# ActuaryProof

> Verified actuarial exam solutions. Every answer earns its proof.

ActuaryProof publishes independently written, verification-gated solutions for official actuarial exam sample-question references. Official question wording is withheld while permission is not granted. Follow each page's official-source link to read the original question.

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## Citation policy

- Cite the exact problem page when using its independent derivation.
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## Exam P: Probability (P) Exam

- [Exam hub](https://actuaryproof.inshihub.com/exam-p/)
- [Topic index](https://actuaryproof.inshihub.com/exam-p/topics/)
- [Official syllabus](https://www.soa.org/globalassets/assets/files/edu/2026/fall/syllabi/2026-09-p-syllabus.pdf)

### General Probability (23–30%)

- [Topic hub](https://actuaryproof.inshihub.com/exam-p/topics/general-probability/)
- [SOA Exam P Sample Question #590](https://actuaryproof.inshihub.com/exam-p/sample-590/) — 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.
- [SOA Exam P Sample Question #594](https://actuaryproof.inshihub.com/exam-p/sample-594/) — 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.
- [SOA Exam P Sample Question #602](https://actuaryproof.inshihub.com/exam-p/sample-602/) — 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.
- [SOA Exam P Sample Question #603](https://actuaryproof.inshihub.com/exam-p/sample-603/) — 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.
- [SOA Exam P Sample Question #605](https://actuaryproof.inshihub.com/exam-p/sample-605/) — 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.
- [SOA Exam P Sample Question #607](https://actuaryproof.inshihub.com/exam-p/sample-607/) — 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.
- [SOA Exam P Sample Question #620](https://actuaryproof.inshihub.com/exam-p/sample-620/) — 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.
- [SOA Exam P Sample Question #640](https://actuaryproof.inshihub.com/exam-p/sample-640/) — 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.
- [SOA Exam P Sample Question #641](https://actuaryproof.inshihub.com/exam-p/sample-641/) — 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.
- [SOA Exam P Sample Question #644](https://actuaryproof.inshihub.com/exam-p/sample-644/) — 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.
- [SOA Exam P Sample Question #645](https://actuaryproof.inshihub.com/exam-p/sample-645/) — 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.
- [SOA Exam P Sample Question #646](https://actuaryproof.inshihub.com/exam-p/sample-646/) — 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.
- [SOA Exam P Sample Question #647](https://actuaryproof.inshihub.com/exam-p/sample-647/) — 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.
- [SOA Exam P Sample Question #653](https://actuaryproof.inshihub.com/exam-p/sample-653/) — 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.
- [SOA Exam P Sample Question #657](https://actuaryproof.inshihub.com/exam-p/sample-657/) — 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.
- [SOA Exam P Sample Question #658](https://actuaryproof.inshihub.com/exam-p/sample-658/) — 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.
- [SOA Exam P Sample Question #663](https://actuaryproof.inshihub.com/exam-p/sample-663/) — 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.
- [SOA Exam P Sample Question #682](https://actuaryproof.inshihub.com/exam-p/sample-682/) — 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.
- [SOA Exam P Sample Question #683](https://actuaryproof.inshihub.com/exam-p/sample-683/) — 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.
- [SOA Exam P Sample Question #685](https://actuaryproof.inshihub.com/exam-p/sample-685/) — 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.
- [SOA Exam P Sample Question #688](https://actuaryproof.inshihub.com/exam-p/sample-688/) — 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.
- [SOA Exam P Sample Question #691](https://actuaryproof.inshihub.com/exam-p/sample-691/) — 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.
- [SOA Exam P Sample Question #694](https://actuaryproof.inshihub.com/exam-p/sample-694/) — 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.
- [SOA Exam P Sample Question #699](https://actuaryproof.inshihub.com/exam-p/sample-699/) — 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.
- [SOA Exam P Sample Question #701](https://actuaryproof.inshihub.com/exam-p/sample-701/) — 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.
- [SOA Exam P Sample Question #723](https://actuaryproof.inshihub.com/exam-p/sample-723/) — 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.
- [SOA Exam P Sample Question #733](https://actuaryproof.inshihub.com/exam-p/sample-733/) — 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.

