0.25^2(6^2)+0.75^2(8^2)=38.25 ### Compute Standardize 15 under the resulting normal distribution. LaTeX: z=\frac{15-13}{\sqrt{38.25}}=0.32338083\ldots LaTeX: \Pr(W>15)=1-\Phi(0.32338083)=0.37320342\ldots ### Answer The blended score exceeds 15 with probability approximately 0.373. LaTeX: \boxed{0.373\quad\text{(B)}} ### Verification record - Main value: 0.37320342158351866 - Main method: Derive the mean and variance of the independent weighted normal sum and evaluate its upper tail. - Independent check: The moment-generating function factors into those of 0.25A and 0.75B, giving normal parameters 13 and 38.25 and the same tail value. - Independent value: 0.37320342158351866 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #672 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-672/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Percentiles and Moments For an exponential lifetime with mean theta, the median is theta times log 2. The stated gap determines theta as 12.3838, so the variance is 153.358 and choice D. ### Independent solution ### Setup Parameterize the exponential lifetime by its mean theta. Its variance is theta squared, so first recover theta from the mean-median gap. LaTeX: \operatorname{E}[T]=\theta,\qquad \operatorname{Var}(T)=\theta^2 ### Model The median solves a one-half survival probability and therefore equals theta times the natural logarithm of two. LaTeX: e^{-m/\theta}=\frac12 LaTeX: m=\theta\ln 2 ### Compute Use the stated difference to solve for the mean and then square that mean. LaTeX: \theta-\theta\ln2=3.80 LaTeX: \theta=\frac{3.80}{1-\ln2}=12.38378714\ldots LaTeX: \operatorname{Var}(T)=\theta^2=153.35818399\ldots ### Answer The variance rounds to 153 at the precision of the choices. LaTeX: \boxed{\operatorname{Var}(T)\approx153\quad\text{(D)}} ### Distractor analysis - A: This is the recovered mean 12.38 rounded to a whole number; the question asks for the variance, which is the square of the mean. - B: This is approximately the square of the stated gap, 3.80² = 14.44, rather than the square of the exponential mean. - C: Equating the gap to the median gives θ = 3.80 / ln(2) and a variance near 30; the gap is actually mean minus median. - E: Approximating 1 − ln(2) by 0.25 gives θ ≈ 15.2 and variance near 231, but that approximation is far too coarse. ### Verification record - Main value: 153.35818398900298 - Main method: Solve theta(1 - log 2) = 3.80 and use the exponential variance theta squared. - Independent check: With theta = 12.38378714, the median is 8.58378714; their difference is 3.80 and theta squared is 153.35818399. - Independent value: 153.35818398900298 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: maintenance ticket duration The resolution time for an automated maintenance ticket is exponential. Its 75th percentile is 3.86294 hours greater than its mean. Calculate the variance, in squared hours, of the resolution time. - A: 38.6 - B: 62.5 - C: 81.0 - D: 100.0 - E: 192.2 The exponential 75th percentile is theta log 4. The stated excess equals theta times (log 4 minus 1), giving theta approximately 10 and variance 100, so choice D. #### Variant solution ### Setup Let theta be the mean resolution time and write the exponential quantile formula. LaTeX: q_p=-\theta\ln(1-p) LaTeX: q_{0.75}=\theta\ln4 ### Model Subtract the mean theta from the 75th percentile and equate the result to the observed gap. LaTeX: q_{0.75}-\operatorname{E}[T]=\theta(\ln4-1)=3.86294 ### Compute Solve for theta and use the exponential variance identity. LaTeX: \theta=\frac{3.86294}{\ln4-1}=9.99999\ldots LaTeX: \operatorname{Var}(T)=\theta^2=99.9998\ldots ### Answer The variance is approximately 100 squared hours. LaTeX: \boxed{100.0\quad\text{(D)}} ### Verification record - Main value: 99.99981277237737 - Main method: Recover the exponential mean from the 75th-percentile excess and square it. - Independent check: At theta = 9.99999064, the 75th percentile is 13.86293064, exactly 3.86294 above the mean to the stated precision. - Independent value: 99.99981277237737 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #673 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-673/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Covariance and Correlation; Linear Combinations The supplied correlation and standard deviations give covariance 1. Expanding the covariance of the two linear combinations yields 2c + 5, so c = -5/2 and choice B. ### Independent solution ### Setup Convert correlation to covariance using the two standard deviations. LaTeX: \sigma_X=1,\qquad \sigma_Y=2 LaTeX: \operatorname{Cov}(X,Y)=\rho\sigma_X\sigma_Y=\frac12(1)(2)=1 ### Model Uncorrelated linear combinations have covariance zero. Expand bilinearly without using the means, which do not affect covariance. LaTeX: \operatorname{Cov}(X+Y,cX+Y)=0 ### Compute Collect the variance and cross-covariance terms that multiply c and the terms that do not. LaTeX: c\operatorname{Var}(X)+\operatorname{Cov}(X,Y)+c\operatorname{Cov}(Y,X)+\operatorname{Var}(Y)=0 LaTeX: c+1+c+4=0 LaTeX: c=-\frac52 ### Answer The required coefficient is negative five halves. LaTeX: \boxed{c=-\frac52\quad\text{(B)}} ### Distractor analysis - A: Substituting c = −3 into the correct covariance gives 2(−3) + 5 = −1, so the two combinations remain correlated. - C: This results from omitting the standalone cross term Cov(X,Y) = 1, which changes the equation from 2c + 5 = 0 to 2c + 4 = 0. - D: Setting the uncentered product expectation equal to zero introduces the nonzero mean of X; that raw-product calculation gives −5/6, but uncorrelated means covariance zero. - E: This value comes from mixing correlation coefficients directly into the linear-combination formula without first converting them to covariance units. ### Verification record - Main value: -2.5 - Main method: Expand the covariance of the two linear combinations after converting correlation to covariance. - Independent check: Using covariance matrix [[1,1],[1,4]], the matrix product [1,1] Sigma [c,1]^T equals 2c + 5 and vanishes at c = -2.5. - Independent value: -2.5 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: paired calibration errors Two calibration errors A and B have variances 4 and 9 and correlation -1/3. Determine the constant k for which the error summaries 2A + B and kA + 3B are uncorrelated. - A: -3.375 - B: -2.500 - C: -1.875 - D: 0 - E: 2.500 The covariance is -2. Expanding the covariance of the two summaries gives 6k + 15, which is zero at k = -2.5 and selects choice B. #### Variant solution ### Setup Recover the covariance from the correlation and the standard deviations 2 and 3. LaTeX: \operatorname{Cov}(A,B)=-\frac13(2)(3)=-2 ### Model Set the covariance between the two proposed error summaries equal to zero. LaTeX: \operatorname{Cov}(2A+B,kA+3B)=0 ### Compute Expand all four covariance contributions and combine like terms. LaTeX: 2k(4)+6(-2)+k(-2)+3(9)=0 LaTeX: 6k+15=0 LaTeX: k=-2.5 ### Answer The summaries are uncorrelated when k is negative 2.5. LaTeX: \boxed{k=-2.500\quad\text{(B)}} ### Verification record - Main value: -2.5 - Main method: Expand the covariance of two linear combinations using variances 4 and 9 and covariance -2. - Independent check: With Sigma = [[4,-2],[-2,9]], the product [2,1] Sigma [k,3]^T is 6k + 15 and equals zero at -2.5. - Independent value: -2.5 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #674 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-674/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Discrete Quantile Selection For a Poisson count with mean 3, the cumulative probabilities through 3 and 4 are 0.6472 and 0.8153. Four is the smallest reimbursement limit meeting 75 percent, so choice C. ### Independent solution ### Setup Let X be the Poisson count with rate 3. A limit of m reimburses every item precisely when X does not exceed m. LaTeX: X\sim\operatorname{Poisson}(3) LaTeX: \Pr(\text{all reimbursed under limit }m)=\Pr(X\le m) ### Model Find the smallest listed limit whose Poisson cumulative probability is at least 0.75. LaTeX: F(m)=e^{-3}\sum_{k=0}^{m}\frac{3^k}{k!} ### Compute The cumulative probability remains below the target through three and first exceeds it at four. LaTeX: F(3)=e^{-3}\left(1+3+\frac{3^2}{2!}+\frac{3^3}{3!}\right)=0.6472318888\ldots LaTeX: F(4)=F(3)+e^{-3}\frac{3^4}{4!}=0.8152632445\ldots ### Answer The plan with a maximum of four is the first one that meets the probability requirement. LaTeX: \boxed{m=4\quad\text{(C)}} ### Distractor analysis - A: A limit of two gives F(2) = 0.42319, far below the required 0.75 probability. - B: A limit of three gives F(3) = 0.64723, so it still fails the required probability. - D: A limit of five gives F(5) = 0.91608 and qualifies, but it is not the lowest-premium qualifying choice because the limit of four already works. - E: A limit of six gives F(6) = 0.96649, but it buys more coverage than needed and therefore cannot be the lowest-premium qualifying plan. ### Verification record - Main value: 0.8152632445237721 - Main method: Evaluate the Poisson(3) CDF and select the first listed limit whose CDF reaches 0.75. - Independent check: Poisson recurrence from p0 = exp(-3) gives F(3) = 0.6472318888 and F(4) = 0.8152632445, bracketing 0.75. - Independent value: 4 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: museum sensor service tier A museum expects the number of emergency sensor replacements in a season to follow a Poisson distribution with mean 2.4. Service tiers 1 through 5 cover at most 1, 2, 3, 4, and 5 replacements, respectively, with price increasing by tier. The museum wants at least a 90 percent probability that every replacement is covered. Which is the least expensive acceptable tier? - A: Tier 1 - B: Tier 2 - C: Tier 3 - D: Tier 4 - E: Tier 5 The Poisson(2.4) CDF is 0.7787 through three and 0.9041 through four. Tier 4 is the first to reach 90 percent, so choice D. #### Variant solution ### Setup Let R be the seasonal replacement count and translate a tier limit into a cumulative probability. LaTeX: R\sim\operatorname{Poisson}(2.4) LaTeX: \Pr(\text{full coverage at tier }m)=\Pr(R\le m) ### Model Search for the smallest limit whose cumulative probability is at least 0.90. LaTeX: F(m)=e^{-2.4}\sum_{k=0}^{m}\frac{2.4^k}{k!} ### Compute Check the two consecutive limits surrounding the target probability. LaTeX: F(3)=0.7787229110\ldots LaTeX: F(4)=0.9041314097\ldots ### Answer Tier 4 is the first tier to clear the 90 percent requirement. LaTeX: \boxed{\text{Tier 4}\quad\text{(D)}} ### Verification record - Main value: 0.9041314096636008 - Main method: Evaluate consecutive Poisson(2.4) cumulative probabilities and select the first tier at or above 0.90. - Independent check: Recursing with p(k+1) = 2.4 p(k)/(k+1) gives F(3) = 0.7787229110 and F(4) = 0.9041314097. - Independent value: 4 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #675 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-675/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Moments The zero-count probability is 0.632, so the Poisson mean is minus the natural logarithm of 0.632. Using the Poisson identity for the second moment gives 0.6694, which selects choice D. ### Independent solution ### Setup Let the count have Poisson mean lambda. Convert the given positive-count probability into its zero-count complement. LaTeX: X\sim\operatorname{Poisson}(\lambda) LaTeX: \Pr(X=0)=1-0.368=0.632 ### Model For a Poisson variable, the zero-count probability identifies lambda, and the second raw moment is the sum of the variance and the squared mean. LaTeX: e^{-\lambda}=0.632 LaTeX: \operatorname{E}[X^2]=\operatorname{Var}(X)+\operatorname{E}[X]^2=\lambda+\lambda^2 ### Compute Solve for the rate and substitute it into the second-moment identity. LaTeX: \lambda=-\ln(0.632)=0.4588658848\ldots LaTeX: \operatorname{E}[X^2]=0.4588658848\ldots+(0.4588658848\ldots)^2 LaTeX: \operatorname{E}[X^2]=0.6694237851\ldots ### Answer The second moment rounds to 0.669. LaTeX: \boxed{\operatorname{E}[X^2]\approx0.669\quad\text{(D)}} ### Distractor analysis - A: This is essentially λ², the second factorial moment E[X(X−1)], rather than the requested raw moment E[X²]. - B: This is approximately 2λ², which applies an exponential-style second-moment formula to a Poisson count. - C: This is λ, which is both the mean and variance of a Poisson count, but not its second raw moment. - E: This can result from reading 0.368 as the zero-count probability, recognizing it as approximately e⁻¹, and then reporting the resulting rate 1 instead of the second raw moment. ### Verification record - Main value: 0.6694237851009437 - Main method: Recover the Poisson rate from the zero-count probability and evaluate lambda plus lambda squared. - Independent check: The factorial-moment identity E[X(X-1)] = lambda^2 gives E[X^2] = E[X(X-1)] + E[X], independently reproducing 0.6694237851. - Independent value: 0.6694237851009437 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: server alert count The number of priority alerts produced by a server during an overnight window follows a Poisson distribution. The probability of no priority alerts is 0.7408. Calculate the second moment of the alert count. - A: 0.090 - B: 0.300 - C: 0.390 - D: 0.741 - E: 0.259 The rate is minus the natural logarithm of 0.7408, approximately 0.3000. Adding the rate and its square gives a second moment of 0.3900, so choice C. #### Variant solution ### Setup Let A denote the overnight alert count and lambda its Poisson mean. LaTeX: A\sim\operatorname{Poisson}(\lambda) LaTeX: \Pr(A=0)=e^{-\lambda}=0.7408 ### Model Determine the rate from the zero-count probability, then use the Poisson second-moment formula. LaTeX: \lambda=-\ln(0.7408) LaTeX: \operatorname{E}[A^2]=\lambda+\lambda^2 ### Compute The chosen zero-count probability corresponds almost exactly to a rate of 0.3. LaTeX: \lambda=0.3000245957\ldots LaTeX: \operatorname{E}[A^2]=0.3000245957\ldots+(0.3000245957\ldots)^2 LaTeX: \operatorname{E}[A^2]=0.3900393536\ldots ### Answer The second moment rounds to 0.390. LaTeX: \boxed{\operatorname{E}[A^2]\approx0.390\quad\text{(C)}} ### Verification record - Main value: 0.3900393536452139 - Main method: Solve lambda = -ln(0.7408) and evaluate the Poisson second raw moment lambda + lambda^2. - Independent check: Adding the independently known Poisson factorial moment lambda^2 to E[A] = lambda gives 0.3900393536. - Independent value: 0.3900393536452139 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #676 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-676/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Poisson Distribution; Sums of Independent Random Variables Independence lets the monthly Poisson variances add across two locations and three months. The total variance is three times the combined monthly mean, or 135, so the answer is B. ### Independent solution ### Setup Represent each location-month count separately. Every component is Poisson, so its variance equals its mean. LaTeX: \operatorname{Var}(X_m)=15,\qquad \operatorname{Var}(Y_m)=30 LaTeX: m=1,2,3 ### Model The six counts are mutually independent, so the variance of their sum is the sum of their variances. LaTeX: T=\sum_{m=1}^{3}(X_m+Y_m) LaTeX: \operatorname{Var}(T)=\sum_{m=1}^{3}\left(\operatorname{Var}(X_m)+\operatorname{Var}(Y_m)\right) ### Compute Each month contributes variance 45, and there are three independent months. LaTeX: \operatorname{Var}(T)=3(15+30)=3(45)=135 ### Answer The variance of the full three-month total is 135. LaTeX: \boxed{\operatorname{Var}(T)=135\quad\text{(B)}} ### Distractor analysis - A: This adds the two location variances for only one month and forgets the three-month period. - C: This squares the smaller monthly mean, but a Poisson variance equals its mean and independent periods must be added. - D: This squares the larger monthly mean, confusing a squared mean with a variance. - E: This adds the squared monthly means, 15² + 30², instead of adding Poisson variances across locations and months. ### Verification record - Main value: 135 - Main method: Add the six independent Poisson component variances across two locations and three months. - Independent check: Poisson superposition makes the complete total Poisson with parameter 3(15 + 30) = 135, whose variance is also 135. - Independent value: 135 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: greenhouse detections A greenhouse uses three independent sensor sectors. In a week, the numbers of unusual detections in the sectors are independent Poisson random variables with means 1.5, 2.5, and 4.0. Counts are also independent from week to week. Calculate the variance of the total number of detections over five weeks. - A: 8 - B: 20 - C: 40 - D: 64 - E: 320 The three sectors contribute a combined weekly variance of 8. Across five independent weeks the total variance is 40, which is choice C. #### Variant solution ### Setup For each sector and week, the Poisson variance equals the corresponding sector mean. LaTeX: \lambda_1=1.5,\qquad \lambda_2=2.5,\qquad \lambda_3=4.0 ### Model Add the independent sector variances within a week, then add the five independent weekly totals. LaTeX: \operatorname{Var}(W)=1.5+2.5+4.0=8 LaTeX: T=W_1+W_2+W_3+W_4+W_5 ### Compute Five weekly totals each contribute variance 8. LaTeX: \operatorname{Var}(T)=5\operatorname{Var}(W)=5(8)=40 ### Answer The five-week total has variance 40. LaTeX: \boxed{\operatorname{Var}(T)=40\quad\text{(C)}} ### Verification record - Main value: 40 - Main method: Add the fifteen independent sector-week Poisson variances. - Independent check: Poisson superposition gives a weekly total with rate 8 and a five-week total with rate 40, so its variance is 40. - Independent value: 40 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #677 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-677/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Mixture Distributions; Percentiles This is a two-component exponential-mixture percentile calculation. Integrating the density and setting the survival probability to one-half gives a median of 0.904609, which rounds to choice B. ### Independent solution ### Setup Integrate the positive-support density to obtain its cumulative distribution function. The two exponential terms carry different integration factors. LaTeX: F(x)=\int_0^x 0.4\left(e^{-0.6t}+e^{-1.2t}\right)\,dt LaTeX: F(x)=1-\frac{2}{3}e^{-0.6x}-\frac{1}{3}e^{-1.2x},\qquad x>0 ### Model At a median m, the CDF and survival probability are both one-half. Substituting y equal to the slower exponential term converts the equation to a quadratic. LaTeX: \frac{2}{3}e^{-0.6m}+\frac{1}{3}e^{-1.2m}=\frac{1}{2} LaTeX: y=e^{-0.6m}\quad\Longrightarrow\quad \frac{2}{3}y+\frac{1}{3}y^2=\frac{1}{2} ### Compute Retain the positive quadratic root because an exponential value must lie between zero and one, then recover m with a logarithm. LaTeX: 2y^2+4y-3=0\quad\Longrightarrow\quad y=\frac{\sqrt{10}-2}{2} LaTeX: m=-\frac{1}{0.6}\log\!\left(\frac{\sqrt{10}-2}{2}\right)=0.9046093341\ldots ### Answer The median rounds to 0.905 at the precision used by the answer list. LaTeX: \boxed{m\approx 0.905\quad\text{(B)}} ### Distractor analysis - A: This value solves approximately f(x)=0.5, confusing the height of a density with the cumulative probability required for a median. - C: This is approximately the median of a single exponential distribution having the same mean as the mixture; a mixture is not generally replaced by that moment-matched exponential. - D: Substitution into the derived CDF gives about 0.611 rather than 0.500, so this candidate lies beyond the median. - E: This value is the mixture mean, 0.4/(0.6)^2+0.4/(1.2)^2=1.389, but a skewed distribution need not have equal mean and median. ### Verification record - Main value: 0.904609334072203 - Main method: Integrate the two exponential density terms, set the survival function equal to one-half, and solve the resulting quadratic in exp(-0.6m). - Independent check: Direct numerical integration of the density from zero to the computed median. - Independent value: 0.49999999999999994 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: routed data-transfer sessions A cloud platform routes each data-transfer session to a standard server with probability 0.25 and to an accelerated server with probability 0.75. Conditional on the route, the completion time in minutes is exponential with rate 0.4 for a standard server and 0.8 for an accelerated server. Calculate the median completion time for a randomly routed session. - A: 0.405 - B: 0.507 - C: 1.014 - D: 1.155 - E: 1.733 The unconditional completion time is a mixture, so its survival function is the route-weighted sum of the two conditional survivals. Solving the median equation gives 1.013663 minutes, selecting choice C. #### Variant solution ### Setup Let T be the session time and condition on the randomly selected server class. LaTeX: \Pr(T>t)=0.25e^{-0.4t}+0.75e^{-0.8t} ### Model For the median m, set the mixture survival probability to one-half and substitute y for the slower exponential factor. LaTeX: 0.25e^{-0.4m}+0.75e^{-0.8m}=0.5 LaTeX: y=e^{-0.4m}\quad\Longrightarrow\quad y+3y^2=2 ### Compute The admissible quadratic root is two-thirds. Convert that survival factor back to time. LaTeX: 3y^2+y-2=0\quad\Longrightarrow\quad y=\frac{2}{3} LaTeX: m=-\frac{\log(2/3)}{0.4}=\frac{\log(3/2)}{0.4}=1.0136627703\ldots ### Answer The median session completes in approximately 1.014 minutes. LaTeX: \boxed{m\approx 1.014\quad\text{(C)}} ### Verification record - Main value: 1.0136627702704109 - Main method: Solve the route-weighted survival equation by a quadratic substitution. - Independent check: Substitution into 1-[0.25 exp(-0.4m)+0.75 exp(-0.8m)] gives the CDF at the proposed median. - Independent value: 0.5 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #678 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-678/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Uniform Distribution; Expectation and Variance This is a uniform-distribution probability centered at the mean. One standard deviation spans a total interval of length 30/sqrt(3), so the requested probability is 1/sqrt(3)=0.577350, which rounds to choice C. ### Independent solution ### Setup For a uniform variable on an interval with endpoints a and b, write its mean and standard deviation. LaTeX: \mu=\frac{a+b}{2}=80 LaTeX: \sigma=\frac{b-a}{\sqrt{12}}=\frac{30}{\sqrt{12}}=5\sqrt{3} ### Model The interval one standard deviation from the mean remains completely inside the distribution's support. Uniform probability is interval length divided by total support length. LaTeX: \Pr(|X-\mu|\le \sigma)=\frac{(\mu+\sigma)-(\mu-\sigma)}{b-a} LaTeX: \Pr(|X-\mu|\le \sigma)=\frac{2\sigma}{30} ### Compute Substitute the uniform standard deviation and simplify. LaTeX: \frac{2(30/\sqrt{12})}{30}=\frac{2}{\sqrt{12}}=\frac{1}{\sqrt{3}} LaTeX: \frac{1}{\sqrt{3}}=0.5773502692\ldots ### Answer The probability rounds to 0.58 at the precision of the listed answers. LaTeX: \boxed{\Pr(|X-\mu|\le \sigma)\approx 0.58\quad\text{(C)}} ### Distractor analysis - A: The value 1/sqrt(12)=0.288675 counts the interval from the mean to only one of the two standard-deviation boundaries. - B: This is close to the complement 1-1/sqrt(3)=0.422650, so it reverses 'within' and 'outside' the centered interval. - D: The familiar 0.68 figure is the normal-distribution probability within one standard deviation; the distribution here is uniform. - E: The value 0.788675 is the one-sided probability above the lower boundary mu-sigma and fails to impose the upper boundary. ### Verification record - Main value: 0.5773502691896258 - Main method: Compute the uniform standard deviation and divide the centered interval's length 2 sigma by the total support length. - Independent check: Integrate the constant density 1/30 directly from 80-5sqrt(3) to 80+5sqrt(3). - Independent value: 0.5773502691896258 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: calibration offsets The calibration offset of a field sensor is uniformly distributed between -4 and 8 degrees. Calculate the probability that the offset is within 0.75 standard deviations of its mean. - A: 0.217 - B: 0.375 - C: 0.547 - D: 0.433 - E: 0.750 For a uniform variable, the centered interval has length twice 0.75 standard deviations. Dividing that length by the 12-degree support gives 0.433013, so choice D is correct. #### Variant solution ### Setup Compute the mean and standard deviation from the endpoints of the uniform support. LaTeX: \mu=\frac{-4+8}{2}=2 LaTeX: \sigma=\frac{8-(-4)}{\sqrt{12}}=\sqrt{12} ### Model A radius of 0.75 standard deviations stays inside the support, so probability is the centered interval length divided by 12. LaTeX: \Pr(|X-2|\le 0.75\sigma)=\frac{2(0.75)\sigma}{12} ### Compute Insert the standard deviation and simplify the length ratio. LaTeX: \frac{1.5\sqrt{12}}{12}=\frac{0.75}{\sqrt{3}} LaTeX: \frac{0.75}{\sqrt{3}}=0.4330127019\ldots ### Answer The required probability is approximately 0.433. LaTeX: \boxed{0.433013\approx 0.433\quad\text{(D)}} ### Verification record - Main value: 0.43301270189221935 - Main method: Divide the width of the centered 0.75-standard-deviation interval by the uniform support width. - Independent check: Direct integration of density 1/12 over [2-0.75sqrt(12), 2+0.75sqrt(12)]. - Independent value: 0.43301270189221935 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #679 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-679/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Normal Distribution; Deductibles and Payments This is a normal-tail inversion after translating a payment threshold back to a loss threshold. The 86th loss percentile is 633.032, so subtracting the 500 payment threshold gives a deductible of 133.032 and choice E. ### Independent solution ### Setup Let X denote the loss and d the deductible. A positive payment on one loss is the excess of X above d. LaTeX: X\sim N(525,100^2),\qquad Y=(X-d)_+ ### Model A payment greater than 500 occurs exactly when the underlying loss exceeds d+500. Convert the upper-tail probability to a cumulative probability. LaTeX: \Pr(Y>500)=\Pr(X>d+500)=0.14 LaTeX: \Pr(X\le d+500)=0.86 ### Compute Standardize the 86th percentile and solve the resulting linear equation for the deductible. LaTeX: \frac{d+500-525}{100}=\Phi^{-1}(0.86)=1.0803193408\ldots LaTeX: d=525+100(1.0803193408\ldots)-500=133.0319341\ldots ### Answer The deductible rounds to 133 in the units of the listed answers. LaTeX: \boxed{d\approx 133\quad\text{(E)}} ### Distractor analysis - A: A value near 36 can result from reading 0.14 as an area from the mean, obtaining z near 0.36, and reporting only the standard-deviation displacement rather than the deductible. - B: The value 39 follows from treating the probability 0.14 itself as a z-score: 525+100(0.14)-500=39. - C: Using z near 0.36 inside the otherwise correct threshold equation produces about 61; the required upper-tail quantile is z=1.0803, not 0.36. - D: Using the lower-tail percentile with z=-1.0803 produces a negative deductible near -83; taking its magnitude loses the direction of the tail. ### Verification record - Main value: 133.0319340814956 - Main method: Translate the payment exceedance to X>d+500 and invert the normal CDF at cumulative probability 0.86. - Independent check: Substitution of the computed deductible into the standardized normal survival probability P[Z>(d-25)/100]. - Independent value: 0.14000000000000007 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: equipment-service reimbursements A service program models each equipment repair cost as normal with mean 850 dollars and standard deviation 120 dollars. A deductible d applies to each repair. Only 8% of repairs generate a reimbursement greater than 300 dollars. Calculate d. - A: 169 - B: 550 - C: 681 - D: 719 - E: 1019 A reimbursement above 300 requires a repair cost above d+300. The 92nd percentile of the repair-cost distribution is 1018.609, leaving a deductible of 718.609 and choice D. #### Variant solution ### Setup Represent the repair cost by X and the reimbursement by the positive part after the deductible. LaTeX: X\sim N(850,120^2),\qquad R=(X-d)_+ ### Model Translate the reimbursement event into a repair-cost event and express the upper tail as a percentile. LaTeX: \Pr(R>300)=\Pr(X>d+300)=0.08 LaTeX: \Pr(X\le d+300)=0.92 ### Compute Use the 92nd standard-normal percentile and isolate d. LaTeX: \Phi^{-1}(0.92)=1.4050715603\ldots LaTeX: d=850+120(1.4050715603\ldots)-300=718.6085872\ldots ### Answer The deductible is approximately 719 dollars. LaTeX: \boxed{d\approx 719\quad\text{(D)}} ### Verification record - Main value: 718.6085872371559 - Main method: Invert the normal CDF at 0.92 and subtract the 300-dollar reimbursement threshold from the resulting repair-cost percentile. - Independent check: Evaluate the normal survival probability at the standardized threshold (d+300-850)/120. - Independent value: 0.0799999999999999 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #680 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-680/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Percentiles This is a shifted-exponential percentile calculation. Setting the CDF to 0.10 gives a standardized excess of log(10/9), so the percentile is d+beta log(10/9), which is choice D. ### Independent solution ### Setup Express the supported variable as its lower endpoint plus an exponential excess with scale beta. LaTeX: X=d+Y,\qquad Y\sim\operatorname{Exp}(\text{scale }\beta) LaTeX: F_X(x)=1-e^{-(x-d)/\beta},\qquad x\ge d ### Model Let q denote the 10th percentile and set the shifted-exponential CDF at q equal to 0.10. LaTeX: 1-e^{-(q-d)/\beta}=0.10 LaTeX: e^{-(q-d)/\beta}=0.90 ### Compute Take logarithms and solve for the percentile on the original scale. LaTeX: \frac{q-d}{\beta}=-\log(0.90)=\log\!\left(\frac{10}{9}\right) LaTeX: q=d+\beta\log\!\left(\frac{10}{9}\right) ### Answer The fixed shift must remain in the percentile, and beta multiplies the dimensionless logarithm. LaTeX: \boxed{q_{0.10}=d+\beta\log(10/9)\quad\text{(D)}} ### Distractor analysis - A: This expression omits the location shift d and also uses log(11/10), which does not come from the required survival probability 0.90. - B: The logarithmic excess beta log(10/9) is correct, but it is measured above the lower endpoint; omitting d leaves the answer on the wrong origin. - C: Adding d fixes the origin, but log(11/10) treats the 10% probability through an incorrect odds-style ratio instead of using -log(0.90). - E: Placing beta in a logarithmic power or denominator does not invert (q-d)/beta and also fails to retain the location shift. ### Verification record - Main value: 0.10536051565782635 - Main method: Invert the shifted-exponential CDF at probability 0.10; the reported numeric artifact is the standardized excess (q-d)/beta. - Independent check: Substitute (q-d)/beta=log(10/9) back into 1-exp[-(q-d)/beta]. - Independent value: 0.10000000000000009 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: sensor recalibration latency A remote sensor's recalibration latency equals a fixed 3-second handshake plus an independent exponential processing time with mean 8 seconds. Calculate the 75th percentile of the total latency. - A: 2.301 - B: 5.301 - C: 11.090 - D: 14.090 - E: 35.000 At the 75th percentile, the exponential processing time has survival probability 0.25. Adding the fixed handshake to 8 log(4) gives 14.090355 seconds, so choice D is correct. #### Variant solution ### Setup Write total latency T as the fixed handshake plus exponential processing time Y. LaTeX: T=3+Y,\qquad Y\sim\operatorname{Exp}(\text{scale }8) ### Model For the 75th percentile q, the remaining upper-tail probability is one-quarter. LaTeX: \Pr(T>q)=e^{-(q-3)/8}=0.25 ### Compute Take logarithms, scale by the exponential mean, and restore the fixed delay. LaTeX: \frac{q-3}{8}=-\log(0.25)=\log 4 LaTeX: q=3+8\log 4=14.0903548890\ldots ### Answer The 75th percentile of total latency is approximately 14.090 seconds. LaTeX: \boxed{q_{0.75}\approx 14.090\quad\text{(D)}} ### Verification record - Main value: 14.090354888959125 - Main method: Invert the exponential survival function at 0.25 and add the fixed three-second latency. - Independent check: Substitute the computed value into the shifted CDF 1-exp[-(q-3)/8]. - Independent value: 0.75 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #681 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-681/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Expectation and Variance; Sums of Independent Random Variables This is a variance-of-an-independent-sum calculation for a three-point count distribution. One individual count has variance 0.1456, so the 64-count total has variance 9.3184 and matches choice D. ### Independent solution ### Setup Let X be one individual's count. Compute its first and second moments directly from its three possible values. LaTeX: \operatorname{E}[X]=0(0.90)+1(0.08)+2(0.02)=0.12 LaTeX: \operatorname{E}[X^2]=0^2(0.90)+1^2(0.08)+2^2(0.02)=0.16 ### Model Convert the two moments to an individual variance. Independence then makes the variance of the group total the sum of 64 equal variances. LaTeX: \operatorname{Var}(X)=\operatorname{E}[X^2]-\operatorname{E}[X]^2 LaTeX: \operatorname{Var}\!\left(\sum_{i=1}^{64}X_i\right)=\sum_{i=1}^{64}\operatorname{Var}(X_i) ### Compute Evaluate the individual variance and multiply by the group size. LaTeX: \operatorname{Var}(X)=0.16-(0.12)^2=0.1456 LaTeX: \operatorname{Var}\!\left(\sum_{i=1}^{64}X_i\right)=64(0.1456)=9.3184 ### Answer The total variance rounds to 9.32. LaTeX: \boxed{9.3184\approx 9.32\quad\text{(D)}} ### Distractor analysis - A: The value 0.1456 is the variance for one individual count; it omits the factor of 64 for the independent group total. - B: The value 2.56 equals 64 times 2(0.02), the expected count contributed by the two-event outcome alone, not a variance. - C: The value 7.68 equals 64E[X], so it is the expected total count rather than its variance. - E: The value 10.24 equals 64E[X^2]; it omits the subtraction of 64(E[X])^2 needed when summing the individual variances. ### Verification record - Main value: 9.3184 - Main method: Compute E[X] and E[X^2] for one count, form Var(X), and add 64 independent identical variances. - Independent check: Use Var(total)=E[sum Xi^2]+2 sum E[Xi]E[Xj]-(64E[X])^2 and simplify with independence. - Independent value: 9.3184 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: network gateway alerts During a maintenance window, each of 50 independent network gateways emits no alerts with probability 0.82, one alert with probability 0.14, and three alerts with probability 0.04. Calculate the variance of the total number of alerts from all gateways. - A: 0.43 - B: 9.62 - C: 21.62 - D: 13.00 - E: 25.00 A gateway's alert count has mean 0.26 and second moment 0.50, giving variance 0.4324. Adding 50 independent gateway variances gives 21.62, which selects choice C. #### Variant solution ### Setup Let G be the alert count from one gateway and calculate its first two moments. LaTeX: \operatorname{E}[G]=0(0.82)+1(0.14)+3(0.04)=0.26 LaTeX: \operatorname{E}[G^2]=0^2(0.82)+1^2(0.14)+3^2(0.04)=0.50 ### Model Find the gateway-level variance, then use independence to add variances across all 50 gateways. LaTeX: \operatorname{Var}(G)=0.50-(0.26)^2 LaTeX: \operatorname{Var}\!\left(\sum_{i=1}^{50}G_i\right)=50\operatorname{Var}(G) ### Compute Evaluate the two variance calculations. LaTeX: \operatorname{Var}(G)=0.4324 LaTeX: 50(0.4324)=21.62 ### Answer The variance of the total alert count is 21.62. LaTeX: \boxed{21.62\quad\text{(C)}} ### Verification record - Main value: 21.62 - Main method: Compute the one-gateway variance from its first two moments and sum 50 independent variances. - Independent check: Expand E[(sum Gi)^2] into diagonal second moments and off-diagonal products, then subtract E[sum Gi]^2. - Independent value: 21.62 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #682 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-682/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Conditional Probability; Binomial Distribution Condition on the pooled alert by dividing the probability of exactly two affected subjects by the probability of at least one affected subject. The ratio is 0.00999966, so choice C is correct. ### Independent solution ### Setup Let X count affected subjects among the three independent trials. The common affected probability is 0.01. LaTeX: X\sim\operatorname{Bin}(3,0.01) ### Model The observed pooled alert is equivalent to X being positive. Because the event X=2 is contained in that alert event, conditional probability reduces to a ratio. LaTeX: \Pr(X=2\mid X\ge 1)=\frac{\Pr(X=2)}{\Pr(X\ge 1)} ### Compute Use the binomial mass in the numerator and a complement in the denominator. LaTeX: \Pr(X=2)=\binom{3}{2}(0.01)^2(0.99)=0.000297 LaTeX: \Pr(X\ge 1)=1-(0.99)^3=0.029701 LaTeX: \frac{0.000297}{0.029701}=0.0099996633\ldots ### Answer The requested conditional probability rounds to 0.0100. LaTeX: \boxed{0.0100\quad\text{(C)}} ### Distractor analysis - A: The value 0.000297 is the unconditional probability of exactly two affected subjects; it omits division by the observed-alert probability. - B: The value is obtained by retaining only one specified arrangement of two affected and one unaffected subject, thereby omitting the binomial factor of three. - D: The value 2(0.01)(0.99)=0.0198 is an exactly-one calculation for only two trials, so it uses neither the three-subject model nor the conditioning event. - E: The value 1-(0.99)^3=0.029701 is the probability of the pooled alert itself, which is the denominator rather than the conditional result. ### Verification record - Main value: 0.009999663310999638 - Main method: Evaluate the conditional binomial ratio P(X=2)/P(X>=1) with n=3 and p=0.01. - Independent check: Enumerate the eight independent affected/unaffected patterns, total the three two-affected patterns, and normalize over the seven nonempty pattern types. - Independent value: 0.009999663310999638 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: mountain relay incident A monitoring dashboard watches four independently operated mountain relays. Each relay has a 4% chance of missing its midnight heartbeat. The dashboard displays an incident whenever one or more heartbeats are absent. Given that an incident is displayed, calculate the probability that exactly two relays missed the heartbeat. - A: 0.00885 - B: 0.00979 - C: 0.05873 - D: 0.06037 - E: 0.15065 The exactly-two probability is 0.00884736 and the incident probability is 0.15065344. Their ratio is 0.0587266, which selects choice C. #### Variant solution ### Setup Let M denote the number of relays that miss the heartbeat. LaTeX: M\sim\operatorname{Bin}(4,0.04) ### Model An incident means M is at least one. Exactly two misses automatically imply that an incident occurred. LaTeX: \Pr(M=2\mid M\ge 1)=\frac{\Pr(M=2)}{1-\Pr(M=0)} ### Compute Evaluate the two binomial quantities and form the conditional ratio. LaTeX: \Pr(M=2)=\binom{4}{2}(0.04)^2(0.96)^2=0.00884736 LaTeX: \Pr(M\ge 1)=1-(0.96)^4=0.15065344 LaTeX: \frac{0.00884736}{0.15065344}=0.0587265714\ldots ### Answer The conditional probability is approximately 0.05873. LaTeX: \boxed{0.05873\quad\text{(C)}} ### Verification record - Main value: 0.05872657139458609 - Main method: Divide the binomial probability of two misses by the complement probability of no misses. - Independent check: Sum the probabilities for all nonempty relay subsets and independently isolate the six subsets having cardinality two. - Independent value: 0.05872657139458609 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #683 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-683/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Conditional Probability; Inclusion–Exclusion The given no-theft probabilities determine a four-cell table: the probability that both households have a theft is 0.10, while the conditioning household has a theft with probability 0.20. Their ratio is 0.50, choice D. ### Independent solution ### Setup Write T_A and T_B for the events that the respective households experience at least one theft. Each complement has probability 0.80. LaTeX: \Pr(T_A)=\Pr(T_B)=0.20 LaTeX: \Pr(T_A^c\cap T_B^c)=0.70 ### Model Recover the probability of both theft events through inclusion–exclusion applied to their complements. LaTeX: \Pr(T_A\cap T_B)=1-\Pr(T_A^c)-\Pr(T_B^c)+\Pr(T_A^c\cap T_B^c) ### Compute Substitute the three supplied probabilities, then divide by the probability of the conditioning event. LaTeX: \Pr(T_A\cap T_B)=1-0.80-0.80+0.70=0.10 LaTeX: \Pr(T_A\mid T_B)=\frac{0.10}{0.20}=0.50 ### Answer Given a theft at household B, the conditional probability of a theft at household A is one-half. LaTeX: \boxed{0.500\quad\text{(D)}} ### Distractor analysis - A: The value 0.20(0.30)=0.060 treats the chance of at least one theft as though it were B's theft probability and also imposes unsupported independence. - B: The value 0.100 is the joint probability of thefts at both households; it is the numerator before conditioning on B's theft. - C: The value 0.200 is A's marginal theft probability and ignores the information about what happened at household B. - E: The ratio 0.70/0.80=0.875 is a conditional no-theft probability, Pr(T_B^c given T_A^c), for the complementary events. ### Verification record - Main value: 0.5 - Main method: Use inclusion–exclusion to recover the joint theft probability, then divide by B's marginal theft probability. - Independent check: Construct the four-cell table with probabilities 0.70, 0.10, 0.10, and 0.10 and read the conditional ratio from B's theft column. - Independent value: 0.5 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: alpine telemetry links A control center receives telemetry through a north link and a south link. Each link is available at the audit time with probability 0.85, and both are available with probability 0.75. If the south link is unavailable, calculate the probability that the north link is also unavailable. - A: 0.050 - B: 0.150 - C: 0.250 - D: 0.333 - E: 0.882 Inclusion–exclusion gives a 0.05 probability that both links are unavailable. The south link is unavailable with probability 0.15, so the conditional probability is 0.05/0.15=1/3, choice D. #### Variant solution ### Setup Let U_N and U_S denote unavailability of the north and south links. LaTeX: \Pr(U_N)=\Pr(U_S)=0.15 LaTeX: \Pr(U_N^c\cap U_S^c)=0.75 ### Model Use the availability information to find the lower-right cell of the two-link status table. LaTeX: \Pr(U_N\cap U_S)=1-0.85-0.85+0.75 ### Compute Normalize the both-unavailable cell by the south-unavailable marginal probability. LaTeX: \Pr(U_N\cap U_S)=0.05 LaTeX: \Pr(U_N\mid U_S)=\frac{0.05}{0.15}=\frac13 ### Answer The required conditional probability is approximately 0.333. LaTeX: \boxed{0.333\quad\text{(D)}} ### Verification record - Main value: 0.3333333333333333 - Main method: Recover the both-unavailable probability by inclusion–exclusion and divide by the south-unavailable marginal. - Independent check: Fill the availability table with cells 0.75, 0.10, 0.10, and 0.05, then calculate 0.05/(0.10+0.05). - Independent value: 0.3333333333333333 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #684 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-684/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Transformations of Random Variables; Deductible Payments; Variance For the payment Y=(X-1000) positive part, direct integration gives E[Y]=6533.33 and E[Y^2]=60,977,777.78. Thus SD(Y)=4277.07, which corresponds to choice B. ### Independent solution ### Setup Let X be the modeled loss and Y the insurer payment after applying the ordinary deductible. LaTeX: X\sim\operatorname{Unif}(0,15000) LaTeX: Y=(X-1000)_+=\max(X-1000,0) ### Model The payment has a point mass at zero, so calculate both raw moments from the original loss density rather than treating Y as an ordinary uniform variable. LaTeX: \mathbb{E}[Y^k]=\frac{1}{15000}\int_{1000}^{15000}(x-1000)^k\,dx,\qquad k=1,2 ### Compute Evaluate the two integrals and center the second moment. LaTeX: \mathbb{E}[Y]=\frac{14000^2}{2(15000)}=6533.3333\ldots LaTeX: \mathbb{E}[Y^2]=\frac{14000^3}{3(15000)}=60977777.7778\ldots LaTeX: \operatorname{Var}(Y)=\mathbb{E}[Y^2]-\mathbb{E}[Y]^2=18293333.3333\ldots LaTeX: \operatorname{SD}(Y)=\sqrt{18293333.3333\ldots}=4277.0706\ldots ### Answer Rounding to the precision of the choices gives a payment standard deviation of 4277. LaTeX: \boxed{4277\quad\text{(B)}} ### Distractor analysis - A: This nearby underestimate is inconsistent with the exact first and second payment moments; it reflects an excessive centering subtraction or a numerical integration error. - C: The value 15000 divided by sqrt(12), approximately 4330, is the standard deviation of the original uniform loss and ignores the deductible transformation. - D: This exceeds the original loss standard deviation even though the payment map x to (x-1000) positive part is 1-Lipschitz, so it fails a useful variance sanity check. - E: The value 6533 is the expected payment, not its standard deviation; it stops after computing the first raw moment. ### Verification record - Main value: 4277.070648625451 - Main method: Integrate the first two moments of the positive-part deductible payment and use Var(Y)=E[Y^2]-E[Y]^2. - Independent check: Represent the payment as zero with probability 1/15 and as Uniform(0,14000) with probability 14/15, then apply the law of total variance. - Independent value: 4277.070648625451 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: workshop repair reimbursement A makerspace reimburses member equipment repairs only for the portion above 2,000 dollars. For an accepted breakdown report, the total repair cost is modeled as uniform from 0 to 12,000 dollars. Calculate the standard deviation of the reimbursement. - A: 2,887 dollars - B: 3,227 dollars - C: 3,464 dollars - D: 4,167 dollars - E: 5,270 dollars The reimbursement has first moment 4166.67 and second moment 27,777,777.78. Its variance is 10,416,666.67, giving a standard deviation of 3227.49 dollars and choice B. #### Variant solution ### Setup Write C for repair cost and R for the reimbursement. LaTeX: C\sim\operatorname{Unif}(0,12000) LaTeX: R=(C-2000)_+ ### Model Integrate over the cost region in which the reimbursement is positive. LaTeX: \mathbb{E}[R^k]=\frac{1}{12000}\int_{2000}^{12000}(c-2000)^k\,dc,\qquad k=1,2 ### Compute Calculate both raw moments before taking the square root of the variance. LaTeX: \mathbb{E}[R]=\frac{10000^2}{2(12000)}=4166.6667\ldots LaTeX: \mathbb{E}[R^2]=\frac{10000^3}{3(12000)}=27777777.7778\ldots LaTeX: \operatorname{Var}(R)=10416666.6667\ldots LaTeX: \operatorname{SD}(R)=3227.4861\ldots ### Answer The reimbursement standard deviation rounds to 3,227 dollars. LaTeX: \boxed{3227\text{ dollars}\quad\text{(B)}} ### Verification record - Main value: 3227.4861218395135 - Main method: Integrate the first two moments of the repair reimbursement and center the second moment. - Independent check: Use a mixture with reimbursement zero with probability 1/6 and Uniform(0,10000) with probability 5/6, then apply total variance. - Independent value: 3227.4861218395135 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #685 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-685/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Conditional Probability; Independence The continuation event occurs unless both spouses die during the period, so its probability is 1-(0.15)(0.05)=0.9925. Dividing the husband's 0.85 survival probability by 0.9925 gives 0.856423, choice B. ### Independent solution ### Setup Let H and W be the events that the husband and wife, respectively, are alive after ten years. LaTeX: \Pr(H)=0.85,\qquad \Pr(W)=0.95 ### Model Payments continue on H union W. Independence makes the probability that both have died the product of their death probabilities. LaTeX: \Pr(H\cup W)=1-\Pr(H^c\cap W^c)=1-(0.15)(0.05) ### Compute Because H is contained in the continuation event, its joint probability with that event is simply Pr(H). LaTeX: \Pr(H\cup W)=0.9925 LaTeX: \Pr(H\mid H\cup W)=\frac{\Pr(H)}{\Pr(H\cup W)} LaTeX: \frac{0.85}{0.9925}=0.8564231738\ldots ### Answer Conditional on continued payments, the husband's survival probability is approximately 0.856. LaTeX: \boxed{0.856\quad\text{(B)}} ### Distractor analysis - A: The value 0.850 is the husband's unconditional survival probability and does not use the information that payments continue. - C: The ratio 0.95/0.9925=0.9572 instead calculates the wife's conditional survival probability under the same continuation event. - D: The value 0.9925 is the probability that payments continue, so it is the conditioning denominator rather than the requested conditional probability. - E: Continued payments establish only that at least one spouse is alive; treating this as proof that the husband is alive ignores the wife-only state. ### Verification record - Main value: 0.8564231738035264 - Main method: Compute the last-survivor continuation probability and normalize the husband's survival event within it. - Independent check: Enumerate the four independent survival states; the continuation states total 0.9925 and those containing the husband total 0.85. - Independent value: 0.8564231738035264 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: redundant coolant pumps A research freezer is kept cold by two redundant coolant pumps. Over the next service interval, pump A has failure probability 0.20 and pump B has failure probability 0.10, independently. At inspection the freezer is still being cooled, which means at least one pump is operating. Calculate the conditional probability that pump A is operating. - A: 0.800 - B: 0.816 - C: 0.900 - D: 0.918 - E: 0.980 Cooling continues with probability 1-(0.20)(0.10)=0.98. Pump A operates with probability 0.80 and that event guarantees cooling, so the conditional probability is 0.80/0.98=0.816327, choice B. #### Variant solution ### Setup Let A and B denote the events that