### Univariate Random Variables (44–50%)

- [Topic hub](https://actuaryproof.inshihub.com/exam-p/topics/univariate-random-variables/)
- [SOA Exam P Sample Question #579](https://actuaryproof.inshihub.com/exam-p/sample-579/) — 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.
- [SOA Exam P Sample Question #583](https://actuaryproof.inshihub.com/exam-p/sample-583/) — 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.
- [SOA Exam P Sample Question #584](https://actuaryproof.inshihub.com/exam-p/sample-584/) — 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.
- [SOA Exam P Sample Question #585](https://actuaryproof.inshihub.com/exam-p/sample-585/) — Intersecting the two interval events leaves 2<Y<3. Integrating the density gives probability 13/54 for that intersection and 13/27 for the conditioning interval 2<Y<4. Their ratio is 1/2, so choice D is correct.
- [SOA Exam P Sample Question #586](https://actuaryproof.inshihub.com/exam-p/sample-586/) — In thousands, a payment above 5 requires the loss excess over the 0.5 deductible to exceed 5/0.8=6.25. The loss must therefore exceed 6.75, whose exponential survival probability is exp(-6.75/6)=0.32465, so choice C is correct.
- [SOA Exam P Sample Question #588](https://actuaryproof.inshihub.com/exam-p/sample-588/) — 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.
- [SOA Exam P Sample Question #589](https://actuaryproof.inshihub.com/exam-p/sample-589/) — 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.
- [SOA Exam P Sample Question #591](https://actuaryproof.inshihub.com/exam-p/sample-591/) — 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.
- [SOA Exam P Sample Question #592](https://actuaryproof.inshihub.com/exam-p/sample-592/) — 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.
- [SOA Exam P Sample Question #593](https://actuaryproof.inshihub.com/exam-p/sample-593/) — 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.
- [SOA Exam P Sample Question #595](https://actuaryproof.inshihub.com/exam-p/sample-595/) — 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.
- [SOA Exam P Sample Question #596](https://actuaryproof.inshihub.com/exam-p/sample-596/) — 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.
- [SOA Exam P Sample Question #597](https://actuaryproof.inshihub.com/exam-p/sample-597/) — 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.
- [SOA Exam P Sample Question #598](https://actuaryproof.inshihub.com/exam-p/sample-598/) — 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.
- [SOA Exam P Sample Question #599](https://actuaryproof.inshihub.com/exam-p/sample-599/) — An exponential variable with mean mu has percentile q_p=-mu ln(1-p). The 75th and 25th percentiles are mu ln 4 and mu ln(4/3), whose difference simplifies to mu ln 3, so choice D is correct.
- [SOA Exam P Sample Question #600](https://actuaryproof.inshihub.com/exam-p/sample-600/) — 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.
- [SOA Exam P Sample Question #601](https://actuaryproof.inshihub.com/exam-p/sample-601/) — 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.
- [SOA Exam P Sample Question #604](https://actuaryproof.inshihub.com/exam-p/sample-604/) — 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.
- [SOA Exam P Sample Question #608](https://actuaryproof.inshihub.com/exam-p/sample-608/) — This is a binomial probability-ratio problem. The zero-to-one mass relation gives failure-to-success odds of 60, and the zero-to-three mass ratio is therefore 60^3 divided by 20, or 10,800, so choice E is correct.
- [SOA Exam P Sample Question #609](https://actuaryproof.inshihub.com/exam-p/sample-609/) — This is a distribution-function normalization followed by an upper-tail calculation. The endpoint condition gives c=1/30, the loss distribution has F(3.2)=0.448, and payment occurs in the complementary event with probability 0.552, so choice D is correct.
- [SOA Exam P Sample Question #610](https://actuaryproof.inshihub.com/exam-p/sample-610/) — This is an inverse limited-expected-value problem for a uniform loss. Equating m-m^2/4000 to 910 gives candidate roots 1400 and 2600, and only 1400 lies within the loss support, so choice C is correct.
- [SOA Exam P Sample Question #611](https://actuaryproof.inshihub.com/exam-p/sample-611/) — 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.
- [SOA Exam P Sample Question #612](https://actuaryproof.inshihub.com/exam-p/sample-612/) — The median identifies the exponential rate as ln(2)/10. Weighting the two disjoint failure-time windows by their respective payments gives 13.06384325, which rounds to choice A.
- [SOA Exam P Sample Question #613](https://actuaryproof.inshihub.com/exam-p/sample-613/) — 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.
- [SOA Exam P Sample Question #615](https://actuaryproof.inshihub.com/exam-p/sample-615/) — 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.
- [SOA Exam P Sample Question #616](https://actuaryproof.inshihub.com/exam-p/sample-616/) — 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.
- [SOA Exam P Sample Question #618](https://actuaryproof.inshihub.com/exam-p/sample-618/) — 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.
- [SOA Exam P Sample Question #619](https://actuaryproof.inshihub.com/exam-p/sample-619/) — 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.