the corresponding pumps operate through the service interval. LaTeX: \Pr(A)=0.80,\qquad \Pr(B)=0.90 ### Model Cooling is the union of the two operating events. Its complement requires both independent failures. LaTeX: \Pr(A\cup B)=1-(0.20)(0.10)=0.98 ### Compute Since an operating pump A necessarily implies cooling, normalize Pr(A) over the cooling event. LaTeX: \Pr(A\mid A\cup B)=\frac{0.80}{0.98}=0.8163265306\ldots ### Answer The conditional probability that pump A operates is approximately 0.816. LaTeX: \boxed{0.816\quad\text{(B)}} ### Verification record - Main value: 0.8163265306122449 - Main method: Divide pump A's operating probability by the probability that at least one independent pump operates. - Independent check: Enumerate states with probabilities 0.72, 0.08, 0.18, and 0.02; among the 0.98 cooling mass, 0.80 includes pump A. - Independent value: 0.8163265306122449 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #686 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-686/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Expected Value; Linearity of Expectation; Insurance Loss Modeling The expected unreduced loss is 30(0.01)+10(0.05)=0.80. A target profit of 0.25 from a premium of 1 leaves 0.75 for expected reimbursement, so the reimbursement share is 0.75/0.80=0.9375, choice E. ### Independent solution ### Setup Let F and L count the two covered event types, and let q be the constant share reimbursed. LaTeX: \mathbb{E}[F]=0.01,\qquad \mathbb{E}[L]=0.05 ### Model Linearity of expectation gives the expected gross loss without requiring any dependence assumption between the event counts. LaTeX: \mathbb{E}[30F+10L]=30\mathbb{E}[F]+10\mathbb{E}[L] ### Compute Set premium less expected reimbursement equal to the target profit and solve for q. LaTeX: \mathbb{E}[30F+10L]=30(0.01)+10(0.05)=0.80 LaTeX: 1-q(0.80)=0.25 LaTeX: q=\frac{0.75}{0.80}=0.9375 ### Answer The company should reimburse 93.75% of each covered loss. LaTeX: \boxed{q=0.9375\quad\text{(E)}} ### Distractor analysis - A: The value 0.20 is the expected profit if all of the 0.80 expected gross loss were reimbursed; it is a profit amount, not the reimbursement share. - B: The ratio 0.25/0.80=0.3125 divides target profit by expected loss, but q must fund the 0.75 reimbursement budget left after target profit. - C: The value 0.46875 results from halving the required reimbursement budget while using a gross-loss total that already includes both covered causes. - D: The value 0.80 is the expected gross loss per policyholder, expressed in premium units, rather than a proportion selected to meet the profit target. ### Verification record - Main value: 0.9375 - Main method: Use linearity to obtain expected gross loss and solve premium minus expected reimbursement equals target profit. - Independent check: At q=0.9375, expected reimbursement is 0.9375(0.80)=0.75, leaving exactly 1-0.75=0.25 expected profit. - Independent value: 0.9375 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: equipment protection cooperative An equipment protection cooperative collects an annual fee of 100 dollars per member. A member is expected to incur 0.30 sensor failures costing 80 dollars each and 0.04 power-surge failures costing 400 dollars each during the year. The cooperative reimburses the same fraction of every covered loss and wants expected surplus of 75 dollars per member, with no other expenses. Calculate that reimbursement fraction. - A: 0.250 - B: 0.375 - C: 0.400 - D: 0.625 - E: 0.750 Expected covered loss is 0.30(80)+0.04(400)=40 dollars. The surplus target leaves 25 dollars for reimbursement, so the common fraction is 25/40=0.625, choice D. #### Variant solution ### Setup Let q denote the common fraction of each loss paid by the cooperative. LaTeX: \text{fee}=100,\qquad \text{target surplus}=75 ### Model Add frequency times severity across the two failure sources to obtain expected covered loss. LaTeX: \mathbb{E}[L]=0.30(80)+0.04(400) ### Compute Equate fee less expected reimbursement to the target surplus. LaTeX: \mathbb{E}[L]=24+16=40 LaTeX: 100-40q=75 LaTeX: q=\frac{25}{40}=0.625 ### Answer The cooperative should reimburse 62.5% of every covered loss. LaTeX: \boxed{q=0.625\quad\text{(D)}} ### Verification record - Main value: 0.625 - Main method: Calculate expected loss from the two frequency-severity products and solve the expected-surplus equation. - Independent check: A 0.625 fraction produces expected reimbursement 0.625(40)=25 dollars and therefore expected surplus 100-25=75 dollars. - Independent value: 0.625 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #687 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-687/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Poisson Distribution; Normal Approximation; Sums of Independent Random Variables The aggregate count has mean and variance 5000. Standardizing 5100 under its normal approximation gives z=sqrt(2), whose upper-tail probability is 0.078650, so choice A is correct. ### Independent solution ### Setup Let T denote the aggregate count. Independent Poisson counts add to another Poisson count, with parameter equal to the sum of the component parameters. LaTeX: T\sim\operatorname{Poisson}(100\cdot 50)=\operatorname{Poisson}(5000) LaTeX: \operatorname{E}[T]=\operatorname{Var}(T)=5000 ### Model Because the aggregate mean is large, approximate T by a normal random variable with the same mean and variance. LaTeX: T\ \dot\sim\ N(5000,5000) LaTeX: Z=\frac{T-5000}{\sqrt{5000}} ### Compute Standardize the stated threshold and evaluate the standard normal upper tail. LaTeX: z=\frac{5100-5000}{\sqrt{5000}}=\sqrt{2}=1.4142136\ldots LaTeX: \Pr(T>5100)\approx 1-\Phi(\sqrt{2})=0.0786496\ldots ### Answer The normal approximation gives an upper-tail probability of about 0.08. LaTeX: \boxed{\Pr(T>5100)\approx 0.08\quad\text{(A)}} ### Distractor analysis - B: A probability of 0.23 would require a standardized gap near 0.74, but the actual gap is 100/sqrt(5000)=1.414. - C: This is approximately the probability between the mean and the threshold, 0.5-0.07865=0.42135, rather than the probability above the threshold. - D: Adding the 100 component standard deviations gives the incorrect aggregate standard deviation 100sqrt(50), producing an upper tail near 0.44; independent variances, not standard deviations, add. - E: Using the aggregate variance 5000 as though it were the standard deviation gives an upper tail near 0.49 and ignores the required square root. ### Verification record - Main value: 0.0786496035251426 - Main method: Sum the independent Poisson parameters, match the aggregate mean and variance with a normal distribution, and evaluate its upper tail at 5100. - Independent check: Applying a continuity correction at 5100.5 gives z=1.4212846 and upper tail 0.0776170, which independently rounds to the same listed choice. - Independent value: 0.07761701495572387 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: migration-night audio uploads A wildlife-monitoring network has 80 independent acoustic sensors. On a migration night, each sensor uploads a Poisson number of audio clips with mean 30. Using a normal approximation, calculate the probability that the network uploads more than 2490 clips in total. - A: 0.0331 - B: 0.4186 - C: 0.4669 - D: 0.5000 - E: 0.9669 The total is approximately normal with mean and variance 2400. Its standardized threshold is 1.8371, giving an upper tail of 0.033096 and choice A. #### Variant solution ### Setup Write U for the total number of uploaded clips and combine the independent Poisson parameters. LaTeX: U\sim\operatorname{Poisson}(80\cdot 30)=\operatorname{Poisson}(2400) LaTeX: \operatorname{E}[U]=\operatorname{Var}(U)=2400 ### Model Use a normal distribution with mean 2400 and standard deviation sqrt(2400) for the large aggregate count. LaTeX: U\ \dot\sim\ N(2400,2400) ### Compute Convert the excess over the mean to standard deviation units and evaluate the upper tail. LaTeX: z=\frac{2490-2400}{\sqrt{2400}}=1.8371173\ldots LaTeX: \Pr(U>2490)\approx 1-\Phi(1.8371173)=0.0330963\ldots ### Answer The probability of exceeding 2490 uploads is approximately 3.31 percent. LaTeX: \boxed{\Pr(U>2490)\approx0.0331\quad\text{(A)}} ### Verification record - Main value: 0.033096289861096716 - Main method: Aggregate the independent Poisson counts and evaluate the matched normal upper tail. - Independent check: A continuity correction at 2490.5 gives an upper tail of 0.0323501, confirming the same answer choice and scale. - Independent value: 0.032350133479706035 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ## SOA Exam P Sample Question #688 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-688/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Counting Methods; Sampling Without Replacement A qualifying four-item subset must have category counts 2, 1, and 1. Counting the three possible doubled categories gives 105 favorable subsets out of 210, so the probability is 1/2 and choice E. ### Independent solution ### Setup All four-item subsets of the ten available items are equally likely. LaTeX: N_{\mathrm{all}}=\binom{10}{4}=210 ### Model To include all three categories in a sample of four, exactly one category must contribute two items and each other category must contribute one. LaTeX: N_{\mathrm{good}}=\binom{5}{2}\binom{3}{1}\binom{2}{1}+\binom{5}{1}\binom{3}{2}\binom{2}{1}+\binom{5}{1}\binom{3}{1}\binom{2}{2} ### Compute Evaluate the three disjoint category-count cases and divide by the total number of subsets. LaTeX: N_{\mathrm{good}}=60+30+15=105 LaTeX: \Pr(\text{all three categories})=\frac{105}{210}=\frac12 ### Answer Exactly half of the equally likely four-item subsets contain every category. LaTeX: \boxed{\frac12\quad\text{(E)}} ### Distractor analysis - A: The value 3/100 is already smaller than the single valid 2-1-1 case with the first category doubled, whose probability is 60/210=2/7, so it severely undercounts favorable subsets. - B: The value 1/24 is also smaller than one complete valid category-count case and cannot include the three disjoint ways to choose the doubled category. - C: A probability of 1/4 would correspond to 52.5 of the 210 equally likely subsets, so it cannot arise from a valid subset count and omits substantial favorable mass. - D: The value 9/25=0.36 is obtained by treating the four selections as independent draws with replacement from category probabilities 0.5, 0.3, and 0.2; the actual sampling is without replacement. ### Verification record - Main value: 0.5 - Main method: Count all multivariate-hypergeometric allocations with category counts 2, 1, and 1 and divide by the total number of four-item subsets. - Independent check: Directly enumerate all 210 four-item index subsets; 105 contain at least one item from each of the three categories. - Independent value: 0.5 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: seed-vault viability panel A seed vault has 5 alpine accessions, 4 coastal accessions, and 3 desert accessions ready for a viability panel. A botanist chooses 5 distinct accessions uniformly at random. Calculate the probability that the panel contains exactly two desert accessions and at least one accession from each of the other two habitats. - A: 0.1136 - B: 0.1515 - C: 0.1953 - D: 0.2652 - E: 0.3182 After choosing two of the three desert accessions, the remaining three positions split between alpine and coastal as 1+2 or 2+1. These cases give 210 of 792 panels, or 0.2652 and choice D. #### Variant solution ### Setup Count the equally likely panels of five distinct accessions. LaTeX: N_{\mathrm{all}}=\binom{12}{5}=792 ### Model Choose two desert accessions. The remaining three selections must consist of one alpine and two coastal accessions or two alpine and one coastal accession. LaTeX: N_{\mathrm{good}}=\binom32\left[\binom51\binom42+\binom52\binom41\right] ### Compute Evaluate the two valid alpine-coastal splits and divide by the total panel count. LaTeX: N_{\mathrm{good}}=3(30+40)=210 LaTeX: \Pr(\text{required composition})=\frac{210}{792}=\frac{35}{132}=0.2651515\ldots ### Answer The required composition occurs with probability approximately 0.2652. LaTeX: \boxed{0.2652\quad\text{(D)}} ### Verification record - Main value: 0.26515151515151514 - Main method: Count the two admissible alpine-coastal splits after fixing exactly two desert accessions. - Independent check: Enumeration of all 792 five-accession subsets finds 210 with exactly two desert labels and both other habitat labels represented. - Independent value: 0.26515151515151514 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #689 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-689/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Conditional Distributions; Law of Total Expectation; Discrete Uniform Distribution Conditional on a first-task duration h, the second duration is uniform on the integers from 1 through h-1 and therefore has mean h/2. Averaging over the three equally likely h values gives 11/6, choice A. ### Independent solution ### Setup Let H denote the first duration and Y the second duration. The three possible values of H have equal probability. LaTeX: \Pr(H=2)=\Pr(H=4)=\Pr(H=5)=\frac13 ### Model For a fixed value h, Y is discrete uniform on 1 through h-1. Its conditional mean is the midpoint of the first and last values. LaTeX: \operatorname{E}[Y\mid H=h]=\frac{1+(h-1)}{2}=\frac h2 ### Compute Apply the law of total expectation to the three conditional means. LaTeX: \operatorname{E}[Y]=\frac13\left(\frac22+\frac42+\frac52\right) LaTeX: \operatorname{E}[Y]=\frac{1+2+2.5}{3}=\frac{11}{6} ### Answer The expected second duration is one and five-sixths hours. LaTeX: \boxed{\operatorname{E}[Y]=\frac{11}{6}=1\frac56\quad\text{(A)}} ### Distractor analysis - B: The value 2 is only the conditional mean when the first duration is 4; it ignores the equally likely first-duration values 2 and 5. - C: The value 17/8 results from unequal weighting of the conditional means 1, 2, and 2.5, even though the three first-duration cases are equally likely. - D: The value 2.5 is the conditional mean in the longest first-duration branch alone, not the average across all three branches. - E: The value 5 2/3 exceeds 4, the largest second duration that can occur, so it cannot be the expectation of the second duration. ### Verification record - Main value: 1.8333333333333333 - Main method: Compute the conditional mean of each discrete-uniform second duration and average the three values using the first-task probabilities. - Independent check: Expand the expectation directly as (1/3) times each branch average: 1, (1+2+3)/3, and (1+2+3+4)/4. - Independent value: 1.8333333333333333 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: nocturnal species detections A field recorder is placed with equal probability in one of three habitat classes whose species capacities are 4, 7, and 10. Conditional on capacity C=c, the number R of distinct nocturnal species detected is uniformly distributed over the integers 0, 1, ..., c. Calculate the expected value of R. - A: 2.333 - B: 3.000 - C: 3.500 - D: 7.000 - E: 10.500 For a capacity c, the uniform count from 0 through c has mean c/2. Averaging the conditional means for capacities 4, 7, and 10 gives 3.5, so choice C is correct. #### Variant solution ### Setup Represent the habitat capacity by C and the observed count by R. LaTeX: \Pr(C=4)=\Pr(C=7)=\Pr(C=10)=\frac13 ### Model The mean of a discrete uniform distribution on the integers from 0 through c is the midpoint c/2. LaTeX: \operatorname{E}[R\mid C=c]=\frac{0+c}{2}=\frac c2 ### Compute Average the three conditional means with their equal habitat probabilities. LaTeX: \operatorname{E}[R]=\frac13\left(\frac42+\frac72+\frac{10}{2}\right) LaTeX: \operatorname{E}[R]=\frac{2+3.5+5}{3}=3.5 ### Answer The recorder detects 3.5 distinct species on average. LaTeX: \boxed{\operatorname{E}[R]=3.500\quad\text{(C)}} ### Verification record - Main value: 3.5 - Main method: Apply total expectation to the midpoint of each conditional discrete-uniform distribution. - Independent check: Directly sum r times its joint probability over all three habitat branches and all integers from 0 through the branch capacity. - Independent value: 3.5 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #690 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-690/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Discrete Random Variables; Sums of Independent Random Variables; Probability Generating Functions A total of two can arise as 2+0+0 or as 1+1+0. Accounting for the arrangements of both disjoint patterns gives 0.120000+0.018375=0.138375, which rounds to choice E. ### Independent solution ### Setup Let X1, X2, and X3 be the independent annual counts for the three selected policies. LaTeX: \Pr(X_i=0)=0.8,\quad \Pr(X_i=1)=0.0875,\quad \Pr(X_i=2)=0.0625 ### Model The sum equals two only through one count of two and two zeros, or through two counts of one and one zero. LaTeX: \Pr(X_1+X_2+X_3=2)=3\Pr(2,0,0)+3\Pr(1,1,0) ### Compute Use independence within each pattern and include its three possible positions. LaTeX: 3(0.0625)(0.8)^2=0.120000 LaTeX: 3(0.0875)^2(0.8)=0.018375 LaTeX: \Pr(X_1+X_2+X_3=2)=0.138375 ### Answer The requested probability rounds to four decimals as 0.1384. LaTeX: \boxed{0.1384\quad\text{(E)}} ### Distractor analysis - A: This is the 1+1+0 contribution alone, 3(0.0875)^2(0.8)=0.018375, and omits the 2+0+0 pattern. - B: The value 0.0230 is approximately 3(0.0875)^2; it counts the positions of two one-claim policies but omits the required zero-claim probability. - C: This is about twice the preceding 1+1 contribution and reflects treating the two identical one-claim outcomes as ordered, while still failing to include the 2+0+0 pattern correctly. - D: This is the 2+0+0 contribution alone, 3(0.0625)(0.8)^2=0.1200, and omits the 1+1+0 pattern. ### Verification record - Main value: 0.138375 - Main method: Partition the total-count event into the disjoint patterns 2+0+0 and 1+1+0 and sum their multinomial arrangements. - Independent check: Extract the coefficient of z^2 from (0.8+0.0875z+0.0625z^2+0.05z^3)^3. - Independent value: 0.138375 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: community help-desk requests For a randomly chosen community center, the number of help-desk requests received before noon is 0, 1, 2, or 3 with probabilities 0.70, 0.20, 0.08, and 0.02. Counts from different centers are independent. Four centers are monitored. Calculate the probability that they receive exactly two requests in total before noon. - A: 0.1098 - B: 0.1176 - C: 0.1600 - D: 0.2274 - E: 0.2778 The only possible patterns are 2+0+0+0 and 1+1+0+0. Their probabilities are 0.10976 and 0.11760, so the total is 0.22736 and choice D. #### Variant solution ### Setup Let R1 through R4 be the independent request counts for the four centers. LaTeX: \Pr(R_i=0)=0.70,\quad\Pr(R_i=1)=0.20,\quad\Pr(R_i=2)=0.08 ### Model A total of two requests consists either of one center with two requests or two centers with one request each. LaTeX: \Pr\!\left(\sum_{i=1}^4R_i=2\right)=4(0.08)(0.70)^3+\binom42(0.20)^2(0.70)^2 ### Compute Evaluate and add the two disjoint pattern probabilities. LaTeX: 4(0.08)(0.70)^3=0.10976 LaTeX: 6(0.20)^2(0.70)^2=0.11760 LaTeX: 0.10976+0.11760=0.22736 ### Answer The probability of exactly two requests is about 0.2274. LaTeX: \boxed{0.2274\quad\text{(D)}} ### Verification record - Main value: 0.22736 - Main method: Sum the multinomial probabilities of the two disjoint count patterns with total two. - Independent check: The coefficient of z^2 in (0.70+0.20z+0.08z^2+0.02z^3)^4 is 0.22736. - Independent value: 0.22736 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #691 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-691/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Set Probability; Complements; Mutually Exclusive Events Complementing the two supplied percentages gives probabilities 0.60 for the three-year event and 0.30 for combined coverage. The availability restrictions make these events disjoint, so their union has probability 0.90 and choice E. ### Independent solution ### Setup Let T be the event of a three-year contract and B the event of combined parts-and-labor coverage. LaTeX: \Pr(T)=1-0.40=0.60 LaTeX: \Pr(B)=1-0.70=0.30 ### Model The availability table permits no three-year contract with combined coverage, so the two events have empty intersection. LaTeX: \Pr(T\cap B)=0 ### Compute Apply the addition rule for the union and use the structural zero for the intersection. LaTeX: \Pr(T\cup B)=\Pr(T)+\Pr(B)-\Pr(T\cap B) LaTeX: \Pr(T\cup B)=0.60+0.30-0=0.90 ### Answer Nine tenths of the contracts satisfy at least one of the two conditions. LaTeX: \boxed{0.90\quad\text{(E)}} ### Distractor analysis - A: This is the probability of the impossible intersection T cap B, but the question asks for the union rather than the intersection. - B: The value 0.18 is 0.60(0.30), obtained by assuming independence and calculating an intersection; the table instead establishes mutual exclusivity. - C: The value 0.28 is (1-0.60)(1-0.30), the probability of neither event under an unjustified independence assumption. - D: The value 0.72 is 0.60+0.30-0.60(0.30), which applies the union formula after incorrectly assuming that the two events are independent. ### Verification record - Main value: 0.9 - Main method: Take complements to obtain the two event probabilities, read their empty intersection from the availability structure, and apply the union formula. - Independent check: Partition all contracts into the disjoint events three-year, combined coverage, and neither; the first two probabilities sum directly to 0.60+0.30. - Independent value: 0.9 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: observatory instrument scheduling Every observation block at a research observatory uses an optical instrument, a radio instrument, or both. Sixty-five percent of blocks use exactly one instrument. Also, 52 percent occur during daylight or twilight. A safety rule prohibits blocks using both instruments from being scheduled at night. Calculate the probability that a randomly selected block is at night or uses both instruments. - A: 0.000 - B: 0.168 - C: 0.338 - D: 0.662 - E: 0.830 Night blocks have probability 0.48 and dual-instrument blocks have probability 0.35. The safety rule makes those events disjoint, so their union is 0.83 and choice E. #### Variant solution ### Setup Let N denote a night block and D a block using both instruments. Convert the supplied complementary percentages. LaTeX: \Pr(N)=1-0.52=0.48 LaTeX: \Pr(D)=1-0.65=0.35 ### Model The scheduling rule excludes every dual-instrument block from the night period. LaTeX: \Pr(N\cap D)=0 ### Compute Add the probabilities of the two disjoint events. LaTeX: \Pr(N\cup D)=0.48+0.35=0.83 ### Answer The requested union contains 83 percent of the scheduled blocks. LaTeX: \boxed{0.830\quad\text{(E)}} ### Verification record - Main value: 0.83 - Main method: Complement the two supplied event probabilities and add them because the scheduling rule makes the events disjoint. - Independent check: The night-only, dual-instrument-only, and neither categories form a partition with probabilities 0.48, 0.35, and 0.17. - Independent value: 0.83 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ## SOA Exam P Sample Question #692 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-692/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Central Limit Theorem; Normal Approximation The aggregate has approximate mean 6250 and standard deviation 200. The threshold is 2.5 standard deviations below the mean, so the upper-tail probability is 0.993790 and choice E. ### Independent solution ### Setup Let S be the sum of the 625 independent payments. Means add, while variances add for independent observations. LaTeX: \operatorname{E}[S]=625(10)=6250 LaTeX: \operatorname{Var}(S)=625(8^2)=40000 ### Model Apply the normal approximation to the aggregate, using the square root of the aggregate variance as its standard deviation. LaTeX: S\ \dot\sim\ N(6250,200^2) LaTeX: Z=\frac{S-6250}{200} ### Compute Standardize the payment threshold and convert the requested upper tail by symmetry of the standard normal distribution. LaTeX: z=\frac{5750-6250}{200}=-2.5 LaTeX: \Pr(S>5750)\approx\Pr(Z>-2.5)=\Phi(2.5)=0.9937903347\ldots ### Answer Because the threshold lies far below the aggregate mean, a probability close to one is also the necessary sanity check. LaTeX: \boxed{\Pr(S>5750)\approx0.9938\quad\text{(E)}} ### Distractor analysis - A: This is the opposite tail, \(\Pr(S\le 5750)\approx\Phi(-2.5)=0.0062\), rather than the probability of exceeding the threshold. - B: A probability below one-half contradicts the fact that 5750 is below the aggregate mean of 6250; it reflects a tail-direction or centering error. - C: This value is only slightly above one-half and corresponds to using a standardized distance near 0.10 instead of the correct distance 2.5. - D: This value corresponds to an upper tail at approximately \(z=-0.40\), understating the concentration created by summing 625 independent payments. ### Verification record - Main value: 0.9937903346742238 - Main method: Normal approximation to the sum with mean 6250 and standard deviation 200. - Independent check: The complementary lower tail is Phi(-2.5) = 0.0062096653, so subtracting it from one gives 0.9937903347. - Independent value: 0.9937903346742238 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: bike-share battery reimbursements A city bike-share program will process 400 independent battery-replacement reimbursements this quarter. Each reimbursement has mean 15 dollars and standard deviation 6 dollars. Using a normal approximation, calculate the probability that the total reimbursement amount exceeds 6240 dollars. - A: 0.0228 - B: 0.1587 - C: 0.5000 - D: 0.9772 - E: 0.9987 The total has approximate mean 6000 and standard deviation 120. The threshold has standardized value 2, giving an upper tail of 0.022750 and choice A. #### Variant solution ### Setup Write T for the total of the 400 independent reimbursements and aggregate the first two moments. LaTeX: \operatorname{E}[T]=400(15)=6000 LaTeX: \operatorname{Var}(T)=400(6^2)=14400 ### Model The normal approximation uses standard deviation 120 for the quarterly total. LaTeX: T\ \dot\sim\ N(6000,120^2) ### Compute Standardize 6240 and evaluate the standard normal upper tail. LaTeX: z=\frac{6240-6000}{120}=2 LaTeX: \Pr(T>6240)\approx1-\Phi(2)=0.0227501319\ldots ### Answer The probability that the quarter exceeds the stated reimbursement budget is about 2.28 percent. LaTeX: \boxed{0.0228\quad\text{(A)}} ### Verification record - Main value: 0.02275013194817921 - Main method: Normal upper tail after aggregating 400 independent reimbursement amounts. - Independent check: The probability of staying at or below 6240 is Phi(2) = 0.9772498681; its complement is 0.0227501319. - Independent value: 0.02275013194817921 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ## SOA Exam P Sample Question #693 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-693/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Memoryless Property Memorylessness converts the stated conditional survival into a 15-year survival probability of 0.40. Scaling the exponential survival to 50 years gives 0.047156, which rounds to choice B. ### Independent solution ### Setup Let T be the exponential future lifetime and write its survival function in terms of the unknown rate. LaTeX: \Pr(T\ge t)=e^{-\lambda t} ### Model The memoryless property says the conditional survival from year 10 to year 25 depends only on the additional 15 years. LaTeX: \Pr(T\ge25\mid T\ge10)=\Pr(T\ge15)=e^{-15\lambda}=0.40 ### Compute Express the 50-year survival as a power of the calibrated 15-year survival probability. LaTeX: e^{-50\lambda}=\left(e^{-15\lambda}\right)^{50/15} LaTeX: \Pr(T\ge50)=0.40^{10/3}=0.0471556032\ldots ### Answer The 50-year survival probability rounds to 0.05 at the precision of the choices. LaTeX: \boxed{\Pr(T\ge50)\approx0.05\quad\text{(B)}} ### Distractor analysis - A: An exponential survival probability is strictly positive at every finite time, so rounding this result all the way to zero is not justified. - C: This is approximately \(0.40^{2.5}\), which treats 50 years as 2.5 blocks of 20 years instead of using the calibrated 15-year block. - D: This is \(0.40^2\), so it accounts for only two 15-year survival blocks, or 30 years, rather than 50 years. - E: This value exceeds \(0.40^2=0.16\), even though 50 years is longer than two calibrated 15-year blocks; it therefore violates monotonicity of survival. ### Verification record - Main value: 0.04715560318259695 - Main method: Scale the calibrated 15-year exponential survival probability to 50 years. - Independent check: Solving lambda = -log(0.40)/15 and evaluating exp(-50 lambda) gives 0.0471556032. - Independent value: 0.04715560318259695 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: underground moisture sensor The operating lifetime M of an underground moisture sensor is exponential. Given that a sensor has operated for at least 6 months, the probability that it operates for at least 18 months is 0.55. Calculate the probability that a new sensor operates for at least 42 months. - A: 0.0503 - B: 0.0915 - C: 0.1234 - D: 0.1664 - E: 0.5500 The conditional statement calibrates a 12-month survival probability of 0.55. Forty-two months is 3.5 such intervals, so the survival probability is 0.123387 and choice C. #### Variant solution ### Setup Represent the exponential survival function by a monthly rate theta. LaTeX: \Pr(M\ge t)=e^{-\theta t} ### Model Memorylessness removes the six months already survived and leaves a 12-month future interval. LaTeX: \Pr(M\ge18\mid M\ge6)=\Pr(M\ge12)=e^{-12\theta}=0.55 ### Compute Write the 42-month survival in units of the calibrated 12-month interval. LaTeX: \Pr(M\ge42)=\left(e^{-12\theta}\right)^{42/12} LaTeX: 0.55^{3.5}=0.1233870023\ldots ### Answer A new sensor reaches 42 months with probability about 0.1234. LaTeX: \boxed{0.1234\quad\text{(C)}} ### Verification record - Main value: 0.12338700232905413 - Main method: Raise the calibrated 12-month survival probability to the 42/12 power. - Independent check: Solving theta = -log(0.55)/12 and evaluating exp(-42 theta) gives 0.1233870023. - Independent value: 0.12338700232905413 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #694 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-694/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Hypergeometric Distribution; Sampling Without Replacement At most two defective selections fails only when all three defectives are among the four selected items. That excluded event has probability 1/30, leaving 29/30 = 0.966667 and choice E. ### Independent solution ### Setup Let X count defectives in a simple random selection of four from a population containing three defectives and seven usable items. LaTeX: X\sim\operatorname{Hypergeometric}(N=10,K=3,n=4) ### Model Because only three defectives exist, the complement of at most two defectives is exactly the event that all three are selected. LaTeX: \Pr(X\le2)=1-\Pr(X=3) ### Compute Count selections containing all three defectives and one of the seven usable items, then divide by all four-item selections. LaTeX: \Pr(X=3)=\frac{\binom33\binom71}{\binom{10}4}=\frac7{210}=\frac1{30} LaTeX: \Pr(X\le2)=1-\frac1{30}=\frac{29}{30}=0.966666\ldots ### Answer The exact without-replacement probability rounds to 0.97. LaTeX: \boxed{\Pr(X\le2)\approx0.97\quad\text{(E)}} ### Distractor analysis - A: This repeats the population defective fraction \(3/10=0.30\) instead of calculating a probability for the four-item selection. - B: The excluded event has probability \(1/30=0.0333\), not 0.33; this choice reflects both taking the complement event and losing a decimal place. - C: Using the ratio \(2/3\) because two of three defectives are allowed ignores the random selection mechanism and is not a hypergeometric probability. - D: Treating the four draws as independent Bernoulli trials with defect probability 0.30 gives \(\Pr(X\le2)=0.9163\), about 0.92; sampling is actually without replacement. ### Verification record - Main value: 0.9666666666666667 - Main method: Complement the hypergeometric probability that all three defectives appear in the four-item selection. - Independent check: Directly summing the favorable counts for 0, 1, and 2 defectives gives (35 + 105 + 63)/210 = 203/210 = 29/30. - Independent value: 0.9666666666666667 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: firmware audit sample A warehouse tray contains 12 control modules, four of which require a firmware update. A technician selects five modules uniformly without replacement for an audit. Calculate the probability that at most three of the selected modules require the update. - A: 0.333 - B: 0.010 - C: 0.667 - D: 0.955 - E: 0.990 The only way to exceed three update-required modules is to select all four of them. That event has probability 1/99, so the requested probability is 98/99 = 0.989899 and choice E. #### Variant solution ### Setup Let Y be the number of update-required modules among five selected from four update-required and eight current modules. LaTeX: Y\sim\operatorname{Hypergeometric}(N=12,K=4,n=5) ### Model The complement of at most three is selecting all four update-required modules, together with one current module. LaTeX: \Pr(Y\le3)=1-\Pr(Y=4) ### Compute Evaluate the single complement count and subtract it from one. LaTeX: \Pr(Y=4)=\frac{\binom44\binom81}{\binom{12}5}=\frac8{792}=\frac1{99} LaTeX: \Pr(Y\le3)=\frac{98}{99}=0.989898\ldots ### Answer The audit contains at most three modules needing an update with probability approximately 0.990. LaTeX: \boxed{0.990\quad\text{(E)}} ### Verification record - Main value: 0.98989898989899 - Main method: Complement the exact hypergeometric probability of selecting all four update-required modules. - Independent check: Summing the favorable counts for 0 through 3 update-required modules gives (56 + 280 + 336 + 112)/792 = 784/792 = 98/99. - Independent value: 0.98989898989899 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ## SOA Exam P Sample Question #695 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-695/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Conditional Probability For a Poisson count, the second moment is lambda squared plus lambda, so the rate is 3. Conditioning the upper count event on at least one count gives 0.842813 and choice C. ### Independent solution ### Setup Let N be Poisson with rate lambda and recover that rate from its stated second moment. LaTeX: \operatorname{E}[N^2]=\operatorname{Var}(N)+\operatorname{E}[N]^2=\lambda+\lambda^2 LaTeX: \lambda^2+\lambda=12 ### Model The nonnegative solution is lambda equal to 3. The requested conditional event uses the probability of two or more counts divided by the probability of one or more. LaTeX: \lambda=3 LaTeX: \Pr(N\ge2\mid N\ge1)=\frac{\Pr(N\ge2)}{\Pr(N\ge1)} ### Compute Use the Poisson probabilities at zero and one to evaluate both the numerator and denominator. LaTeX: \Pr(N\ge2)=1-e^{-3}-3e^{-3}=1-4e^{-3} LaTeX: \Pr(N\ge1)=1-e^{-3} LaTeX: \frac{1-4e^{-3}}{1-e^{-3}}=0.8428129105\ldots ### Answer The conditional probability rounds to 0.84. LaTeX: \boxed{\Pr(N\ge2\mid N\ge1)\approx0.84\quad\text{(C)}} ### Distractor analysis - A: This is below the unconditioned value \(\Pr(N\ge2)=0.80085\), even though conditioning on the smaller space \(N\ge1\) must increase the probability. - B: This is the unconditioned numerator \(\Pr(N\ge2)=1-4e^{-3}=0.80085\); it omits division by \(\Pr(N\ge1)\). - D: Using \(\lambda=\sqrt{12}\) treats the second moment as only \(\lambda^2\); that wrong rate gives a conditional probability about 0.888. - E: Solving the wrong equation \(\lambda^2-\lambda=12\) gives \(\lambda=4\), whose conditional probability is about 0.925 rather than the value for the correct rate. ### Verification record - Main value: 0.8428129105262321 - Main method: Recover the Poisson rate from its second moment and evaluate the conditional tail ratio. - Independent check: Within the event N >= 1, the complement is N = 1; computing 1 - 3 exp(-3)/(1 - exp(-3)) gives 0.8428129105. - Independent value: 0.8428129105262321 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: overnight support queue The number Q of urgent support tickets arriving during an overnight shift is Poisson. Its second moment is 20. Calculate the probability that at least two tickets arrive, given that at least one ticket arrives. - A: 0.800 - B: 0.908 - C: 0.925 - D: 0.948 - E: 0.982 The second-moment identity gives a Poisson rate of 4. Dividing the probability of at least two tickets by the probability of at least one gives 0.925371 and choice C. #### Variant solution ### Setup Use the Poisson second-moment formula to identify the overnight rate. LaTeX: \lambda^2+\lambda=20 LaTeX: \lambda=4 ### Model The event of two or more tickets is contained in the conditioning event of one or more tickets. LaTeX: \Pr(Q\ge2\mid Q\ge1)=\frac{1-\Pr(Q=0)-\Pr(Q=1)}{1-\Pr(Q=0)} ### Compute Insert the zero-count and one-count probabilities at rate 4. LaTeX: \Pr(Q\ge2\mid Q\ge1)=\frac{1-5e^{-4}}{1-e^{-4}} LaTeX: \frac{1-5e^{-4}}{1-e^{-4}}=0.9253705585\ldots ### Answer Given that the queue is nonempty, the chance that it contains at least two tickets is about 0.925. LaTeX: \boxed{0.925\quad\text{(C)}} ### Verification record - Main value: 0.9253705585449039 - Main method: Poisson conditional tail ratio after solving the second-moment equation for rate 4. - Independent check: Compute the complement inside the nonempty queue: 1 - 4 exp(-4)/(1 - exp(-4)) = 0.9253705585. - Independent value: 0.9253705585449039 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #696 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-696/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Normal Distribution; Percentiles and Interquartile Range; Independence Independence makes the one-year positive-profit probability the square root of 0.9025, or 0.95. This determines the standard deviation, and the normal interquartile width is 205.03, selecting choice B. ### Independent solution ### Setup Let X be one annual profit with mean 250 and standard deviation sigma. Use independence to recover the one-year probability of a positive profit. LaTeX: X\sim N(250,\sigma^2) LaTeX: \Pr(X_1>0,X_2>0)=\Pr(X>0)^2=0.9025 ### Model The one-year probability is 0.95. Standardizing zero shows that 250 divided by sigma is the 95th standard normal percentile. LaTeX: \Pr(X>0)=\sqrt{0.9025}=0.95 LaTeX: \Phi\!\left(\frac{250}{\sigma}\right)=0.95 ### Compute Solve for sigma and multiply it by the distance between the standard normal 75th and 25th percentiles. LaTeX: \sigma=\frac{250}{z_{0.95}}=\frac{250}{1.644853627}=151.98920798\ldots LaTeX: \operatorname{IQR}=(z_{0.75}-z_{0.25})\sigma LaTeX: \operatorname{IQR}=2(0.6744897502)(151.98920798)=205.0303258\ldots ### Answer The normal profit distribution has an interquartile range of about 205. LaTeX: \boxed{\operatorname{IQR}\approx205\quad\text{(B)}} ### Distractor analysis - A: This is approximately the recovered standard deviation 151.99; the interquartile range is \(1.34898\sigma\), not \(\sigma\). - C: Treating 0.9025 as the one-year positive-profit probability, instead of taking its square root, gives an IQR about 260. - D: This uses a standard-normal half-width near 0.915 rather than the quartile value 0.67449, so it measures a wider central interval. - E: This similarly substitutes a half-width near 1.21 for the normal quartile, producing a much wider interval than the middle 50 percent. ### Verification record - Main value: 205.0303258430853 - Main method: Recover the normal standard deviation from the independent two-year positive probability and apply the normal quartile constants. - Independent check: The recovered quartiles are 250 - 0.6744897502 sigma and 250 + 0.6744897502 sigma; subtracting them gives 205.0303258. - Independent value: 205.0303258430853 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: cleanroom inspection score A cleanroom's hourly particle-control score is normally distributed with mean 72. Scores recorded during independently selected inspection hours have the same unknown standard deviation. The probability that both of two selected scores are below 90 is 0.9216. Calculate the interquartile range of one hourly score. - A: 6.9 - B: 10.3 - C: 13.9 - D: 17.1 - E: 20.6 One selected score is below 90 with probability 0.96. This gives a standard deviation of 10.282, and multiplying by the normal interquartile factor yields 13.870, so choice C. #### Variant solution ### Setup Let C be one inspection score with standard deviation tau and recover its one-score probability from independence. LaTeX: C\sim N(72,\tau^2) LaTeX: \Pr(C<90)=\sqrt{0.9216}=0.96 ### Model Standardizing the score threshold identifies the 18-point distance from the mean as the 96th standard normal percentile times tau. LaTeX: \Phi\!\left(\frac{90-72}{\tau}\right)=0.96 LaTeX: \tau=\frac{18}{z_{0.96}} ### Compute Evaluate the standard deviation, then apply the symmetric normal quartile spacing. LaTeX: \tau=\frac{18}{1.750686071}=10.28168345\ldots LaTeX: \operatorname{IQR}=2z_{0.75}\tau LaTeX: \operatorname{IQR}=2(0.6744897502)(10.28168345)=13.86978020\ldots ### Answer The middle half of hourly inspection scores spans about 13.9 points. LaTeX: \boxed{13.9\quad\text{(C)}} ### Verification record - Main value: 13.8697801997657 - Main method: Recover the normal score scale from the independent two-inspection event and apply the normal interquartile factor. - Independent check: The 25th and 75th percentiles are 65.06510990 and 78.93489010; their difference is 13.86978020. - Independent value: 13.8697801997657 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #697 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-697/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Independence The equipment remains operational only when both independent exponential components exceed the five-year threshold. Multiplying their survival probabilities gives 0.054114, which rounds to choice B. ### Independent solution ### Setup Let P and S be the two component lifetimes. Convert their respective means of 3 and 4 years to exponential rates. LaTeX: P\sim\operatorname{Exp}(1/3),\qquad S\sim\operatorname{Exp}(1/4) LaTeX: \Pr(P>t)=e^{-t/3},\qquad \Pr(S>t)=e^{-t/4} ### Model A series system survives the target duration precisely when both components survive it. Independence turns the joint survival probability into a product. LaTeX: \Pr(\min(P,S)\ge 5)=\Pr(P\ge 5)\Pr(S\ge 5) ### Compute Insert the two exponential survival probabilities and combine their exponents. LaTeX: \Pr(\min(P,S)\ge 5)=e^{-5/3}e^{-5/4} LaTeX: e^{-5/3-5/4}=e^{-35/12}=0.0541137662\ldots ### Answer The system survival probability is about 0.05 at the precision of the listed answers. LaTeX: \boxed{\Pr(\min(P,S)\ge 5)\approx 0.05\quad\text{(B)}} ### Distractor analysis - A: A continuous lifetime has probability zero of equaling one exact time, but the requested event is survival through a threshold, not equality at that instant. - C: The value e^{-5/3}=0.188876 uses only the component with mean 3 and omits the second required component. - D: The value e^{-5/4}=0.286505 uses only the component with mean 4 and omits the first required component. - E: The value (1-e^{-5/3})(1-e^{-5/4})=0.578733 is the probability that both components have failed by year 5, the opposite joint event. ### Verification record - Main value: 0.054113766222821595 - Main method: Product of the two independent exponential survival probabilities at five years. - Independent check: The minimum of independent exponentials has rate 1/3 + 1/4 = 7/12, so its five-year survival is exp[-5(7/12)]. - Independent value: 0.054113766222821595 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: remote greenhouse controls A remote greenhouse stays within its safe temperature range only while both a circulation fan and a thermostat remain operational. Their independent operating times are exponential with means 6 months and 10 months. Calculate the probability that both devices last through an 8-month unattended season. - A: 0.118 - B: 0.264 - C: 0.406 - D: 0.449 - E: 0.594 Both independent devices must survive eight months, so their exponential survival probabilities multiply. The result is 0.118442, which selects choice A. #### Variant solution ### Setup Represent the fan and thermostat lifetimes by F and T and write their survival functions from the stated means. LaTeX: \Pr(F>t)=e^{-t/6},\qquad \Pr(T>t)=e^{-t/10} ### Model Safe operation for the entire season requires the intersection of the two eight-month survival events. LaTeX: \Pr(F>8,T>8)=\Pr(F>8)\Pr(T>8) ### Compute Multiply the independent survival probabilities and combine the rates. LaTeX: \Pr(F>8,T>8)=e^{-8/6}e^{-8/10} LaTeX: e^{-32/15}=0.1184418290\ldots ### Answer The greenhouse remains protected for the full season with probability approximately 0.118. LaTeX: \boxed{0.118442\approx 0.118\quad\text{(A)}} ### Verification record - Main value: 0.11844182901380369 - Main method: Product of independent exponential survival probabilities at eight months. - Independent check: The first device failure has exponential rate 1/6 + 1/10 = 4/15; its survival through month 8 is exp(-32/15). - Independent value: 0.11844182901380369 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ## SOA Exam P Sample Question #698 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-698/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Sums of Independent Random Variables The stated ratio of the one-count and zero-count masses fixes each Poisson mean at 2. Adding three independent counts gives a Poisson mean of 6, whose probability of at least two events is 0.982649, matching choice E. ### Independent solution ### Setup Let lambda be the common mean for one location and translate the supplied relationship between its first two Poisson masses. LaTeX: \Pr(X=0)=e^{-\lambda},\qquad \Pr(X=1)=\lambda e^{-\lambda} LaTeX: \lambda e^{-\lambda}=2e^{-\lambda} ### Model Cancel the positive exponential factor to find the individual mean, then superpose the three independent Poisson counts. LaTeX: \lambda=2 LaTeX: T=X_1+X_2+X_3\sim\operatorname{Poisson}(3\lambda)=\operatorname{Poisson}(6) ### Compute Use the complement of total counts zero and one. LaTeX: \Pr(T\ge 2)=1-\Pr(T=0)-\Pr(T=1) LaTeX: \Pr(T\ge 2)=1-e^{-6}(1+6)=0.9826487348\ldots ### Answer Rounding the computed probability to three decimals gives the final listed value. LaTeX: \boxed{\Pr(T\ge 2)\approx 0.983\quad\text{(E)}} ### Distractor analysis - A: The value 18e^{-6}=0.044618 is the probability that the combined count equals exactly two, not that it is at least two. - B: The value e^{-6}(1+6+18)=0.061969 is the probability that the combined count is at most two, which reverses the tail and includes the boundary. - C: The value 1-e^{-2}(1+2)=0.593994 is the at-least-two probability for one location and fails to combine all three counts. - D: The value 1-e^{-6}(1+6+18)=0.938031 is the probability of at least three events, an off-by-one threshold error. ### Verification record - Main value: 0.9826487347633355 - Main method: Recover the individual Poisson mean from the mass ratio, add the three means, and complement total counts zero and one. - Independent check: Direct independence expansion gives P(T=0)=(e^-2)^3=e^-6 and P(T=1)=3(2e^-2)(e^-2)^2=6e^-6. - Independent value: 0.9826487347633355 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: satellite receiver alerts Four satellite receivers record independent Poisson numbers of false alerts during a diagnostic interval, all with the same positive mean. At one receiver, the probability of two alerts is one-half the probability of one alert. Calculate the probability of at least three alerts across all four receivers. - A: 0.080 - B: 0.195 - C: 0.238 - D: 0.762 - E: 0.908 The within-receiver mass ratio gives a mean of 1, so the four-receiver total is Poisson with mean 4. Its probability of at least three alerts is 0.761897, selecting choice D. #### Variant solution ### Setup Write the one- and two-alert masses for a Poisson variable with positive mean lambda. LaTeX: \Pr(X=1)=\lambda e^{-\lambda},\qquad \Pr(X=2)=\frac{\lambda^2e^{-\lambda}}{2} ### Model Apply the stated mass ratio and then add the independent receiver counts. LaTeX: \frac{\lambda^2e^{-\lambda}}{2}=\frac{1}{2}\lambda e^{-\lambda}\quad\Longrightarrow\quad \lambda=1 LaTeX: S\sim\operatorname{Poisson}(4) ### Compute Subtract the probabilities of zero, one, and two alerts from one. LaTeX: \Pr(S\ge 3)=1-e^{-4}\left(1+4+\frac{4^2}{2}\right) LaTeX: \Pr(S\ge 3)=0.7618966944\ldots ### Answer The required four-receiver tail probability rounds to 0.762. LaTeX: \boxed{\Pr(S\ge 3)\approx 0.762\quad\text{(D)}} ### Verification record - Main value: 0.7618966944464557 - Main method: Poisson mass-ratio inversion, four-count superposition, and a three-term complement. - Independent check: SciPy Poisson survival function evaluated above 2 with mean 4. - Independent value: 0.7618966944464556 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #699 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-699/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Bayes' Theorem; Poisson Distribution Bayes' theorem combines the 20% prior class share with the two class-specific Poisson likelihoods for a count of two. The posterior probability is 0.170298, which matches choice B. ### Independent solution ### Setup Let A denote membership in the lower-rate class and let N be the observed annual count. Record the class proportions and Poisson means. LaTeX: \Pr(A)=0.20,\qquad \Pr(A^c)=0.80 LaTeX: N\mid A\sim\operatorname{Poisson}(1),\qquad N\mid A^c\sim\operatorname{Poisson}(3) ### Model Evaluate the likelihood of exactly two events in each class and place those likelihoods into Bayes' formula. LaTeX: \Pr(N=2\mid A)=\frac{e^{-1}}{2},\qquad \Pr(N=2\mid A^c)=\frac{3^2e^{-3}}{2} LaTeX: \Pr(A\mid