- [SOA Exam P Sample Question #621](https://actuaryproof.inshihub.com/exam-p/sample-621/) — 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.
- [SOA Exam P Sample Question #622](https://actuaryproof.inshihub.com/exam-p/sample-622/) — 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.
- [SOA Exam P Sample Question #623](https://actuaryproof.inshihub.com/exam-p/sample-623/) — 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.
- [SOA Exam P Sample Question #625](https://actuaryproof.inshihub.com/exam-p/sample-625/) — 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.
- [SOA Exam P Sample Question #626](https://actuaryproof.inshihub.com/exam-p/sample-626/) — 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.
- [SOA Exam P Sample Question #628](https://actuaryproof.inshihub.com/exam-p/sample-628/) — 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.
- [SOA Exam P Sample Question #629](https://actuaryproof.inshihub.com/exam-p/sample-629/) — 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.
- [SOA Exam P Sample Question #630](https://actuaryproof.inshihub.com/exam-p/sample-630/) — 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.
- [SOA Exam P Sample Question #632](https://actuaryproof.inshihub.com/exam-p/sample-632/) — 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.
- [SOA Exam P Sample Question #639](https://actuaryproof.inshihub.com/exam-p/sample-639/) — 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.
- [SOA Exam P Sample Question #642](https://actuaryproof.inshihub.com/exam-p/sample-642/) — 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.
- [SOA Exam P Sample Question #643](https://actuaryproof.inshihub.com/exam-p/sample-643/) — 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.
- [SOA Exam P Sample Question #648](https://actuaryproof.inshihub.com/exam-p/sample-648/) — 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.
- [SOA Exam P Sample Question #649](https://actuaryproof.inshihub.com/exam-p/sample-649/) — Integrating the density gives cumulative distribution 1 minus exp(-4x squared). Setting this equal to one-half yields a median of 0.416277, which rounds to choice C.
- [SOA Exam P Sample Question #654](https://actuaryproof.inshihub.com/exam-p/sample-654/) — 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.
- [SOA Exam P Sample Question #655](https://actuaryproof.inshihub.com/exam-p/sample-655/) — 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.
- [SOA Exam P Sample Question #659](https://actuaryproof.inshihub.com/exam-p/sample-659/) — 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.
- [SOA Exam P Sample Question #660](https://actuaryproof.inshihub.com/exam-p/sample-660/) — 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.
- [SOA Exam P Sample Question #661](https://actuaryproof.inshihub.com/exam-p/sample-661/) — 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.
- [SOA Exam P Sample Question #664](https://actuaryproof.inshihub.com/exam-p/sample-664/) — 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.
- [SOA Exam P Sample Question #665](https://actuaryproof.inshihub.com/exam-p/sample-665/) — 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.
- [SOA Exam P Sample Question #666](https://actuaryproof.inshihub.com/exam-p/sample-666/) — 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.
- [SOA Exam P Sample Question #667](https://actuaryproof.inshihub.com/exam-p/sample-667/) — 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.
- [SOA Exam P Sample Question #668](https://actuaryproof.inshihub.com/exam-p/sample-668/) — 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.
- [SOA Exam P Sample Question #669](https://actuaryproof.inshihub.com/exam-p/sample-669/) — 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.
- [SOA Exam P Sample Question #670](https://actuaryproof.inshihub.com/exam-p/sample-670/) — 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.
- [SOA Exam P Sample Question #672](https://actuaryproof.inshihub.com/exam-p/sample-672/) — 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.
- [SOA Exam P Sample Question #674](https://actuaryproof.inshihub.com/exam-p/sample-674/) — 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.
- [SOA Exam P Sample Question #675](https://actuaryproof.inshihub.com/exam-p/sample-675/) — 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.
- [SOA Exam P Sample Question #677](https://actuaryproof.inshihub.com/exam-p/sample-677/) — 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.
- [SOA Exam P Sample Question #678](https://actuaryproof.inshihub.com/exam-p/sample-678/) — 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.
- [SOA Exam P Sample Question #679](https://actuaryproof.inshihub.com/exam-p/sample-679/) — 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.
- [SOA Exam P Sample Question #680](https://actuaryproof.inshihub.com/exam-p/sample-680/) — 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.
- [SOA Exam P Sample Question #684](https://actuaryproof.inshihub.com/exam-p/sample-684/) — 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.
- [SOA Exam P Sample Question #693](https://actuaryproof.inshihub.com/exam-p/sample-693/) — 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.
- [SOA Exam P Sample Question #695](https://actuaryproof.inshihub.com/exam-p/sample-695/) — 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.