N=2)=\frac{0.20(e^{-1}/2)}{0.20(e^{-1}/2)+0.80(9e^{-3}/2)} ### Compute Calculate the weighted likelihood in the numerator and the total mixture likelihood in the denominator. LaTeX: \Pr(A,N=2)=0.0367879441\ldots LaTeX: \Pr(N=2)=0.0367879441\ldots+0.1792334461\ldots=0.2160213902\ldots LaTeX: \Pr(A\mid N=2)=0.1702976917\ldots ### Answer The requested posterior class probability rounds to 0.170. LaTeX: \boxed{\Pr(A\mid N=2)\approx 0.170\quad\text{(B)}} ### Distractor analysis - A: The value 0.20(e^{-1}/2)=0.036788 is the joint numerator P(A and N=2), not the posterior after division by the mixture likelihood. - C: The value e^{-1}/2=0.183940 is the within-class likelihood of two events and omits both the prior and Bayes normalization. - D: The value 0.200 is the original class proportion; it ignores the information supplied by the observed count. - E: The value (e^{-1}/2)/(e^{-1}/2+9e^{-3}/2)=0.450853 treats the two classes as equally likely before observing the count. ### Verification record - Main value: 0.1702976917055533 - Main method: Bayes' theorem with the population shares as priors and the two Poisson probability masses as likelihoods. - Independent check: Posterior odds equal prior odds times the likelihood ratio: (1/4)(e^2/9)=e^2/36, then convert odds to probability. - Independent value: 0.1702976917055533 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: sensor warning classification A warehouse uses two sensor models. Model R makes up 30% of installed sensors and produces a Poisson number of warning signals per week with mean 0.5. Model S makes up the remainder and has mean 2. Given that a randomly selected sensor produced exactly one warning last week, calculate the probability that it is model R. - A: 0.091 - B: 0.300 - C: 0.303 - D: 0.324 - E: 0.528 Weighting the two one-warning likelihoods by the 30% and 70% model shares gives a posterior model-R probability of 0.324407, so choice D is correct. #### Variant solution ### Setup Let W be the weekly warning count and write the prior model shares and conditional means. LaTeX: \Pr(R)=0.30,\qquad \Pr(S)=0.70 LaTeX: W\mid R\sim\operatorname{Poisson}(0.5),\qquad W\mid S\sim\operatorname{Poisson}(2) ### Model Use the one-warning mass under each model as the likelihood in Bayes' formula. LaTeX: \Pr(W=1\mid R)=0.5e^{-0.5},\qquad \Pr(W=1\mid S)=2e^{-2} LaTeX: \Pr(R\mid W=1)=\frac{0.30(0.5e^{-0.5})}{0.30(0.5e^{-0.5})+0.70(2e^{-2})} ### Compute Evaluate the weighted likelihoods and normalize. LaTeX: \Pr(R,W=1)=0.0909795990\ldots LaTeX: \Pr(W=1)=0.0909795990\ldots+0.1894693965\ldots=0.2804489955\ldots LaTeX: \Pr(R\mid W=1)=0.3244069347\ldots ### Answer The posterior probability is approximately 0.324. LaTeX: \boxed{\Pr(R\mid W=1)\approx 0.324\quad\text{(D)}} ### Verification record - Main value: 0.3244069346675137 - Main method: Bayesian updating with model shares and model-specific Poisson one-count masses. - Independent check: Use posterior odds (3/7)[(0.5e^-0.5)/(2e^-2)] = 3e^1.5/28 and convert the odds to a probability. - Independent value: 0.3244069346675137 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #700 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-700/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Sums of Independent Random Variables; Cumulative Probabilities Adding the 25 independent Poisson counts gives a manuscript total with mean 5. Summing its masses from zero through three produces 0.265026, which rounds to choice E. ### Independent solution ### Setup Let X_j be the count on item j and let T be the count across all 25 items. LaTeX: X_j\sim\operatorname{Poisson}(0.20),\qquad j=1,\ldots,25 LaTeX: T=\sum_{j=1}^{25}X_j ### Model A sum of independent Poisson variables is Poisson with mean equal to the sum of the component means. LaTeX: T\sim\operatorname{Poisson}(25\times 0.20)=\operatorname{Poisson}(5) ### Compute Three or fewer includes each total from zero through three, so add all four corresponding masses. LaTeX: \Pr(T\le 3)=e^{-5}\sum_{k=0}^{3}\frac{5^k}{k!} LaTeX: \Pr(T\le 3)=e^{-5}\left(1+5+\frac{25}{2}+\frac{125}{6}\right)=0.2650259153\ldots ### Answer The cumulative probability rounds to 0.265. LaTeX: \boxed{\Pr(T\le 3)\approx 0.265\quad\text{(E)}} ### Distractor analysis - A: Using an aggregate mean of 10 instead of 5 gives P(Poisson(10) at most 3)=0.010336, reflecting a doubled-exposure error. - B: The value e^{-5}(1+5+25/2)=0.124652 is P(T at most 2), which drops the allowed boundary value 3. - C: The value 5^3e^{-5}/3!=0.140374 is P(T=3) only, rather than the cumulative probability through 3. - D: The value e^{-5}(5+25/2+125/6)=0.258288 is P(1 at most T at most 3), which incorrectly excludes the zero-count outcome. ### Verification record - Main value: 0.2650259152973617 - Main method: Poisson superposition to mean 5 followed by summation of masses for totals zero through three. - Independent check: Starting from p0=e^-5, use the recurrence p(k+1)=5p(k)/(k+1) and add the first four recursively generated masses. - Independent value: 0.2650259152973617 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: telescope calibration hits A telescope records cosmic-ray hits during 12 separate calibration windows. Counts in different windows are independent, and each window's count is Poisson with mean 0.25. Calculate the probability that no more than two hits are recorded over the full calibration session. - A: 0.050 - B: 0.199 - C: 0.224 - D: 0.373 - E: 0.423 The 12 independent counts combine to a Poisson total with mean 3. Adding the zero-, one-, and two-hit masses gives 0.423190, so choice E is correct. #### Variant solution ### Setup Let H be the total number of hits in all calibration windows. LaTeX: H=\sum_{j=1}^{12}H_j,\qquad H_j\sim\operatorname{Poisson}(0.25) ### Model Superposition preserves the Poisson form and adds the 12 means. LaTeX: H\sim\operatorname{Poisson}(12\times 0.25)=\operatorname{Poisson}(3) ### Compute No more than two hits means a total of zero, one, or two. LaTeX: \Pr(H\le 2)=e^{-3}\left(1+3+\frac{3^2}{2}\right) LaTeX: \Pr(H\le 2)=0.4231900811\ldots ### Answer The full-session probability is approximately 0.423. LaTeX: \boxed{\Pr(H\le 2)\approx 0.423\quad\text{(E)}} ### Verification record - Main value: 0.42319008112684353 - Main method: Sum 12 independent Poisson means and evaluate the CDF through two. - Independent check: Poisson recurrence from p0=e^-3 gives p1=3p0 and p2=(3/2)p1; their sum reproduces the result. - Independent value: 0.42319008112684353 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ## SOA Exam P Sample Question #701 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-701/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Discrete Random Variables; Geometric Series Normalizing the infinite probability sequence gives k=2. Summing the even-indexed terms as a geometric series then yields 1/4, which is choice B. ### Independent solution ### Setup Use exhaustiveness to require the probabilities over all positive indices to sum to one. LaTeX: 1=\sum_{i=1}^{\infty}k\left(\frac13\right)^i ### Model Evaluate the full geometric series to determine its normalizing constant. LaTeX: \sum_{i=1}^{\infty}\left(\frac13\right)^i=\frac{1/3}{1-1/3}=\frac12 LaTeX: 1=\frac{k}{2}\quad\Longrightarrow\quad k=2 ### Compute The even indices are 2j for positive integers j, producing a second geometric series with ratio 1/9. LaTeX: \Pr(\text{even index})=2\sum_{j=1}^{\infty}\left(\frac13\right)^{2j} LaTeX: \Pr(\text{even index})=2\frac{1/9}{1-1/9}=\frac14 ### Answer The total probability carried by the even-indexed outcomes is one quarter. LaTeX: \boxed{\Pr(\text{even index})=\frac14\quad\text{(B)}} ### Distractor analysis - A: The raw even-term series (1/9)/(1-1/9)=1/8 omits the required normalizing constant k=2. - C: Using k=3, as if the first term alone had to equal one, gives 3(1/8)=3/8 instead of normalizing the entire sequence. - D: Assigning probability 1/2 to each parity assumes even and odd indices are symmetric, but the geometrically decreasing masses favor early odd indices. - E: The value 3/4 is the complementary probability of an odd index, not the requested even-index probability. ### Verification record - Main value: 0.25 - Main method: Normalize the geometric mass sequence and then sum its even-index subsequence. - Independent check: Rewrite the normalized law as P(I=i)=(1-r)r^(i-1) with r=1/3; its even-index probability is r/(1+r)=1/4. - Independent value: 0.25 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: observatory maintenance tiers An autonomous observatory assigns each maintenance request to a positive integer tier T. For a normalizing constant c, the assignment probabilities satisfy P(T=n)=c(2/5)^n for n=1,2,3,... . Calculate the probability that the selected tier number is divisible by 3. - A: 8/117 - B: 4/39 - C: 10/39 - D: 1/3 - E: 25/39 The full geometric series gives c=3/2. Applying that constant to the every-third-term series yields 4/39, so choice B is correct. #### Variant solution ### Setup Normalize the tier probabilities over all positive integers. LaTeX: 1=c\sum_{n=1}^{\infty}\left(\frac25\right)^n ### Model Evaluate the geometric series and solve for c. LaTeX: \sum_{n=1}^{\infty}\left(\frac25\right)^n=\frac{2/5}{1-2/5}=\frac23 LaTeX: c=\frac32 ### Compute Tier numbers divisible by three have indices n=3j. Their probabilities form a series with ratio (2/5)^3. LaTeX: \Pr(T\equiv 0\pmod 3)=\frac32\sum_{j=1}^{\infty}\left(\frac25\right)^{3j} LaTeX: \Pr(T\equiv 0\pmod 3)=\frac32\frac{8/125}{1-8/125}=\frac4{39} ### Answer The probability assigned to every third tier is 4/39. LaTeX: \boxed{\Pr(T\equiv 0\pmod 3)=\frac4{39}\quad\text{(B)}} ### Verification record - Main value: 0.10256410256410256 - Main method: Normalize the geometric sequence and sum the subsequence whose indices are multiples of three. - Independent check: For normalized geometric masses with r=2/5, the multiple-of-three probability is r^2/(1+r+r^2)=4/39. - Independent value: 0.10256410256410256 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #702 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-702/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Joint Distributions; Covariance and Correlation This is a correlation calculation for two binary random variables. The joint table gives covariance 0.0175 and marginal variances 0.1275 and 0.2475, so the correlation is approximately 0.098513 and rounds to choice B. ### Independent solution ### Setup Read the two Bernoulli success probabilities and the joint-success probability from the joint distribution. LaTeX: \Pr(X=1)=0.05+0.10=0.15 LaTeX: \Pr(Y=1)=0.45+0.10=0.55 LaTeX: \mathbb{E}[XY]=\Pr(X=1,Y=1)=0.10 ### Model For a Bernoulli variable, the variance is p(1-p). Center the product moment to obtain the covariance. LaTeX: \operatorname{Var}(X)=0.15(0.85)=0.1275 LaTeX: \operatorname{Var}(Y)=0.55(0.45)=0.2475 LaTeX: \operatorname{Cov}(X,Y)=0.10-(0.15)(0.55)=0.0175 ### Compute Divide the covariance by the product of the two marginal standard deviations. LaTeX: \rho_{X,Y}=\frac{0.0175}{\sqrt{(0.1275)(0.2475)}} LaTeX: \rho_{X,Y}=0.0985134105\ldots ### Answer The correlation rounds to three decimals as 0.099. LaTeX: \boxed{\rho_{X,Y}\approx 0.099\quad\text{(B)}} ### Distractor analysis - A: The covariance is 0.0175, which rounds to 0.018. Reporting covariance without standardizing it produces choice A. - C: Using the uncentered product moment gives 0.10 divided by sqrt(0.15 times 0.55), approximately 0.348; correlation requires centered moments. - D: Dividing covariance by the product of the variances, without taking the square root, gives approximately 0.555 and selects D. - E: Applying one square root to the entire erroneous ratio Cov(X,Y)/(Var(X)Var(Y)) gives approximately 0.745 rather than the correlation formula. ### Verification record - Main value: 0.09851341054817479 - Main method: Bernoulli marginal variances and the centered joint-product moment, followed by covariance standardization. - Independent check: Binary-table phi coefficient: (p11 p00 - p10 p01) / sqrt(p1. p0. p.1 p.0). - Independent value: 0.09851341054817479 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: observatory maintenance signals On a randomly selected night, let A indicate that an observatory's tracking motor needs recalibration and let B indicate that its cooling monitor sends an alert. Historical frequencies are 0.54 for neither event, 0.08 for recalibration only, 0.18 for an alert only, and 0.20 for both events. Calculate the correlation between A and B. - A: 0.094 - B: 0.429 - C: 0.613 - D: 1.971 - E: -0.429 The marginal event probabilities are 0.28 and 0.38, while the joint-event probability is 0.20. Standardizing covariance 0.0936 gives correlation 0.429480, so choice B is correct. #### Variant solution ### Setup Form the two marginal event probabilities from the four operating states. LaTeX: \Pr(A=1)=0.08+0.20=0.28 LaTeX: \Pr(B=1)=0.18+0.20=0.38 LaTeX: \mathbb{E}[AB]=0.20 ### Model Use Bernoulli variances and subtract the product of the marginal means from the joint moment. LaTeX: \operatorname{Var}(A)=0.28(0.72)=0.2016 LaTeX: \operatorname{Var}(B)=0.38(0.62)=0.2356 LaTeX: \operatorname{Cov}(A,B)=0.20-(0.28)(0.38)=0.0936 ### Compute Standardize the covariance by the two marginal standard deviations. LaTeX: \rho_{A,B}=\frac{0.0936}{\sqrt{(0.2016)(0.2356)}} LaTeX: \rho_{A,B}=0.4294799974\ldots ### Answer The positive association between the two maintenance signals has correlation approximately 0.429. LaTeX: \boxed{\rho_{A,B}\approx0.429\quad\text{(B)}} ### Verification record - Main value: 0.4294799973855088 - Main method: Marginal Bernoulli moments and covariance standardization. - Independent check: The 2-by-2 phi formula using determinant 0.20(0.54)-0.08(0.18) and all four marginal probabilities. - Independent value: 0.4294799973855088 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #703 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-703/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Binomial Distribution; Waiting-Time Probabilities Model the independent claim indicators as Bernoulli trials and equate the two specified waiting-time probabilities. Cancellation leaves 7p(1-p)=1, whose admissible root is approximately 0.172673 and therefore choice C. ### Independent solution ### Setup Let p be the claim probability and let q=1-p. Translate each stopping event into a Bernoulli-sequence probability. LaTeX: q=1-p LaTeX: \Pr(\text{first claim at trial }6)=q^5p LaTeX: \Pr(\text{second claim at trial }8)=\binom{7}{1}p^2q^6 ### Model Equate the probabilities and cancel factors that are positive for an interior probability. LaTeX: q^5p=7p^2q^6 LaTeX: 1=7pq=7p(1-p) ### Compute Solve the resulting quadratic and apply the stated restriction that p is below one half. LaTeX: 7p^2-7p+1=0 LaTeX: p=\frac{7\pm\sqrt{21}}{14} LaTeX: p=\frac{7-\sqrt{21}}{14}=0.1726731646\ldots ### Answer The smaller root rounds to 0.173 and matches choice C. LaTeX: \boxed{p\approx0.173\quad\text{(C)}} ### Distractor analysis - A: Dropping the remaining factor 1-p after cancellation gives 7p=1 and therefore p=1/7, which rounds to 0.143. - B: Counting only six possible positions for the earlier claim and also dropping the remaining failure factor gives p=1/6, or 0.167. - D: Using six rather than seven possible positions but retaining the failure factor leads to 6p(1-p)=1; its smaller root is approximately 0.211. - E: Confusing the five failures before the first claim with the number of possible earlier-claim positions gives 5p(1-p)=1 and the smaller root 0.276. ### Verification record - Main value: 0.17267316464601143 - Main method: Negative-binomial event probabilities reduced to the quadratic equation 7p(1-p)=1. - Independent check: Substitution of the smaller root into both original event probabilities. - Independent value: both probabilities = 0.06692839427810378 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: seal-inspection calibration A packaging line produces independently sealed cartons, each of which fails an inspection with an unknown probability p below 0.5. The chance that exactly two of the next five cartons fail is equal to the chance that none of the next three cartons fail. Determine p. - A: 0.100 - B: 0.224 - C: 0.258 - D: 0.316 - E: 0.408 The two probabilities share a factor of (1-p) cubed. Cancelling it leaves 10p squared equal to one, so p is approximately 0.316228 and choice D is correct. #### Variant solution ### Setup Write each inspection statement as a binomial probability, using q=1-p for a passing carton. LaTeX: \Pr(\text{two failures in five})=\binom{5}{2}p^2q^3 LaTeX: \Pr(\text{no failures in three})=q^3 ### Model Equate the two expressions and cancel q cubed, which is positive because p is below one half. LaTeX: 10p^2q^3=q^3 LaTeX: 10p^2=1 ### Compute Take the admissible positive square root. LaTeX: p=\frac{1}{\sqrt{10}}=0.3162277660\ldots ### Answer Rounding to three decimals selects choice D. LaTeX: \boxed{p\approx0.316\quad\text{(D)}} ### Verification record - Main value: 0.31622776601683794 - Main method: Cancellation of the common no-failure factor in two binomial event probabilities. - Independent check: Direct evaluation at p=1/sqrt(10): both event probabilities equal (1-p)^3. - Independent value: both probabilities = 0.31969392534780233 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #704 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-704/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Binomial Distribution; Independence First convert two independent individual survivals into a couple-level success probability of 0.79 squared, or 0.6241. A binomial upper tail for seven couples then gives 0.925226, which selects choice D. ### Independent solution ### Setup A couple counts as a success only when both of its members survive. Independence turns that intersection into a product. LaTeX: q=\Pr(\text{both members survive})=(0.79)^2=0.6241 ### Model Let N count successful couples. Independence across all people makes the seven couple indicators independent Bernoulli variables. LaTeX: N\sim\operatorname{Binomial}(7,0.6241) LaTeX: \Pr(N\ge3)=1-\Pr(N\le2) ### Compute Evaluate the short lower tail and take its complement. LaTeX: \Pr(N\le2)=\sum_{k=0}^{2}\binom{7}{k}(0.6241)^k(0.3759)^{7-k} LaTeX: \Pr(N\le2)=0.07477428837\ldots LaTeX: \Pr(N\ge3)=0.9252257116\ldots ### Answer The requested probability rounds to 0.925. LaTeX: \boxed{\Pr(N\ge3)\approx0.925\quad\text{(D)}} ### Distractor analysis - A: An off-by-one interpretation that effectively requires four or more successful couples pushes the result into this lower range; the event begins at three. - B: Mixing the individual survival probability 0.79 with the couple probability 0.6241 across different binomial terms understates the coherent upper tail. - C: A rough normal approximation without a continuity correction depresses the answer; the seven-trial binomial tail is short enough to evaluate exactly. - E: Counting two or more successful couples instead gives 0.986615, near the high distractor after coarse rounding; the threshold is at least three. ### Verification record - Main value: 0.9252257116317639 - Main method: Couple-level success probability 0.79 squared followed by the complement of a Binomial(7, 0.6241) lower tail. - Independent check: Seven-step Bernoulli probability recursion, summing the terminal masses for counts 3 through 7. - Independent value: 0.9252257116317638 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: observatory cooling clusters A remote observatory operates eight independent cooling clusters. Each cluster contains three fans, and every fan completes the overnight cycle with probability 0.90 independently of every other fan. A cluster is fully operational only if all three of its fans complete the cycle. Calculate the probability that at least five clusters are fully operational. - A: 0.229 - B: 0.626 - C: 0.855 - D: 0.962 - E: 0.995 A cluster succeeds with probability 0.90 cubed, or 0.729. The Binomial(8, 0.729) probability of at least five successes is approximately 0.855108, so choice C is correct. #### Variant solution ### Setup Combine the three independent fan outcomes into one cluster-level success probability. LaTeX: q=(0.90)^3=0.729 ### Model Let K count fully operational clusters among the eight independent clusters. LaTeX: K\sim\operatorname{Binomial}(8,0.729) LaTeX: \Pr(K\ge5)=1-\Pr(K\le4) ### Compute Sum the lower five binomial masses and subtract from one. LaTeX: \Pr(K\ge5)=1-\sum_{k=0}^{4}\binom{8}{k}(0.729)^k(0.271)^{8-k} LaTeX: \Pr(K\ge5)=0.8551076438\ldots ### Answer The probability rounds to 0.855 and selects choice C. LaTeX: \boxed{\Pr(K\ge5)\approx0.855\quad\text{(C)}} ### Verification record - Main value: 0.8551076438114125 - Main method: Complement of the Binomial(8, 0.729) cumulative probability through four. - Independent check: Direct sum of the four binomial masses for counts 5, 6, 7, and 8. - Independent value: 0.8551076438114122 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #705 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-705/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Continuous Random Variables; Insurance Payment Variables The policyholder retains 40% of each loss, so the payment threshold corresponds to an original loss below 3. Integrating the given density from 0 to 3 gives 13/56, approximately 0.232143, and selects choice E. ### Independent solution ### Setup Let X denote the original loss and U the unreimbursed amount. Convert the payment condition back to a condition on X. LaTeX: U=(1-0.60)X=0.40X LaTeX: U<1.20\iff X<\frac{1.20}{0.40}=3 ### Model Use the loss density over the transformed interval. LaTeX: \Pr(U<1.20)=\int_0^3\frac{1}{42}\left(\frac{x}{2}+1\right)^2\,dx ### Compute Integrate after recognizing the linear inner function and evaluate at both bounds. LaTeX: \int\frac{1}{42}\left(\frac{x}{2}+1\right)^2dx=\frac{1}{63}\left(\frac{x}{2}+1\right)^3 LaTeX: \Pr(U<1.20)=\frac{1}{63}\left[\left(\frac{5}{2}\right)^3-1\right] LaTeX: \Pr(U<1.20)=\frac{13}{56}=0.2321428571\ldots ### Answer Rounding to three decimals gives 0.232. LaTeX: \boxed{\Pr(U<1.20)\approx0.232\quad\text{(E)}} ### Distractor analysis - A: Using 60% as the retained share gives a cutoff of 2, and then missing the factor of two from substitution halves the resulting 1/9 to about 0.056. - B: Treating the reimbursed 60% as the policyholder's retained share changes the cutoff to 2 and gives F_X(2)=1/9, approximately 0.111. - C: The value 0.116 is one half of the correct probability; it results from omitting the derivative factor when substituting u=x/2+1. - D: Evaluating the density at the cutoff gives f_X(3)=6.25/42, approximately 0.149, but a probability requires integrating over the interval. ### Verification record - Main value: 0.23214285714285715 - Main method: Payment-variable threshold conversion followed by the exact antiderivative of the stated density. - Independent check: Expand the density as (x squared/4 + x + 1)/42 and integrate the polynomial term by term from 0 to 3. - Independent value: 0.23214285714285715 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: ceramic restoration credit The replacement cost X, in thousands of dollars, for a custom ceramic installation has density f(x)=(x+2)/30 for 0500)=\frac{1400-500}{1400-200}=\frac{3}{4} LaTeX: \mathbb{E}[X-500\mid X>500]=\frac{0+900}{2}=450 ### Compute Multiply the probability of crossing the trigger by the average excess once it is crossed. LaTeX: \mathbb{E}[R]=\frac{3}{4}(450)=337.50 ### Answer The reserve's expected reimbursement is 337.50 dollars. LaTeX: \boxed{\mathbb{E}[R]=\$337.50\quad\text{(C)}} ### Verification record - Main value: 337.5 - Main method: Exceedance probability times conditional mean excess for a uniform repair cost. - Independent check: Direct integration gives integral from 500 to 1400 of (x-500)/1200 dx = 900^2/2400 = 337.5. - Independent value: 337.5 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #708 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-708/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Continuous Random Variables; Piecewise Density Functions; Percentiles and Quantiles The first density segment contains probability 0.40, so the median lies in the next segment. Accumulating another 0.10 at density 0.30 requires one-third of a year beyond 0.5, giving a median of 5/6 = 0.833333 year and choice B. ### Independent solution ### Setup A continuous median m satisfies F(m) = 0.5. First locate the density segment in which the cumulative probability reaches one-half. LaTeX: \Pr(X\le 0.5)=0.80(0.5)=0.40 ### Model Because 0.40 is below one-half, continue through the following segment, where the density is constant at 0.30. LaTeX: F(m)=0.40+0.30(m-0.5),\qquad 0.51)=\frac{\int_1^3 x/(1+x^2)\,dx}{\int_0^3 x/(1+x^2)\,dx} LaTeX: \int \frac{x}{1+x^2}\,dx=\frac{1}{2}\ln(1+x^2) ### Compute Evaluate the logarithmic antiderivative at the two sets of limits and simplify the ratio. LaTeX: \Pr(X>1)=\frac{\tfrac12[\ln(10)-\ln(2)]}{\tfrac12[\ln(10)-\ln(1)]} LaTeX: \Pr(X>1)=\frac{\ln(5)}{\ln(10)}=0.6989700043\ldots ### Answer The requested upper-tail