- [SOA Exam P Sample Question #696](https://actuaryproof.inshihub.com/exam-p/sample-696/) — 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.
- [SOA Exam P Sample Question #697](https://actuaryproof.inshihub.com/exam-p/sample-697/) — 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.
- [SOA Exam P Sample Question #698](https://actuaryproof.inshihub.com/exam-p/sample-698/) — 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.
- [SOA Exam P Sample Question #700](https://actuaryproof.inshihub.com/exam-p/sample-700/) — 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.
- [SOA Exam P Sample Question #703](https://actuaryproof.inshihub.com/exam-p/sample-703/) — 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.
- [SOA Exam P Sample Question #704](https://actuaryproof.inshihub.com/exam-p/sample-704/) — 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.
- [SOA Exam P Sample Question #705](https://actuaryproof.inshihub.com/exam-p/sample-705/) — 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.
- [SOA Exam P Sample Question #706](https://actuaryproof.inshihub.com/exam-p/sample-706/) — Write both percentile conditions with the exponential CDF and eliminate the percentile level. The two survival values sum to one, giving y=-10 log(1-exp(-x/10)), which is choice A.
- [SOA Exam P Sample Question #707](https://actuaryproof.inshihub.com/exam-p/sample-707/) — Write the claim payment as the positive part of the loss after the deductible. Integrating its triangular tail gives an expected payment of (500-d)^2/1000; setting this equal to 90 yields d = 200, so the verified answer is choice B.
- [SOA Exam P Sample Question #708](https://actuaryproof.inshihub.com/exam-p/sample-708/) — 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.
- [SOA Exam P Sample Question #709](https://actuaryproof.inshihub.com/exam-p/sample-709/) — The unknown proportionality constant cancels when the upper-tail integral is divided by the integral over the full support. The ratio is ln(5)/ln(10) = 0.698970, which rounds to 0.70 and matches choice D.
- [SOA Exam P Sample Question #710](https://actuaryproof.inshihub.com/exam-p/sample-710/) — 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.
- [SOA Exam P Sample Question #711](https://actuaryproof.inshihub.com/exam-p/sample-711/) — 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.
- [SOA Exam P Sample Question #712](https://actuaryproof.inshihub.com/exam-p/sample-712/) — 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.
- [SOA Exam P Sample Question #713](https://actuaryproof.inshihub.com/exam-p/sample-713/) — Integrate the density shape and divide by its total integral. Setting the resulting CDF equal to one-half gives m=sqrt(sqrt(26)-1)=2.0246, which matches choice A.
- [SOA Exam P Sample Question #714](https://actuaryproof.inshihub.com/exam-p/sample-714/) — Integrating after the substitution u=ln x gives F(x)=(ln x/ln 80)^2. Inverting F(x)=0.75 yields x=80^{sqrt(0.75)}=44.4759, corresponding to choice C.
- [SOA Exam P Sample Question #715](https://actuaryproof.inshihub.com/exam-p/sample-715/) — Weight each office's relative frequencies by its share of combined sales. The combined cumulative probability is 0.41 through 160 and 0.51 through 165, so the median is 165 and choice B is correct.
- [SOA Exam P Sample Question #718](https://actuaryproof.inshihub.com/exam-p/sample-718/) — Two adjacent Poisson modes at 2 and 3 force the associated rate to equal 3, and the stated mean ratio therefore makes the other rate 12. A Poisson distribution with integer rate 12 has modes 11 and 12, so the verified answer is choice D.
- [SOA Exam P Sample Question #719](https://actuaryproof.inshihub.com/exam-p/sample-719/) — 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.
- [SOA Exam P Sample Question #720](https://actuaryproof.inshihub.com/exam-p/sample-720/) — 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.
- [SOA Exam P Sample Question #722](https://actuaryproof.inshihub.com/exam-p/sample-722/) — 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.
- [SOA Exam P Sample Question #725](https://actuaryproof.inshihub.com/exam-p/sample-725/) — 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.
- [SOA Exam P Sample Question #727](https://actuaryproof.inshihub.com/exam-p/sample-727/) — 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.
- [SOA Exam P Sample Question #728](https://actuaryproof.inshihub.com/exam-p/sample-728/) — 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.
- [SOA Exam P Sample Question #729](https://actuaryproof.inshihub.com/exam-p/sample-729/) — 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.
- [SOA Exam P Sample Question #730](https://actuaryproof.inshihub.com/exam-p/sample-730/) — 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.
- [SOA Exam P Sample Question #731](https://actuaryproof.inshihub.com/exam-p/sample-731/) — 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.
- [SOA Exam P Sample Question #732](https://actuaryproof.inshihub.com/exam-p/sample-732/) — 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.
- [SOA Exam P Sample Question #734](https://actuaryproof.inshihub.com/exam-p/sample-734/) — 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.
- [SOA Exam P Sample Question #736](https://actuaryproof.inshihub.com/exam-p/sample-736/) — 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.