probability rounds to 0.70. LaTeX: \boxed{\Pr(X>1)\approx 0.70\quad\text{(D)}} ### Distractor analysis - A: The value 0.15 is close to one-half of log base 10 of 2. It both keeps an antiderivative factor that should cancel and reports a lower-tail quantity instead of the requested upper tail. - B: Dropping the numerator x changes the kernel to 1/(1+x^2); its normalized upper tail is about 0.3712, which points to B but solves a different density. - C: Replacing the correct logarithmic ratio 10/2 = 5 by (3^2-1)/(1+1^2) = 4 gives log base 10 of 4 = 0.6021. The plus one in the antiderivative cannot be changed to a subtraction. - E: Keeping an extra one-half in the lower-tail probability gives 0.5 log base 10 of 2 = 0.1505; taking its complement then gives 0.8495, near E, but the one-half cancels from the normalized ratio. ### Verification record - Main value: 0.6989700043360187 - Main method: Ratio of exact kernel integrals over (1,3) and (0,3), simplifying to ln(5)/ln(10). - Independent check: Normalize with c = 2/ln(10); the lower-tail probability is c times one-half ln(2) = ln(2)/ln(10) = 0.3010299957, whose complement is 0.6989700043. - Independent value: 0.6989700043360187 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: acoustic signal intensity A marine recorder assigns a dimensionless intensity score S between 0 and 4 to each detected acoustic event. The score has a continuous density proportional to s on that interval. Calculate the probability that a detected event has an intensity score greater than 3. - A: 0.0625 - B: 0.2500 - C: 0.4375 - D: 0.5625 - E: 0.7500 Normalizing a density proportional to s gives f(s) = s/8. Its probability above 3 is the ratio (4 squared - 3 squared)/4 squared = 7/16 = 0.4375, so choice C is correct. #### Variant solution ### Setup Introduce the normalizing constant for the linearly increasing density. LaTeX: f(s)=cs,\qquad 03)=\int_3^4\frac{s}{8}\,ds LaTeX: \Pr(S>3)=\frac{4^2-3^2}{16}=\frac{7}{16}=0.4375 ### Answer The upper-tail probability is 0.4375. LaTeX: \boxed{\Pr(S>3)=0.4375\quad\text{(C)}} ### Verification record - Main value: 0.4375 - Main method: Normalize the linear density and integrate it from 3 to 4. - Independent check: Area ratios cancel c directly: integral from 3 to 4 of s divided by integral from 0 to 4 of s equals (16-9)/16 = 7/16. - Independent value: 0.4375 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #710 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-710/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Uniform Distribution; Percentiles and Quantiles; Expected Value For a continuous uniform distribution, both the mean and median are the interval midpoint, while the 75th percentile is one-quarter of the lower endpoint plus three-quarters of the upper endpoint. Their sum is (5a + 7b)/4, which is choice B. ### Independent solution ### Setup Use the standard mean and quantile formulas for a continuous uniform random variable on an interval with endpoints a and b. LaTeX: X\sim\operatorname{Uniform}(a,b),\qquad q_p=a+p(b-a) ### Model The mean and median coincide at the midpoint, and the upper quartile follows from the quantile formula. LaTeX: x=\frac{a+b}{2},\qquad y=\frac{a+b}{2} LaTeX: z=q_{0.75}=a+\frac{3}{4}(b-a)=\frac{a+3b}{4} ### Compute Add the three location measures using a common denominator. LaTeX: x+y+z=(a+b)+\frac{a+3b}{4} LaTeX: x+y+z=\frac{5a+7b}{4} ### Answer The required symbolic sum is the expression listed under choice B. LaTeX: \boxed{\frac{5a+7b}{4}\quad\text{(B)}} ### Distractor analysis - A: The expression (a+7b)/4 results if the two midpoint terms are replaced by b/2 each, dropping all of their lower-endpoint contribution. - C: Using 0.75(a+b) for the upper quartile instead of a + 0.75(b-a) changes z from (a+3b)/4 to (3a+3b)/4 and produces (7a+7b)/4. - D: The expression (7a+9b)/4 equals the correct sum plus another midpoint (a+b)/2, so it double-counts one of the identical mean-or-median terms. - E: The expression (9a+7b)/4 assigns four extra units of lower-endpoint weight. It also fails the translation check: moving both endpoints by c should move the sum by exactly 3c. ### Verification record - Main value: (5a+7b)/4 - Main method: Add the uniform mean, median, and 75th-percentile formulas and collect endpoint coefficients. - Independent check: For the representative interval [2,10], the mean, median, and 75th percentile are 6, 6, and 8, summing to 20; (5(2)+7(10))/4 also equals 20. - Independent value: 20 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: courier delay threshold A courier service models a route's delay T as uniformly distributed between 6 and 18 minutes. A delay is labeled slow when it exceeds the 65th percentile of T. Calculate the amount by which that percentile exceeds the average delay. - A: -1.8 minutes - B: 0.9 minute - C: 1.2 minutes - D: 1.8 minutes - E: 7.8 minutes The 65th percentile is 6 + 0.65(12) = 13.8 minutes, while the uniform mean is 12 minutes. The requested excess is 1.8 minutes, so choice D is correct. #### Variant solution ### Setup Record the interval width and the uniform quantile formula. LaTeX: T\sim\operatorname{Uniform}(6,18),\qquad 18-6=12 LaTeX: q_p=6+12p ### Model Evaluate the slow-delay threshold and the average delay separately. LaTeX: q_{0.65}=6+0.65(12)=13.8 LaTeX: \mathbb{E}[T]=\frac{6+18}{2}=12 ### Compute Subtract the average from the percentile as requested. LaTeX: q_{0.65}-\mathbb{E}[T]=13.8-12=1.8 ### Answer The 65th percentile is 1.8 minutes above the average. LaTeX: \boxed{1.8\text{ minutes}\quad\text{(D)}} ### Verification record - Main value: 1.8 - Main method: Uniform 65th-percentile formula minus the interval midpoint. - Independent check: The CDF at 13.8 is (13.8-6)/12 = 0.65, and direct integration of the uniform density gives mean 12; their difference is 1.8. - Independent value: 1.8 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #711 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-711/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Memoryless Property; Conditional Probability After conditioning on survival through year 2, exponential memorylessness restarts the waiting clock. The original years 5 and 8 become residual times 3 and 6, so the required probability is exp(-3/4) - exp(-6/4) = 0.249236, which rounds to choice C. ### Independent solution ### Setup Let T be the waiting time and convert the stated mean to the exponential rate. LaTeX: T\sim\operatorname{Exponential}(\lambda),\qquad \lambda=\frac{1}{4} ### Model Conditioning on no event before year 2 leaves an independent exponential residual wait with the same rate. LaTeX: (T-2\mid T>2)\sim\operatorname{Exponential}\!\left(\frac14\right) LaTeX: \Pr(52)=\Pr(32) ### Compute Subtract exponential survival probabilities at the two residual-time boundaries. LaTeX: \Pr(52)=e^{-3/4}-e^{-6/4} LaTeX: =0.4723665527-0.2231301601=0.2492363926 ### Answer The conditional probability rounds to 0.25. LaTeX: \boxed{0.25\quad\text{(C)}} ### Distractor analysis - A: Using rate 1/2 instead of 1/4 gives exp(-5/2) - exp(-4) = 0.06377, near A. The two claim-free years are conditioning information, not a replacement for the stated mean. - B: The unconditional probability exp(-5/4) - exp(-8/4) = 0.15117 rounds to 0.15; it ignores the information that the waiting time has already exceeded two years. - D: Rounding exp(-3/4) to 0.5 and exp(-6/4) to 0.2 before subtracting gives 0.30. Such coarse intermediate rounding materially changes the result. - E: The value 0.75 is approximately the complement of the correct interval probability. The question asks for occurrence inside the interval, not outside it. ### Verification record - Main value: 0.24923639259258487 - Main method: Apply exponential memorylessness and subtract residual-time survival probabilities at 3 and 6 years. - Independent check: Direct conditional probability gives [S(5)-S(8)]/S(2) = [exp(-5/4)-exp(-8/4)]/exp(-2/4) = 0.2492363926. - Independent value: 0.24923639259258487 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: remote pump repair A remote irrigation pump requires an exponentially distributed amount of technician time with mean 10 hours. After 4 hours of work, the repair is still incomplete. Given that information, calculate the probability that the repair finishes between 7 and 12 hours after work began. - A: 0.1954 - B: 0.2592 - C: 0.2915 - D: 0.4493 - E: 0.7085 Memorylessness turns the requested original-time window into a residual-time window from 3 to 8 hours. Its probability is exp(-0.3) - exp(-0.8) = 0.291489, so choice C is correct. #### Variant solution ### Setup Let R denote the total repair duration and record its rate and the elapsed-work condition. LaTeX: R\sim\operatorname{Exponential}(0.1),\qquad R>4 ### Model The residual repair time after hour 4 remains exponential with rate 0.1. LaTeX: (R-4\mid R>4)\sim\operatorname{Exponential}(0.1) LaTeX: 74)=e^{-0.1(3)}-e^{-0.1(8)} LaTeX: =0.7408182207-0.4493289641=0.2914892566 ### Answer The conditional completion probability is approximately 0.2915. LaTeX: \boxed{0.2915\quad\text{(C)}} ### Verification record - Main value: 0.2914892565644963 - Main method: Use exponential memorylessness and subtract residual survival probabilities at 3 and 8 hours. - Independent check: Direct conditioning gives [exp(-0.7)-exp(-1.2)]/exp(-0.4) = 0.2914892566. - Independent value: 0.2914892565644963 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #712 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-712/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Survival Probabilities Express five-year survival as q. The supplied interval probability forces q(1-q)=0.25 and hence q=0.5; the requested interval probability is then q^2-q^4=0.1875, corresponding to choice C. ### Independent solution ### Setup Represent survival over one five-year block by a single quantity. LaTeX: q=e^{-5\lambda}=\Pr(X>5),\qquad 04)=1-0.30=0.70 ### Model For an exponential lifetime, survival over equal blocks multiplies. LaTeX: \Pr(T>8)=q^2,\qquad \Pr(T>12)=q^3 ### Compute Subtract survival through month 12 from survival through month 8. LaTeX: \Pr(81,\qquad \frac{p_{11}}{p_{10}}=\frac{10.5}{11}<1 LaTeX: \operatorname{Mode}(N)=\lfloor 10.5\rfloor=10 ### Answer The unique modal packet-failure count is 10. LaTeX: \boxed{10\text{ only}\quad\text{(D)}} ### Verification record - Main value: 10 - Main method: Infer rate 10.5 from the consecutive Poisson probability ratio and apply the noninteger-rate mode rule. - Independent check: Direct enumeration of Poisson log masses at rate 10.5 finds a unique global maximum at count 10. - Independent value: 10 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #719 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-719/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Moments The Poisson mean-variance identity turns the two supplied relationships into lambda_X minus lambda_Y equal to 3 and lambda_X equal to twice lambda_Y, so lambda_Y is 3. Therefore E[Y squared] equals lambda_Y squared plus lambda_Y, or 12, matching choice D. ### Independent solution ### Setup Assign rates to the two Poisson variables and use the fact that each rate is both its mean and its variance. LaTeX: X\sim\operatorname{Poisson}(\lambda_X),\quad Y\sim\operatorname{Poisson}(\lambda_Y) LaTeX: \mathbb{E}[X]=\operatorname{Var}(X)=\lambda_X,\quad \mathbb{E}[Y]=\operatorname{Var}(Y)=\lambda_Y ### Model Translate the mean-difference and variance-ratio information into two equations for the rates. LaTeX: \lambda_X-\lambda_Y=3 LaTeX: \lambda_X=2\lambda_Y ### Compute Solve for the target rate, then apply the Poisson second-moment identity. LaTeX: 2\lambda_Y-\lambda_Y=3\quad\Longrightarrow\quad\lambda_Y=3 LaTeX: \mathbb{E}[Y^2]=\operatorname{Var}(Y)+\mathbb{E}[Y]^2 LaTeX: \mathbb{E}[Y^2]=3+3^2=12 ### Answer The second raw moment is 12, the value listed under choice D. LaTeX: \boxed{\mathbb{E}[Y^2]=12\quad\text{(D)}} ### Distractor analysis - A: The value 3 is E[Y] and Var(Y); it omits the squared-mean term needed for the second raw moment. - B: The value 6 results from using 2 lambda_Y, which doubles the variance but is not the identity for E[Y squared]. - C: The value 9 is lambda_Y squared, equivalently E[Y(Y-1)], and leaves out the additional lambda_Y term. - E: The value 15 comes from lambda_Y squared plus 2 lambda_Y, so the mean contribution has been counted twice. ### Verification record - Main value: 12 - Main method: Solve the Poisson rate equations from the mean difference and variance ratio, then use E[Y^2] = lambda_Y^2 + lambda_Y. - Independent check: Factorial moments give E[Y(Y-1)] = lambda_Y^2 = 9 and E[Y] = 3; adding them yields E[Y^2] = 12. - Independent value: 12 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: cryogenic-detector count A cryogenic detector's number N of spurious pulses during a test window follows a Poisson distribution. Calibration establishes that the factorial second moment E[N(N - 1)] equals 16. Calculate E[(N + 1)^2]. - A: 16 - B: 20 - C: 21 - D: 25 - E: 29 For a Poisson variable, E[N(N - 1)] equals lambda squared, so the nonnegative rate is 4. Expanding the requested square and using E[N squared] = 20 gives 29, making E correct. #### Variant solution ### Setup Use the Poisson factorial second-moment identity to identify the rate. LaTeX: \mathbb{E}[N(N-1)]=\lambda^2=16 LaTeX: \lambda=4 ### Model Convert the factorial moment to the ordinary second moment. LaTeX: \mathbb{E}[N^2]=\mathbb{E}[N(N-1)]+\mathbb{E}[N] LaTeX: \mathbb{E}[N^2]=16+4=20 ### Compute Expand the shifted square and take expectations term by term. LaTeX: (N+1)^2=N^2+2N+1 LaTeX: \mathbb{E}[(N+1)^2]=20+2(4)+1=29 ### Answer The shifted count has second moment 29. LaTeX: \boxed{\mathbb{E}[(N+1)^2]=29\quad\text{(E)}} ### Verification record - Main value: 29 - Main method: Recover the Poisson rate from the factorial second moment, convert to E[N^2], and expand the shifted square. - Independent check: Treat N+1 directly: Var(N+1) + E[N+1]^2 = 4 + 5^2 = 29. - Independent value: 29 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ## SOA Exam P Sample Question #720 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-720/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Normal Distribution; Percentiles; Coefficient of Variation Solving the two normal-quantile equations at z_0.05 = -1.644854 and z_0.99 = 2.326348 gives mean 19.733110 and standard deviation 27.195800. Their requested absolute ratio is 1.378181, which rounds to 1.38 and matches choice E. ### Independent solution ### Setup Write each supplied percentile as the mean plus its standard-normal quantile times the standard deviation. LaTeX: q_p=\mu+z_p\sigma LaTeX: z_{0.05}=-1.644853627,\qquad z_{0.99}=2.326347874 ### Model Substitute the two given percentile values to obtain a linear system for the mean and standard deviation. LaTeX: -25=\mu-1.644853627\sigma LaTeX: 83=\mu+2.326347874\sigma ### Compute Subtract the equations to find the standard deviation, then back-substitute for the mean and form the requested ratio. LaTeX: \sigma=\frac{83-(-25)}{2.326347874-(-1.644853627)}=27.19579955 LaTeX: \mu=-25+1.644853627(27.19579955)=19.73310953 LaTeX: \left|\frac{\sigma}{\mu}\right|=\frac{27.19579955}{19.73310953}=1.378181148 ### Answer The ratio rounds to 1.38, the value listed under choice E. LaTeX: \boxed{\left|\sigma/\mu\right|\approx 1.38\quad\text{(E)}} ### Distractor analysis - A: Dropping the minus sign from the lower percentile and solving with 25 instead of -25 gives an absolute ratio about 0.298, which rounds to 0.30. - B: Pairing the lower observation with +2.326 and the upper observation with +1.645, then ignoring the resulting negative standard deviation, gives an absolute ratio about 0.461. - C: Using positive signs for both standard scores while keeping their order gives an absolute ratio about 0.555, which rounds to 0.56. - D: Taking the ratio of the two standard-score magnitudes, 1.644854 divided by 2.326348, gives about 0.707 or 0.71; that is not sigma divided by mu. ### Verification record - Main value: 1.3781811482801893 - Main method: Solve the two linear normal-quantile equations using high-precision inverse-normal values and evaluate the absolute standard-deviation-to-mean ratio. - Independent check: Back-substitution reconstructs the supplied percentiles: mu + z_0.05 sigma = -25 and mu + z_0.99 sigma = 83 to floating-point precision. - Independent value: 1.3781811482801893 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: resin-strength percentiles A composite resin's strength index R is normally distributed with a positive mean and coefficient of variation 0.20. Its percentile corresponding to standard-normal score z = 1 is 52. Calculate the value at the symmetric percentile corresponding to z = -1. - A: 34.7 - B: 41.6 - C: 43.3 - D: 52.0 - E: 62.4 The named percentiles correspond to z = 1 and z = -1, while sigma = 0.20 mu. Solving mu + sigma = 52 gives mu = 43.3333 and sigma = 8.6667, so the lower percentile is 34.6667 and choice A. #### Variant solution ### Setup Translate the coefficient of variation into a relationship between the positive mean and standard deviation. LaTeX: \frac{\sigma}{\mu}=0.20\quad\Longrightarrow\quad\sigma=0.20\mu ### Model The 84.1345th percentile has standard score 1, so use its value to solve for the distribution parameters. LaTeX: 52=\mu+\sigma=\mu+0.20\mu=1.20\mu LaTeX: \mu=43.3333333,\qquad \sigma=8.6666667 ### Compute The 15.8655th percentile has standard score -1, so evaluate the symmetric lower point. LaTeX: q_{0.158655}=\mu-\sigma LaTeX: q_{0.158655}=43.3333333-8.6666667=34.6666667 ### Answer The lower resin-strength percentile rounds to 34.7. LaTeX: \boxed{q_{0.158655}\approx 34.7\quad\text{(A)}} ### Verification record - Main value: 34.666666666666664 - Main method: Combine sigma = 0.20 mu with the z=1 upper percentile, then evaluate the symmetric z=-1 percentile. - Independent check: Normal symmetry gives q_0.158655 = 2mu - q_0.841345 = 2(43.3333333) - 52 = 34.6666667. - Independent value: 34.666666666666664 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ## SOA Exam P Sample Question #721 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-721/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Linear Combinations; Normal Distribution; Percentiles Independence makes the aggregate variance equal to 49 times 700 squared, so the aggregate standard deviation is 4,900 while its mean is 122,500. Adding 1.281552 standard deviations gives 128.779603 thousand, which rounds to 129 and matches choice D. ### Independent solution ### Setup Let S be the sum of all 49 independent normal amounts. A sum of independent normal variables is normal. LaTeX: S=\sum_{i=1}^{49}X_i,\qquad S\sim N(\mu_S,\sigma_S^2) LaTeX: \mu_S=122{,}500 ### Model Add the independent component variances; the differing component means do not affect the aggregate variance. LaTeX: \sigma_S^2=\sum_{i=1}^{49}\operatorname{Var}(X_i)=49(700^2)=24{,}010{,}000 LaTeX: \sigma_S=\sqrt{24{,}010{,}000}=4{,}900 ### Compute Use the 90th-percentile standard-normal value and convert the result to thousands. LaTeX: z_{0.90}=1.281551566 LaTeX: q_{0.90}=122{,}500+1.281551566(4{,}900)=128{,}779.6027 LaTeX: q_{0.90}=128.7796027\text{ thousand} ### Answer The requested percentile rounds to 129 thousand, which is choice D. LaTeX: \boxed{q_{0.90}\approx 129\text{ thousand}\quad\text{(D)}} ### Distractor analysis - A: Rounding the aggregate mean 122.5 thousand to 123 ignores the positive 90th-percentile shift entirely. - B: A total of 125 thousand is only 0.5102 aggregate standard deviations above the mean, far below the required z_0.90 = 1.2816. - C: Using a one-standard-deviation shift gives 122.5 + 4.9 = 127.4 thousand, which rounds to 127 instead of using the 90th-percentile z-value. - E: Using the 95th-percentile value 1.6449 gives 130.56 thousand, which rounds to 131 and answers a different percentile. ### Verification record - Main value: 128.77960267116853 - Main method: Add independent normal variances, take the aggregate standard deviation, and evaluate the normal 90th-percentile formula in thousands. - Independent check: Standardizing 128.779602671 thousand with mean 122.5 and standard deviation 4.9 gives 1.281551566; the standard-normal CDF is 0.900000. - Independent value: 128.77960267116853 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: satellite calibration score Three independent satellite calibration readings are A ~ N(40, 3^2), B ~ N(25, 4^2), and C ~ N(10, 2^2). Engineers combine them into the score S = 2A - B + 0.5C. Calculate the 95th percentile of S, rounded to the nearest whole number. - A: 60 - B: 67 - C: 69 - D: 72 - E: 78 The weighted score is normal with mean 60 and variance 2 squared times 9 plus 16 plus 0.5 squared times 4, which equals 53. Its 95th percentile is 60 + 1.644854 square root of 53 = 71.974715, rounding to 72 and choice D. #### Variant solution ### Setup A linear combination of independent normal readings is itself normal. LaTeX: S=2A-B+0.5C LaTeX: S\sim N(\mu_S,\sigma_S^2) ### Model Apply the coefficients linearly to means and quadratically to independent variances. LaTeX: \mu_S=2(40)-25+0.5(10)=60 LaTeX: \sigma_S^2=2^2(3^2)+(-1)^2(4^2)+0.5^2(2^2)=36+16+1=53 ### Compute Use the positive 95th-percentile standard-normal value with the score's standard deviation. LaTeX: z_{0.95}=1.644853627 LaTeX: q_{0.95}=60+1.644853627\sqrt{53}=71.97471516 ### Answer The requested score percentile rounds to 72. LaTeX: \boxed{q_{0.95}\approx 72\quad\text{(D)}} ### Verification record - Main value: 71.97471515598833 - Main method: Calculate the weighted linear-combination mean and variance, then apply the standard-normal 95th-percentile multiplier. - Independent check: The covariance-matrix check w' diag(9,16,4) w with w=(2,-1,0.5) returns variance 53; standardizing 71.974715156 gives normal CDF 0.950000. - Independent value: 71.97471515598833 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #722 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-722/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Expectation and Variance Partition the Poisson count into the four payout categories, then compute both the first and second payout moments. Subtracting the squared mean gives a variance of approximately 138,861, which rounds to choice B. ### Independent solution ### Setup Let N denote the count and C the resulting payment. Express C as a discrete transformation of N so that its moments can be calculated category by category. LaTeX: C=\begin{cases}0,&N=0,\\200,&N=1,\\500,&N=2,\\1000,&N\ge 3.\end{cases} ### Model For a Poisson rate of 2, the first two positive masses are equal and the upper category is obtained by complement. LaTeX: p_0=e^{-2},\qquad p_1=2e^{-2},\qquad p_2=2e^{-2} LaTeX: p_{3+}=1-(p_0+p_1+p_2)=1-5e^{-2} ### Compute Evaluate the two raw moments using the same probability categories, then apply the variance identity. LaTeX: \mathbb{E}[C]=200p_1+500p_2+1000p_{3+}=512.7929803\ldots LaTeX: \mathbb{E}[C^2]=200^2p_1+500^2p_2+1000^2p_{3+}=401818.0481\ldots LaTeX: \operatorname{Var}(C)=\mathbb{E}[C^2]-\mathbb{E}[C]^2=138861.4074\ldots ### Answer Rounding the variance to the requested thousand selects the second answer choice. LaTeX: \boxed{\operatorname{Var}(C)\approx 139000\quad\text{(B)}} ### Distractor analysis - A: Moving the highest payout category from three-or-more events to four-or-more gives about 110,917, a low result in A's range caused by dropping the entire N=3 mass. - C: A value near 166,000 results if the correct second moment is paired with an understated mean near 486. The first and second moments must come from the same payout mapping. - D: Omitting the 500 payment attached to N=2 while retaining the other categories gives approximately 191,676, which rounds to choice D. - E: Inflating the highest payment to 1200 raises the variance to about 210,623, pulling the result toward E; the top category pays 1000 in the reference calculation. ### Verification record - Main value: 138861.40739978827 - Main method: Closed-form first and second moments over the four Poisson payout categories - Independent check: Direct SciPy enumeration of the transformed Poisson support through count 39 - Independent value: 138861.40739978832 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: cloud incident reserve A cloud service models its weekly number of severity-one incidents by a Poisson random variable with mean 1.5. Its incident reserve is 300 dollars for exactly one incident, 700 dollars for exactly two, 1200 dollars for three or more, and zero otherwise. Calculate the variance of the weekly reserve. - A: 128,015 - B: 151,755 - C: 172,846 - D: 234,156 - E: 428,384 Compute the category probabilities for a Poisson rate of 1.5 and use them for both reserve moments. The resulting variance is approximately 172,846, so choice C is correct. #### Variant solution ### Setup Let R be the reserve and partition the incident count into zero, one, two, and at least three. LaTeX: R=\begin{cases}0,&N=0,\\300,&N=1,\\700,&N=2,\\1200,&N\ge3.\end{cases} ### Model Use the Poisson mass function for the first three counts and a complement for the final category. LaTeX: p_0=e^{-1.5},\quad p_1=1.5e^{-1.5},\quad p_2=\frac{1.5^2}{2}e^{-1.5} LaTeX: p_{3+}=1-e^{-1.5}\left(1+1.5+\frac{1.5^2}{2}\right) ### Compute Form the two moments and subtract the squared mean. LaTeX: \mathbb{E}[R]=300p_1+700p_2+1200p_{3+}=505.5073765\ldots LaTeX: \mathbb{E}[R^2]=300^2p_1+700^2p_2+1200^2p_{3+}=428383.6364\ldots LaTeX: \operatorname{Var}(R)=428383.6364\ldots-(505.5073765\ldots)^2=172845.9287\ldots ### Answer The variance rounds to the value shown in choice C. LaTeX: \boxed{\operatorname{Var}(R)\approx172846\quad\text{(C)}} ### Verification record - Main value: 172845.92869271265 - Main method: Closed-form reserve moments from Poisson category probabilities - Independent check: Direct SciPy enumeration of the transformed Poisson support through count 39 - Independent value: 172845.92869271268 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #723 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-723/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Law of Total Probability; Mixture Probabilities Express the low- and high-group death probabilities as multiples of the medium-group probability, then apply the law of total probability. Solving the weighted equation gives approximately 0.009863, which corresponds to choice B. ### Independent solution ### Setup Let q denote the one-year death probability for a member of the medium group. Translate the two relative-risk statements into probabilities expressed in terms of q. LaTeX: q_M=q,\qquad q_L=\frac{q}{2},\qquad q_H=4q ### Model The remaining population share is 0.15, so the overall probability is the weighted average of the three group probabilities. LaTeX: 0.15\left(\frac{q}{2}\right)+0.55q+0.30(4q)=0.018 ### Compute Collect the coefficients of q and solve the single linear equation. LaTeX: (0.075+0.55+1.20)q=1.825q=0.018 LaTeX: q=\frac{0.018}{1.825}=0.0098630137\ldots ### Answer The medium-group probability rounds to the value displayed in choice B. LaTeX: \boxed{q_M\approx0.010\quad\text{(B)}} ### Distractor analysis - A: This is the low-group probability, q_L=q_M/2=0.0049315..., rounded to 0.005; it answers for the wrong category. - C: Using a high-group multiplier of 2 instead of 4 gives q_M=0.018/1.225=0.014694..., which rounds to choice C. - D: This repeats the population-wide death probability 0.018 without resolving the mixture into its three group-specific rates. - E: If q_M were 0.021, the stated weights and multipliers would produce an overall probability of 1.825(0.021)=0.038325, not 0.018. ### Verification record - Main value: 0.009863013698630137 - Main method: Algebraic solution of the total-probability mixture in terms of the medium-group rate - Independent check: Solve first for the low-group baseline and then double it; direct weighted substitution returns 0.018 - Independent value: 0.009863013698630137 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: field-sensor failure classes A sensor fleet is divided among sheltered, standard, and harsh sites in proportions 0.20, 0.50, and 0.30. A standard-site sensor is three times as likely to fail during a year as a sheltered-site sensor, while a harsh-site sensor is twice as likely to fail as a standard-site sensor. The annual failure probability for a randomly selected sensor is 0.028. Calculate the annual failure probability at a standard site. - A: 0.80% - B: 1.40% - C: 2.40% - D: 2.80% - E: 4.80% Write the sheltered and harsh failure rates as one-third and twice the standard-site rate. The weighted fleet equation gives a standard-site probability of 0.024, so the answer is C. #### Variant solution ### Setup Let s be the standard-site annual failure probability and express the other two rates relative to it. LaTeX: q_{\mathrm{sheltered}}=\frac{s}{3},\qquad q_{\mathrm{standard}}=s,\qquad q_{\mathrm{harsh}}=2s ### Model Average the three rates with the fleet proportions. LaTeX: 0.20\left(\frac{s}{3}\right)+0.50s+0.30(2s)=0.028 ### Compute The coefficient is 7/6, so the standard-site rate follows directly. LaTeX: \left(\frac{1}{15}+\frac{1}{2}+\frac{3}{5}\right)s=\frac{7}{6}s=0.028 LaTeX: s=0.028\left(\frac{6}{7}\right)=0.024 ### Answer The standard-site annual failure probability is listed in choice C. LaTeX: \boxed{s=0.024\quad\text{(C)}} ### Verification record - Main value: 0.024 - Main method: Weighted total-probability equation expressed in terms of the standard-site rate - Independent check: Recover the sheltered baseline as 0.008, scale the two other rates, and substitute all three into the fleet average - Independent value: 0.024 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #724 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-724/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Linear Combinations of Independent Random Variables; Normal Distribution; Percentiles Model the company total as normal with mean proportional to n and standard deviation proportional to the square root of n, then impose the 90th-percentile equation. Its positive root gives n approximately 64, and the integer answer is choice D. ### Independent solution ### Setup Let S_n be the sum for n independent employees. Add the means and variances before taking the standard deviation. LaTeX: S_n\sim N\!\left(2500n,\,700^2n\right) LaTeX: \mathbb{E}[S_n]=2500n,\qquad \operatorname{SD}(S_n)=700\sqrt{n} ### Model A normal 90th percentile is the mean plus z_{0.90} standard deviations. LaTeX: 167168=2500n+z_{0.90}(700\sqrt{n}),\qquad z_{0.90}=1.2815516\ldots ### Compute Set x equal to the square root of n and solve the resulting quadratic. The tiny deviation from 64 comes from using the full normal quantile while the target value reflects table rounding. LaTeX: 2500x^2+897.0861\ldots x-167168=0 LaTeX: x=7.9997875\ldots,\qquad n=x^2=63.9966007\ldots LaTeX: \text{Using }z_{0.90}=1.28:\quad 2500(64)+1.28(700)(8)=167168 ### Answer The population count must be an integer, and the matching listed value is 64. LaTeX: \boxed{n=64\quad\text{(D)}} ### Distractor analysis - A: Adding individual standard deviations linearly gives 167168 = n(2500 + 1.28155 times 700), or n=49.21, which leads to A instead of using a square-root scaling. - B: For n=54 the reconstructed 90th percentile is only about 141,592, so B fails a direct substitution check by more than 25,000. - C: For n=59 the reconstructed percentile is about 154,391, still roughly 12,777 below the supplied total; stopping at this nearby trial value is premature. - E: Using a minus sign for the upper percentile gives n=69.87, which points to E. The 90th percentile lies above, not below, the mean. ### Verification record - Main value: 64 - Main method: Positive-root solution of the normal-sum percentile equation; continuous root 63.9966007301 - Independent check: Back-substitution at n=64 gives standardized value 1.28 and cumulative probability 0.8997274320 - Independent value: 64 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: incubator energy total The daily energy use of each of n laboratory incubators is normally distributed with mean 40 kWh and standard deviation 6 kWh. Uses by different incubators are independent. The 95th percentile of their combined daily energy use is 4098.69 kWh. Calculate n. - A: 82 - B: 99 - C: 100 - D: 102 - E: 105 The combined use has mean 40n and standard deviation 6 times the square root of n. Solving its 95th-percentile equation gives n essentially equal to 100, so choice C is correct. #### Variant solution ### Setup Let T_n be the combined energy use and add the independent normal means and variances. LaTeX: T_n\sim N\!\left(40n,\,36n\right) LaTeX: \operatorname{SD}(T_n)=6\sqrt{n} ### Model Use the standard normal 95th percentile to connect the total to n. LaTeX: 4098.69=40n+1.6448536\ldots(6\sqrt{n}) ### Compute With x equal to the square root of n, solve the positive quadratic root and then square it. LaTeX: 40x^2+9.8691218\ldots x-4098.69=0 LaTeX: x=9.9999985\ldots,\qquad n=x^2=99.9999699\ldots ### Answer The integer number of incubators is 100. LaTeX: \boxed{n=100\quad\text{(C)}} ### Verification record - Main value: 100 - Main method: Positive-root solution of the independent-normal-sum percentile equation; continuous root 99.9999699305 - Independent check: Direct recomposition at n=100 gives a 95th percentile of 4098.6912176171 kWh - Independent value: 100 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #725 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-725/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Tail Probabilities Take the ratio of the adjacent Poisson masses to identify the rate as 4. The complement through count 2 then gives an upper-tail probability of approximately 0.761897, corresponding to choice E. ### Independent solution ### Setup Let M be the Poisson count with unknown rate lambda and write its probability mass function. LaTeX: \Pr(M=k)=e^{-\lambda}\frac{\lambda^k}{k!} ### Model For adjacent Poisson masses, their ratio removes the common exponential factor and most powers. LaTeX: \frac{\Pr(M=4)}{\Pr(M=3)}=\frac{\lambda}{4} LaTeX: \Pr(M=3)=\Pr(M=4)\quad\Longrightarrow\quad\frac{\lambda}{4}=1 ### Compute Substitute the identified rate and calculate the requested upper tail by complement. LaTeX: \lambda=4 LaTeX: \Pr(M\ge3)=1-\Pr(M\le2) LaTeX: \Pr(M\ge3)=1-e^{-4}\left(1+4+\frac{4^2}{2}\right)=1-13e^{-4}=0.7618966944\ldots ### Answer The tail probability rounds to the value shown in choice E. LaTeX: \boxed{\Pr(M\ge3)\approx0.762\quad\text{(E)}} ### Distractor analysis - A: This is the single mass P(M=3)=0.195367..., not the probability of the whole upper tail. - B: This is the complement P(M <= 2)=0.238103..., so it reverses the requested event. - C: This is P(M <= 3)=0.433470..., obtained by adding the mass at 3 to choice B rather than taking the upper tail. - D: This is P(M >= 4)=0.566530..., caused by shifting the inclusive cutoff from 3 to 4. ### Verification record - Main value: 0.7618966944464557 - Main method: Adjacent-mass ratio to obtain lambda=4, followed by an exact Poisson complement - Independent check: SciPy Poisson survival function evaluated at cutoff 2 with rate 4 - Independent value: 0.7618966944464557 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: telescope calibration alerts An autonomous telescope models its number of calibration alerts in one night by a Poisson distribution. The probability of exactly two alerts equals the probability of exactly three alerts. Calculate the probability that no more than one alert occurs during a night. - A: 0.0498 - B: 0.1991 - C: 0.2240 - D: 0.4232 - E: 0.8009 Equality of the masses at 2 and 3 forces the Poisson rate to equal 3. Summing the zero- and one-alert probabilities gives approximately 0.1991, so the answer is B. #### Variant solution ### Setup Let N be the nightly alert count with Poisson rate lambda. LaTeX: \Pr(N=k)=e^{-\lambda}\frac{\lambda^k}{k!} ### Model Use the ratio of the two adjacent masses to solve for the rate. LaTeX: \frac{\Pr(N=3)}{\Pr(N=2)}=\frac{\lambda}{3}=1 LaTeX: \lambda=3 ### Compute No more than one alert includes exactly zero and exactly one. LaTeX: \Pr(N\le1)=\Pr(N=0)+\Pr(N=1) LaTeX: \Pr(N\le1)=e^{-3}+3e^{-3}=4e^{-3}=0.1991482735\ldots ### Answer The probability rounds to choice B. LaTeX: \boxed{\Pr(N\le1)\approx0.1991\quad\text{(B)}} ### Verification record - Main value: 0.1991482734714558 - Main method: Adjacent-mass ratio to obtain lambda=3, followed by direct summation of the first two masses - Independent check: SciPy Poisson cumulative distribution function at 1 with rate 3 - Independent value: 0.1991482734714558 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #726 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-726/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Joint Distributions; Conditional Distributions; Expectation and Variance First obtain the marginal distribution of Y and the conditional distribution of Y given X=0, then compute each variance from its first two moments. Their ratio is 279/140, approximately 1.99286, so the correct choice is D. ### Independent solution ### Setup Sum across each row of the joint table to obtain the marginal probabilities for Y. LaTeX: \Pr(Y=0,1,2,3)=\frac{1}{28}(6,8,8,6) LaTeX: \mathbb{E}[Y]=\frac{42}{28}=\frac{3}{2},\qquad \mathbb{E}[Y^2]=\frac{94}{28}=\frac{47}{14} ### Model Use the X=0 column and divide by its column total to form the required conditional distribution. LaTeX: \Pr(Y=0,1,2,3\mid X=0)=\frac{1}{6}(3,2,1,0) LaTeX: \mathbb{E}[Y\mid X=0]=\frac{4}{6}=\frac{2}{3},\qquad \mathbb{E}[Y^2\mid X=0]=\frac{6}{6}=1 ### Compute Convert each pair of raw moments into a variance and divide in the requested order. LaTeX: \operatorname{Var}(Y)=\frac{47}{14}-\left(\frac{3}{2}\right)^2=\frac{31}{28} LaTeX: \operatorname{Var}(Y\mid X=0)=1-\left(\frac{2}{3}\right)^2=\frac{5}{9} LaTeX: \frac{\operatorname{Var}(Y)}{\operatorname{Var}(Y\mid X=0)}=\frac{31/28}{5/9}=\frac{279}{140}=1.992857\ldots ### Answer The ratio rounds to the value listed in choice D. LaTeX: \boxed{1.99\quad\text{(D)}} ### Distractor analysis - A: This is the conditional variance alone, 5/9=0.55556..., rounded to 0.56; it omits the requested division. - B: This is the conditional second moment E[Y^2 | X=0]=1 before subtracting the square of the conditional mean. - C: This is the marginal variance alone, 31/28=1.10714..., rounded to 1.11; it is only the numerator. - E: This is the marginal second moment E[Y^2]=47/14=3.35714..., rounded to 3.36, rather than a ratio of variances. ### Verification record - Main value: 1.992857142857142 - Main method: Exact marginal and conditional moment calculation, yielding (31/28)/(5/9) - Independent check: NumPy enumeration of the complete 4-by-4 joint probability matrix and normalized X=0 column - Independent value: 1.9928571428571424 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: network links and alert bursts At a monitoring center, X is the number of backup links available at the start of a shift and Y is the number of alert bursts during that shift; each takes values 0, 1, or 2. With rows indexed by Y and columns by X, the joint probability matrix is (1/20)[[4,2,1],[2,4,2],[1,2,2]]. Calculate Var(Y)/Var(Y given X=1). - A: 0.50 - B: 0.59 - C: 0.81 - D: 1.18 - E: 1.40 The row totals give a marginal variance of 0.59, while normalizing the middle column gives a conditional variance of 0.50. Dividing them yields 1.18, which is choice D. #### Variant solution ### Setup Sum each row of the joint matrix to obtain the distribution of Y. LaTeX: \Pr(Y=0,1,2)=(0.35,0.40,0.25) LaTeX: \mathbb{E}[Y]=0.90,\qquad \mathbb{E}[Y^2]=1.40 ### Model The X=1 column has weights 2, 4, and 2, so normalize them by their total of 8. LaTeX: \Pr(Y=0,1,2\mid X=1)=(0.25,0.50,0.25) LaTeX: \mathbb{E}[Y\mid X=1]=1,\qquad \mathbb{E}[Y^2\mid X=1]=1.5 ### Compute Calculate both variances and form the ratio. LaTeX: \operatorname{Var}(Y)=1.40-(0.90)^2=0.59 LaTeX: \operatorname{Var}(Y\mid X=1)=1.50-1^2=0.50 LaTeX: \frac{0.59}{0.50}=1.18 ### Answer The ratio is listed in choice D. LaTeX: \boxed{1.18\quad\text{(D)}} ### Verification record - Main value: 1.18 - Main method: Exact marginal and conditional moments from the authored 3-by-3 probability matrix - Independent check: NumPy enumeration of all nine joint outcomes and normalization of the X=1 column - Independent value: 1.1799999999999997 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #727 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-727/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Insurance Payment Variables For an exponential loss, the expected excess above a deductible equals the mean times the survival probability at that deductible. Applying that identity gives a mean of approximately 1790.190, which matches choice B. ### Independent solution ### Setup Represent the insurer's payment as the positive part of the loss after the deductible. LaTeX: d=0.4\beta,\qquad Y=(X-d)_+ ### Model Use the exponential survival function and memoryless property to factor the expected payment into the chance of crossing the deductible and the mean residual loss. LaTeX: \mathbb{E}[Y]=\Pr(X>d)\,\mathbb{E}[X-d\mid X>d] LaTeX: \Pr(X>d)=e^{-d/\beta}=e^{-0.4},\qquad \mathbb{E}[X-d\mid X>d]=\beta ### Compute Equate the stop-loss expectation to the supplied expected benefit and solve for the exponential mean. LaTeX: 1200=\beta e^{-0.4} LaTeX: \beta=1200e^{0.4}=1790.189637\ldots ### Answer The nearest listed amount is 1790. LaTeX: \boxed{\beta\approx 1790.190\quad\text{(B)}} ### Distractor analysis - A: Multiplying 1200 by the survival factor gives 1200e^{-0.4}=804.38; the factor must divide the expected payment when solving for the mean. - C: Using 1-0.4=0.6 as though it were an exponential survival probability gives 1200/0.6=2000, but the correct survival value is e^{-0.4}. - D: Dividing the correct 1790.19 result by a further 0.6 gives 2983.65; that applies an unsupported retained-loss fraction after the deductible has already been handled. - E: Using the probability of falling below the deductible gives 1200/(1-e^{-0.4})=3639.89, which conditions on the wrong side of the threshold. ### Verification record - Main value: 1790.1896371695243 - Main method: Exponential stop-loss identity beta times exp(-d/beta) - Independent check: SciPy quadrature of the excess-loss integral followed by Brent root solving - Independent value: 1790.1896371695245 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ### Original variant: environmental cleanup reimbursement An environmental response fund reimburses 80% of the portion of a chemical-cleanup cost above a trigger. The cleanup cost C is exponential with mean theta, the trigger is 0.2 theta, and the expected reimbursement per incident is 1000 dollars. Determine theta. - A: 819 - B: 1250 - C: 1527 - D: 1908 - E: 6896 Coinsurance multiplies the exponential excess-loss expectation by 0.8. Solving the resulting equation gives theta approximately 1526.753, so choice C is correct. #### Variant solution ### Setup Write the fund's reimbursement using the trigger and the 80% reimbursement rate. LaTeX: R=0.8(C-0.2\theta)_+ ### Model For an exponential cost, the expected excess above 0.2 times the mean is the mean multiplied by the corresponding survival factor. LaTeX: \mathbb{E}[R]=0.8\theta e^{-0.2} ### Compute Set the modeled reimbursement equal to 1000 and isolate the mean. LaTeX: 1000=0.8\theta e^{-0.2} LaTeX: \theta=\frac{1000e^{0.2}}{0.8}=1526.753448\ldots ### Answer Rounding to the listed whole-dollar values selects choice C. LaTeX: \boxed{\theta\approx 1527\quad\text{(C)}} ### Verification record - Main value: 1526.7534477002123 - Main method: Coinsured exponential stop-loss expectation - Independent check: SciPy quadrature of the reimbursement integral followed by Brent root solving - Independent value: 1526.7534477002123 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #728 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-728/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Sums of Independent Random Variables The zero-count probability determines the one-hour Poisson mean, and independent hourly counts add their means over the requested interval. The resulting probability is approximately 0.735758, matching choice A. ### Independent solution ### Setup Let lambda denote the common Poisson mean for one hour and recover it from the probability of a zero count. LaTeX: e^{-\lambda}=0.60653 LaTeX: \lambda=-\log(0.60653)=0.5000010877\ldots ### Model The sum across two independent hours is Poisson with mean equal to the sum of the hourly means. LaTeX: T\sim\operatorname{Poisson}(2\lambda) LaTeX: \mu_T=2\lambda=1.0000021754\ldots ### Compute Fewer than two arrivals means that the two-hour total is either zero or one. LaTeX: \Pr(T<2)=\Pr(T=0)+\Pr(T=1) LaTeX: \Pr(T<2)=e^{-\mu_T}(1+\mu_T)=0.7357580821\ldots ### Answer The computed probability rounds to the first listed value. LaTeX: \boxed{\Pr(T<2)\approx 0.736\quad\text{(A)}} ### Distractor analysis - B: The value 0.78694 is 2(1-0.60653); it doubles the chance of a nonempty hour and treats that expected number of nonempty hours as the requested count probability. - C: The value 0.803265 comes from 0.60653+0.5(1-0.60653), which mixes a zero-count probability with an invalid proxy for a one-count probability. - D: The value 0.909796 is e^{-lambda}(1+lambda), the fewer-than-two probability for one hour; it fails to aggregate both hours. - E: The value 0.919698 is e^{-1}(1+1+1/2), which includes totals of two even though the event is strictly fewer than two. ### Verification record - Main value: 0.7357580820705719 - Main method: Poisson zero-mass inversion and independent-count superposition - Independent check: SciPy Poisson CDF at one using the recovered two-hour mean - Independent value: 0.7357580820705719 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: backup security alerts During an automated overnight backup, security alerts in each 15-minute block follow independent Poisson distributions with mean 0.4. A monitoring window contains three consecutive blocks. Calculate the probability that the window contains at least two alerts. - A: 0.301 - B: 0.337 - C: 0.361 - D: 0.663 - E: 0.699 Three independent blocks produce a Poisson total with mean 1.2. Subtracting the zero- and one-alert probabilities from one gives approximately 0.337373, so choice B is correct. #### Variant solution ### Setup Identify the mean in each block and the number of independent blocks in the monitoring window. LaTeX: \lambda_{\text{block}}=0.4,\qquad m=3 ### Model Add the independent Poisson counts to obtain the distribution of the window total. LaTeX: S\sim\operatorname{Poisson}(3\times 0.4)=\operatorname{Poisson}(1.2) ### Compute Use the complement of zero or one alert. LaTeX: \Pr(S\ge 2)=1-e^{-1.2}(1+1.2) LaTeX: \Pr(S\ge 2)=0.3373727338\ldots ### Answer The closest listed probability is choice B. LaTeX: \boxed{\Pr(S\ge 2)\approx 0.337\quad\text{(B)}} ### Verification record - Main value: 0.33737273379315524 - Main method: Poisson superposition followed by a two-term complement - Independent check: SciPy Poisson survival function at one - Independent value: 0.3373727337931553 - Independent check agrees: yes - Official answer letter: B - Gate agreement: pass ## SOA Exam P Sample Question #729 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-729/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Discrete Random Variables; Conditional Probability Normalizing the geometric-form probability function fixes its constant and reveals a continuation ratio of one third. The required conditional probability is 8/9, or approximately 0.888889, which matches choice E. ### Independent solution ### Setup Normalize the infinite probability sequence before forming the conditional event. LaTeX: 1=6\sum_{n=0}^{\infty}3^{-(n+c)} ### Model Evaluate the geometric series to determine the constant and rewrite the masses in standard geometric form. LaTeX: 1=6\cdot 3^{-c}\frac{1}{1-1/3}=9\cdot 3^{-c} LaTeX: c=2,\qquad p_n=\frac{2}{3}\left(\frac{1}{3}\right)^n ### Compute The numerator consists of counts three and four, while the denominator is the full tail beginning at three. LaTeX: p_3+p_4=\frac{2}{81}+\frac{2}{243}=\frac{8}{243} LaTeX: \Pr(N\ge 3)=\left(\frac{1}{3}\right)^3=\frac{1}{27} LaTeX: \Pr(3\le N<5\mid N\ge3)=\frac{8/243}{1/27}=\frac{8}{9} ### Answer The conditional probability rounds to the final listed value. LaTeX: \boxed{\frac{8}{9}=0.888889\ldots\quad\text{(E)}} ### Distractor analysis - A: The value 0.296296 equals 8/27, which is also p_1+p_2; it shifts the two-count window downward and leaves it unconditional. - B: With the correct numerator 0.0329218, a result of 0.434 would require a conditioning denominator near 0.07586 instead of the geometric tail 1/27=0.037037, indicating a mis-summed tail. - C: The value 0.590 is inconsistent with the tail ratio p_{n+1}/p_n=1/3; after conditioning at three, the chance of remaining within the next two masses must be 1-(1/3)^2=0.888889. - D: A result of 0.704 would use an effective denominator near 0.04676, so it has included probability outside the conditioning event or otherwise failed to renormalize by the exact 1/27 tail. ### Verification record - Main value: 0.8888888888888888 - Main method: Geometric-series normalization and exact conditional-tail ratio - Independent check: Direct floating-point summation of normalized masses through n=999 - Independent value: 0.8888888888888891 - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: greenhouse sensor recalibration A greenhouse controller repeats a sensor calibration until it succeeds. Let K be the number of failed attempts before success, with probability P(K=k)=12/4^(k+r) for k=0,1,2,..., where r makes this a valid distribution. Given that at least one attempt failed, calculate the probability that no more than two attempts failed. - A: 0.234 - B: 0.750 - C: 0.938 - D: 0.984 - E: 0.996 Normalization gives r=2 and a geometric continuation ratio of one fourth. Conditional on at least one failure, allowing one or two failures has probability 15/16=0.9375, so choice C is correct. #### Variant solution ### Setup Use the total probability requirement to identify r. LaTeX: 1=12\sum_{k=0}^{\infty}4^{-(k+r)} ### Model Evaluate the geometric sum and express the resulting distribution with its continuation ratio. LaTeX: 1=12\cdot4^{-r}\frac{1}{1-1/4}=16\cdot4^{-r} LaTeX: r=2,\qquad \Pr(K=k)=\frac{3}{4}\left(\frac{1}{4}\right)^k ### Compute After conditioning on one or more failures, the excess count restarts with the same geometric ratio. LaTeX: \Pr(1\le K\le2\mid K\ge1)=1-\left(\frac14\right)^2 LaTeX: \Pr(1\le K\le2\mid K\ge1)=\frac{15}{16}=0.9375 ### Answer The rounded probability is listed under choice C. LaTeX: \boxed{0.9375\approx0.938\quad\text{(C)}} ### Verification record - Main value: 0.9375 - Main method: Exact geometric normalization and conditional residual-tail calculation - Independent check: Direct summation of the normalized probability sequence through k=999 - Independent value: 0.9375 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #730 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-730/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Percentiles and Quantiles Exponential quantiles convert the stated percentile gap into the distribution mean through a logarithmic ratio. The resulting probability is approximately 0.776868, corresponding to choice A. ### Independent solution ### Setup Write the exponential quantile function in terms of the mean beta. LaTeX: q_p=-\beta\log(1-p) ### Model Subtract the median from the 75th percentile and simplify the logarithms. LaTeX: q_{0.75}-q_{0.50}=\beta\bigl(\log4-\log2\bigr)=\beta\log2 LaTeX: \beta\log2=1.3863 ### Compute Recover the mean and evaluate the exponential distribution function at three. LaTeX: \beta=\frac{1.3863}{\log2}=2.000008135\ldots LaTeX: \Pr(X<3)=1-e^{-3/\beta}=0.7768684784\ldots ### Answer Rounding to three decimals selects choice A. LaTeX: \boxed{\Pr(X<3)\approx0.777\quad\text{(A)}} ### Distractor analysis - B: The value 0.793 implies an exponential mean of about 1.9047 through beta=-3/log(1-p); that mean does not reproduce the given percentile gap. - C: The value 0.810 implies a mean near 1.8064, again contradicting beta log(2)=1.3863. - D: The value 0.864 implies a mean near 1.5037 and results from an incorrect quantile-spacing conversion. - E: Treating 1.3863 itself as the mean gives 1-e^{-3/1.3863}=0.885, but the supplied number is a percentile difference, not beta. ### Verification record - Main value: 0.776868478449211 - Main method: Closed-form exponential quantile difference and CDF evaluation - Independent check: SciPy exponential PPF root solve followed by SciPy CDF - Independent value: 0.7768684784492111 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: protective coating lifetime The time T until a protective coating develops its first microfracture is exponential. Laboratory records show that the 90th percentile exceeds the median by 12.8755 hours. Calculate the probability that a coating develops a microfracture within six hours. - A: 0.372 - B: 0.472 - C: 0.528 - D: 0.658 - E: 0.777 The 90th-percentile-to-median gap equals theta log(5), so the exponential mean is approximately eight hours. Evaluating the six-hour CDF gives approximately 0.527634, which selects choice C. #### Variant solution ### Setup Express the two relevant quantiles using the exponential mean theta. LaTeX: q_{0.90}=-\theta\log(0.10),\qquad q_{0.50}=-\theta\log(0.50) ### Model Take their difference and reduce the logarithmic ratio. LaTeX: q_{0.90}-q_{0.50}=\theta\log\left(\frac{0.50}{0.10}\right)=\theta\log5 LaTeX: \theta\log5=12.8755 ### Compute Solve for the mean and evaluate the probability of failure by six hours. LaTeX: \theta=\frac{12.8755}{\log5}=7.999997950\ldots LaTeX: \Pr(T<6)=1-e^{-6/\theta}=0.5276335380\ldots ### Answer The probability rounds to the value under choice C. LaTeX: \boxed{\Pr(T<6)\approx0.528\quad\text{(C)}} ### Verification record - Main value: 0.5276335380453884 - Main method: Closed-form exponential quantile ratio and CDF - Independent check: SciPy exponential PPF root solve followed by SciPy CDF - Independent value: 0.5276335380453885 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #731 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-731/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Normal Distribution; Linear Transformations The normal percentile identifies an input variance of 25, after which the variance condition fixes the positive scale factor at one. The required output mean then gives an intercept of 30, matching choice D. ### Independent solution ### Setup Standardize the reported percentile to estimate the standard deviation of the raw-score distribution. LaTeX: z_{0.80}=0.841621\ldots LaTeX: \sigma_X=\frac{24.20-20}{z_{0.80}}=4.99037\ldots ### Model Use the integer-variance condition and the moment rules for a positive affine transformation. LaTeX: \operatorname{Var}(X)=25 LaTeX: \operatorname{Var}(Y)=a^2\operatorname{Var}(X)=25 LaTeX: a>0\quad\Longrightarrow\quad a=1 ### Compute Translate the stated relationship between the input variance and output expectation, then solve the affine mean equation. LaTeX: \operatorname{Var}(X)=\frac12\mathbb{E}[Y]\quad\Longrightarrow\quad\mathbb{E}[Y]=50 LaTeX: 50=\mathbb{E}[aX+b]=a\mathbb{E}[X]+b=20+b LaTeX: b=30 ### Answer The intercept is the positive value listed under choice D. LaTeX: \boxed{b=30\quad\text{(D)}} ### Distractor analysis - A: With the verified scale a=1, b=-34.72 would make E[Y]=20-34.72=-14.72 rather than the required 50. - B: The sign-error value b=-30 gives E[Y]=-10; solving 50=20+b requires adding 30, not subtracting it. - C: Setting b=0 preserves the raw mean at 20 and ignores the condition that E[Y]=50. - E: The value b=34.72 would produce E[Y]=54.72, so it is incompatible with the mean constraint even though its sign is positive. ### Verification record - Main value: 30 - Main method: Normal percentile standardization plus affine mean and variance identities - Independent check: SciPy normal CDF check at sigma=5 and direct transformed-moment substitution - Independent value: 30 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ### Original variant: acoustic calibration offset An acoustic laboratory's uncalibrated noise index R is normally distributed with average 40, and 90% of readings do not exceed 52.8155. The lab publishes S=cR+d. Instrument records state that the variance of R is an integer, while calibration requires c to be positive, the standard deviation of S to be 15, and the average of S to be 180. Determine the programmed offset d. - A: -120 - B: 60 - C: 100 - D: 120 - E: 180 The 90th percentile identifies an input standard deviation of 10, so the positive output scale is 15/10=1.5. Matching the required output average then gives d=120, which is choice D. #### Variant solution ### Setup Use the 90th-percentile z value to recover the spread of the uncalibrated index. LaTeX: z_{0.90}=1.2815516\ldots LaTeX: \sigma_R=\frac{52.8155-40}{z_{0.90}}=9.99999\ldots ### Model The integer variance is therefore 100, and a positive affine scale multiplies standard deviation by c. LaTeX: \operatorname{Var}(R)=100 LaTeX: 15=\operatorname{SD}(S)=c\operatorname{SD}(R)=10c LaTeX: c=1.5 ### Compute Apply the output-average requirement to determine the remaining offset. LaTeX: 180=\mathbb{E}[S]=c\mathbb{E}[R]+d LaTeX: 180=1.5(40)+d LaTeX: d=120 ### Answer The laboratory must program the offset shown in choice D. LaTeX: \boxed{d=120\quad\text{(D)}} ### Verification record - Main value: 120 - Main method: Normal percentile inversion followed by affine standard-deviation and mean equations - Independent check: SciPy normal CDF validation and direct substitution of both transformed moments - Independent value: 120 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #732 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-732/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Central Limit Theorem; Sampling Distributions Apply the central limit theorem to the sample average, using an exponential standard deviation equal to its mean. Standardizing the requested cutoff gives z = 1.6 and an upper-tail probability of 0.054799, so the answer is C. ### Independent solution ### Setup An exponential observation with the supplied mean has the same numerical standard deviation. LaTeX: \mu=5,\qquad \sigma=5,\qquad n=64 ### Model For a sample this large, approximate the standardized sample average by a standard normal random variable. LaTeX: \overline X\approx N\!\left(5,\frac{25}{64}\right) LaTeX: \operatorname{SE}(\overline X)=\frac{5}{\sqrt{64}}=0.625 ### Compute Convert the cutoff to a standard-normal score and read the probability to its right. LaTeX: z=\frac{6-5}{0.625}=1.6 LaTeX: \Pr(\overline X>6)\approx 1-\Phi(1.6)=0.0547992917\ldots ### Answer The matching listed probability is choice C. LaTeX: \boxed{\Pr(\overline X>6)\approx 0.0548\quad\text{(C)}} ### Distractor analysis - A: This is far too small for a cutoff only 1.6 standard errors above the mean; it is consistent with using the raw-observation scale in the denominator incorrectly. - B: This resembles the tail beyond about 2.56 rather than 1.6, which can result from dividing by the variance of the mean instead of its standard deviation. - D: This corresponds to a much smaller standardized distance and is tempting when the square-root adjustment for the sample size is only partly applied. - E: This is close to one-half and ignores how far the requested cutoff lies above the center of the sampling distribution. ### Verification record - Main value: 0.054799291699558 - Main method: Central-limit standardization followed by the standard-normal survival function - Independent check: Direct complementary-error-function evaluation, 0.5 erfc(1.6 / sqrt(2)) - Independent value: 0.054799291699558 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: plankton detection intervals A marine laboratory models the time between successive plankton detections by an exponential distribution with mean 8 minutes. During a calibration study, the laboratory records 100 independent intervals. Using the central limit theorem, approximate the probability that their average exceeds 9 minutes. - A: 0.0062 - B: 0.0528 - C: 0.1056 - D: 0.3944 - E: 0.8944 The average of 100 exponential intervals has approximate standard error 0.8 minute. A 9-minute average is 1.25 standard errors above the mean, giving an upper tail of about 0.1056. #### Variant solution ### Setup Record the exponential mean, its equal standard deviation, and the study size. LaTeX: \mu=\sigma=8,\qquad n=100 ### Model Use the central limit theorem for the average of the independent intervals. LaTeX: \overline T\approx N\!\left(8,\frac{64}{100}\right) LaTeX: \operatorname{SE}(\overline T)=\frac{8}{10}=0.8 ### Compute Standardize 9 minutes and evaluate the right-hand normal tail. LaTeX: z=\frac{9-8}{0.8}=1.25 LaTeX: \Pr(\overline T>9)\approx 1-\Phi(1.25)=0.1056497737\ldots ### Answer Choice C is the nearest listed value. LaTeX: \boxed{0.1056\quad\text{(C)}} ### Verification record - Main value: 0.10564977366685528 - Main method: CLT standardization and standard-normal survival probability at z = 1.25 - Independent check: Direct complementary-error-function evaluation, 0.5 erfc(1.25 / sqrt(2)) - Independent value: 0.10564977366685528 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #733 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-733/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: General Probability; Conditional Probability; Set Probability Translate both conditional percentages into equations for the same intersection, which makes the dental-insurance probability 0.625 times the medical-insurance probability. Inclusion–exclusion then makes the requested union-to-medical ratio 1.375, corresponding to A. ### Independent solution ### Setup Let M and D denote the two insurance events and express every probability relative to the size of M. LaTeX: \Pr(D\mid M)=0.25,\qquad \Pr(M\mid D)=0.40 ### Model Both conditional statements describe the same overlap, so equate their two representations. LaTeX: \Pr(M\cap D)=0.25\Pr(M)=0.40\Pr(D) LaTeX: \frac{\Pr(D)}{\Pr(M)}=\frac{0.25}{0.40}=0.625 ### Compute Apply inclusion–exclusion and divide through by the medical-insurance probability. LaTeX: \frac{\Pr(M\cup D)}{\Pr(M)}=1+\frac{\Pr(D)}{\Pr(M)}-\frac{\Pr(M\cap D)}{\Pr(M)} LaTeX: =1+0.625-0.25=1.375 ### Answer The ratio is the value shown in choice A. LaTeX: \boxed{1.375\quad\text{(A)}} ### Distractor analysis - B: This adds the relative size of the dental group to one but forgets to subtract the shared members. - C: This treats the overlap as an additional group instead of a group already counted in both event totals. - D: This can arise after reversing one conditional denominator and then combining incompatible bases. - E: This exceeds even the result of adding the two relative group sizes without a valid overlap correction, signaling that the conditional percentages were inverted. ### Verification record - Main value: 1.375 - Main method: Conditional-probability identities followed by inclusion–exclusion - Independent check: Integer cohort check: 200 medical, 125 dental, 50 both, so 275 / 200 - Independent value: 1.375 - Independent check agrees: yes - Official answer letter: A - Gate agreement: pass ### Original variant: media subscriptions In a reader panel, 30% of audiobook subscribers also subscribe to a documentary service, while 45% of documentary-service subscribers also subscribe to audiobooks. Let U be the probability that a randomly chosen panelist uses at least one of the two services, and let A be the probability that the panelist subscribes to audiobooks. Calculate U/A. - A: 0.300 - B: 0.667 - C: 1.067 - D: 1.367 - E: 1.667 The common overlap is 0.30A and also 0.45 times the documentary probability, so the documentary group is two-thirds the size of A. Inclusion–exclusion gives U/A = 1 + 2/3 - 0.30 = 1.3667. #### Variant solution ### Setup Use A and D for the two subscription events and place both conditional facts on a common scale. LaTeX: \Pr(A\cap D)=0.30\Pr(A)=0.45\Pr(D) ### Model Solve for the documentary probability as a multiple of the audiobook probability. LaTeX: \frac{\Pr(D)}{\Pr(A)}=\frac{0.30}{0.45}=\frac{2}{3} ### Compute Divide the inclusion–exclusion identity by the audiobook probability. LaTeX: \frac{U}{\Pr(A)}=1+\frac{2}{3}-0.30 LaTeX: \frac{U}{\Pr(A)}=1.366666\ldots ### Answer The ratio rounds to choice D. LaTeX: \boxed{1.367\quad\text{(D)}} ### Verification record - Main value: 1.3666666666666667 - Main method: Conditional overlap equations and inclusion–exclusion - Independent check: Cohort check: 90 audiobook, 60 documentary, 27 both; union 123 and 123 / 90 - Independent value: 1.3666666666666667 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ## SOA Exam P Sample Question #734 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-734/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Exponential Distribution; Percentiles For an exponential lifetime with mean θ, the variance is θ² and the percentile rank of a value x is 100F(x). Evaluating the exponential CDF at x = θ² gives 100(1 - e^{-θ}), which is choice E. ### Independent solution ### Setup Write the distribution function and variance using the mean parameter. LaTeX: F(x)=1-e^{-x/\theta}\quad(x\ge 0),\qquad \operatorname{Var}(X)=\theta^2 ### Model The requested percentile rank is the distribution function evaluated at the stated lifetime value, multiplied by 100. LaTeX: \frac{k}{100}=F(\theta^2) ### Compute Substitute θ² into the exponential CDF and simplify its exponent. LaTeX: F(\theta^2)=1-e^{-\theta^2/\theta}=1-e^{-\theta} LaTeX: k=100\left(1-e^{-\theta}\right) ### Answer The symbolic expression matches choice E. LaTeX: \boxed{k=100\left(1-e^{-\theta}\right)\quad\text{(E)}} ### Distractor analysis - A: This keeps only an exponential survival term and also squares the parameter in the exponent rather than dividing the lifetime by the mean. - B: This reports a survival percentage instead of a percentile and therefore omits the leading one from the CDF. - C: This replaces the parameter-dependent standardized lifetime by the fixed value one. - D: This retains θ² in the exponent after substitution, missing the cancellation in θ²/θ. ### Verification record - Main value: 100(1 − exp(−θ)) - Main method: Evaluate the exponential CDF at the variance theta squared - Independent check: Quantile inversion: solve -theta log(1-k/100) = theta squared - Independent value: 100(1 − exp(−θ)) - Independent check agrees: yes - Official answer letter: E - Gate agreement: pass ### Original variant: coating maintenance percentile A protective coating has an exponentially distributed service life with mean 4 months. A maintenance team replaces the coating at 6 months if it has not already failed. The 6-month mark is approximately what percentile of the service-life distribution? - A: 22.3rd - B: 39.3rd - C: 77.7th - D: 86.5th - E: 95.0th Standardizing 6 months by the 4-month mean gives 1.5. The exponential CDF is 1 - e^{-1.5} = 0.77687, so 6 months is about the 77.7th percentile. #### Variant solution ### Setup Use the exponential CDF with the stated mean as its scale. LaTeX: F(x)=1-e^{-x/4} ### Model A percentile rank is 100 times the cumulative probability at the specified service life. LaTeX: k=100F(6) ### Compute Evaluate the standardized time and the corresponding cumulative probability. LaTeX: F(6)=1-e^{-6/4}=1-e^{-1.5} LaTeX: F(6)=0.7768698399\ldots ### Answer The nearest percentile is listed under C. LaTeX: \boxed{77.7\text{th percentile}\quad\text{(C)}} ### Verification record - Main value: 0.7768698398515702 - Main method: Exponential CDF evaluation at 6 with mean 4 - Independent check: Numerical integration of (1/4) exp(-x/4) from 0 to 6 - Independent value: 0.7768698398515702 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #735 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-735/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Multivariate Random Variables; Joint Distributions; Expected Value Marginalizing the joint mass function over the theft coordinate gives fire-loss probabilities 17/60, 20/60, and 23/60, so E[X] = 2.1. The policyholder retains 44% of that loss, producing 0.924 and choice C. ### Independent solution ### Setup Only the fire coordinate enters the requested loss, so first remove the other coordinate by summation. LaTeX: f_X(x)=\sum_{y=1}^{3}\frac{x+y^2}{60} ### Model Evaluate the marginal mass for each possible fire-loss value. LaTeX: f_X(x)=\frac{3x+14}{60} LaTeX: f_X(1)=\frac{17}{60},\quad f_X(2)=\frac{20}{60},\quad f_X(3)=\frac{23}{60} ### Compute Find the fire-loss expectation and multiply by the fraction not reimbursed. LaTeX: \mathbb E[X]=\frac{1(17)+2(20)+3(23)}{60}=2.1 LaTeX: \mathbb E[\text{unreimbursed fire loss}]=(1-0.56)(2.1)=0.924 ### Answer The expected retained amount is choice C. LaTeX: \boxed{0.924\quad\text{(C)}} ### Distractor analysis - A: This is consistent with weighting the marginal probabilities without the loss values, which understates the expected fire loss. - B: This results from an incomplete marginal total, typically omitting part of the highest-loss row or column. - D: This applies the 56% reimbursement rate as though it were the retained fraction. - E: This overweights the larger fire-loss values before applying the retained percentage. ### Verification record - Main value: 0.924 - Main method: Marginalize the joint mass function, compute E[X], then apply the retained fraction - Independent check: Direct double sum of 0.44 x p(x,y) over all nine support points - Independent value: 0.924 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: automated fault diagnosis A repair center observes X electrical faults and Y cosmetic faults in a device, where each count can be 0, 1, or 2 and their joint probability mass is p(x,y) = (x + 2y + 1)/36. Each electrical fault normally requires 50 minutes of diagnosis, but an automated bench completes 60% of that work. Calculate the expected number of technician minutes spent diagnosing electrical faults. - A: 10.000 - B: 20.000 - C: 23.333 - D: 35.000 - E: 58.333 Summing the joint mass over cosmetic faults gives f_X(x) = (x + 3)/12 and E[X] = 7/6. Technicians perform 40% of 50 minutes per electrical fault, so the expected manual time is 20(7/6) = 23.333 minutes. #### Variant solution ### Setup Marginalize over the cosmetic-fault count because the labor calculation depends only on X. LaTeX: f_X(x)=\sum_{y=0}^{2}\frac{x+2y+1}{36} ### Model Simplify the marginal and compute the expected electrical-fault count. LaTeX: f_X(x)=\frac{3x+9}{36}=\frac{x+3}{12} LaTeX: \mathbb E[X]=0\left(\frac{3}{12}\right)+1\left(\frac{4}{12}\right)+2\left(\frac{5}{12}\right)=\frac{7}{6} ### Compute The technician supplies the 40% of diagnosis time not handled by the bench. LaTeX: \text{manual minutes per fault}=0.40(50)=20 LaTeX: \mathbb E[\text{manual minutes}]=20\left(\frac{7}{6}\right)=\frac{70}{3} ### Answer The expected technician time rounds to choice C. LaTeX: \boxed{23.333\text{ minutes}\quad\text{(C)}} ### Verification record - Main value: 23.333333333333332 - Main method: Marginal expectation E[X] = 7/6 followed by 20 minutes of manual work per fault - Independent check: Direct double sum of 20 x p(x,y) over the nine support points - Independent value: 23.333333333333332 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ## SOA Exam P Sample Question #736 Canonical page: https://actuaryproof.inshihub.com/exam-p/sample-736/ Official source: https://www.soa.org/globalassets/assets/files/edu/2026/spring/questions-solutions/2026-05-exam-p-sample-questions.pdf Topics: Univariate Random Variables; Poisson Distribution; Tail Probabilities Use the Poisson second-moment identity to identify the rate, then evaluate the upper tail by complement. The verified value is approximately 0.323324, corresponding to choice C. ### Independent solution ### Setup Write the Poisson mean and variance in terms of its rate parameter. LaTeX: \mathbb{E}[X]=\lambda,\qquad \operatorname{Var}(X)=\lambda ### Model Connect the given second moment to the rate through the variance identity. LaTeX: \mathbb{E}[X^2]=\operatorname{Var}(X)+\mathbb{E}[X]^2=\lambda+\lambda^2 LaTeX: \lambda^2+\lambda=6 ### Compute Only the nonnegative root is admissible for a Poisson rate. Compute the requested tail by subtracting the first three masses. LaTeX: (\lambda-2)(\lambda+3)=0\quad\Longrightarrow\quad\lambda=2 LaTeX: \Pr(X\ge 3)=1-\sum_{k=0}^{2}e^{-2}\frac{2^k}{k!} LaTeX: \Pr(X\ge 3)=1-5e^{-2}=0.3233235838\ldots ### Answer The computed value rounds to the value listed under choice C. LaTeX: \boxed{\Pr(X\ge 3)\approx 0.323324\quad\text{(C)}} ### Distractor analysis - A: This is the probability of four or more events, so it results from shifting the tail cutoff upward by one. - B: This keeps only the probability of exactly three events instead of adding the entire upper tail. - D: This is produced by using 3 as the Poisson rate before evaluating the same tail. - E: This is the probability of two or more events, so it results from shifting the cutoff downward by one. ### Verification record - Main value: 0.3233235838169365 - Main method: Solve the Poisson second-moment equation, then evaluate the upper tail by complement - Independent check: Recursive Poisson masses p(k+1) = 2p(k)/(k+1), subtracted from one through k = 2 - Independent value: 0.3233235838169365 - Independent check agrees: yes - Official answer letter: C - Gate agreement: pass ### Original variant: observatory alert count An observatory models the number of false instrument alerts during a 12-hour shift with a Poisson random variable N. The probability of no alerts during a shift is e^{-1.6}. Calculate the probability that at most two alerts occur during a shift. - A: 0.202 - B: 0.323 - C: 0.525 - D: 0.783 - E: 0.921 The zero-count probability identifies the rate as 1.6. Summing the masses for zero, one, and two gives approximately 0.783. #### Variant solution ### Setup For a Poisson count, the zero-count probability is the exponential of the negative rate. LaTeX: \Pr(N=0)=e^{-\lambda} ### Model Match the supplied zero-count probability to the Poisson formula. LaTeX: e^{-\lambda}=e^{-1.6}\quad\Longrightarrow\quad\lambda=1.6 ### Compute Add the probabilities of zero, one, and two alerts. LaTeX: \Pr(N\le 2)=e^{-1.6}\left(1+1.6+\frac{1.6^2}{2}\right) LaTeX: \Pr(N\le 2)=3.88e^{-1.6}=0.783358\ldots ### Answer The closest listed value is choice D. LaTeX: \boxed{\Pr(N\le 2)\approx 0.783\quad\text{(D)}} ### Verification record - Main value: 0.7833584898192629 - Main method: Sum the Poisson masses for zero, one, and two events at rate 1.6 - Independent check: Poisson recurrence beginning with p(0) = exp(-1.6) - Independent value: 0.7833584898192629 - Independent check agrees: yes - Official answer letter: D - Gate agreement: pass ActuaryProof is not affiliated with, endorsed by, or sponsored by the Society of Actuaries. 'SOA' and exam names are used solely to identify the exams for which these study materials are relevant. All solutions and variant problems are original works.