### Multivariate Random Variables (23–30%)

- [Topic hub](https://actuaryproof.inshihub.com/exam-p/topics/multivariate-random-variables/)
- [SOA Exam P Sample Question #577](https://actuaryproof.inshihub.com/exam-p/sample-577/) — 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.
- [SOA Exam P Sample Question #578](https://actuaryproof.inshihub.com/exam-p/sample-578/) — 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.
- [SOA Exam P Sample Question #580](https://actuaryproof.inshihub.com/exam-p/sample-580/) — 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.
- [SOA Exam P Sample Question #581](https://actuaryproof.inshihub.com/exam-p/sample-581/) — 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.
- [SOA Exam P Sample Question #582](https://actuaryproof.inshihub.com/exam-p/sample-582/) — 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.
- [SOA Exam P Sample Question #587](https://actuaryproof.inshihub.com/exam-p/sample-587/) — 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.
- [SOA Exam P Sample Question #606](https://actuaryproof.inshihub.com/exam-p/sample-606/) — 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.
- [SOA Exam P Sample Question #614](https://actuaryproof.inshihub.com/exam-p/sample-614/) — 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.
- [SOA Exam P Sample Question #617](https://actuaryproof.inshihub.com/exam-p/sample-617/) — 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.
- [SOA Exam P Sample Question #624](https://actuaryproof.inshihub.com/exam-p/sample-624/) — 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.
- [SOA Exam P Sample Question #627](https://actuaryproof.inshihub.com/exam-p/sample-627/) — 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.
- [SOA Exam P Sample Question #631](https://actuaryproof.inshihub.com/exam-p/sample-631/) — 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.
- [SOA Exam P Sample Question #633](https://actuaryproof.inshihub.com/exam-p/sample-633/) — 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.
- [SOA Exam P Sample Question #634](https://actuaryproof.inshihub.com/exam-p/sample-634/) — 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.
- [SOA Exam P Sample Question #635](https://actuaryproof.inshihub.com/exam-p/sample-635/) — 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.
- [SOA Exam P Sample Question #636](https://actuaryproof.inshihub.com/exam-p/sample-636/) — 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.
- [SOA Exam P Sample Question #637](https://actuaryproof.inshihub.com/exam-p/sample-637/) — This is a central-limit approximation for a sum of 303 independent daily counts. The standardized cutoff is 0.044191, giving an upper-tail probability of 0.482376 and selecting choice B.
- [SOA Exam P Sample Question #638](https://actuaryproof.inshihub.com/exam-p/sample-638/) — 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.
- [SOA Exam P Sample Question #650](https://actuaryproof.inshihub.com/exam-p/sample-650/) — 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.
- [SOA Exam P Sample Question #651](https://actuaryproof.inshihub.com/exam-p/sample-651/) — 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.
- [SOA Exam P Sample Question #652](https://actuaryproof.inshihub.com/exam-p/sample-652/) — 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.
- [SOA Exam P Sample Question #656](https://actuaryproof.inshihub.com/exam-p/sample-656/) — 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.
- [SOA Exam P Sample Question #662](https://actuaryproof.inshihub.com/exam-p/sample-662/) — 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.
- [SOA Exam P Sample Question #671](https://actuaryproof.inshihub.com/exam-p/sample-671/) — 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.
- [SOA Exam P Sample Question #673](https://actuaryproof.inshihub.com/exam-p/sample-673/) — 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.
- [SOA Exam P Sample Question #676](https://actuaryproof.inshihub.com/exam-p/sample-676/) — 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.
- [SOA Exam P Sample Question #681](https://actuaryproof.inshihub.com/exam-p/sample-681/) — 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.
- [SOA Exam P Sample Question #686](https://actuaryproof.inshihub.com/exam-p/sample-686/) — 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.
- [SOA Exam P Sample Question #687](https://actuaryproof.inshihub.com/exam-p/sample-687/) — 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.
- [SOA Exam P Sample Question #689](https://actuaryproof.inshihub.com/exam-p/sample-689/) — 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.
- [SOA Exam P Sample Question #690](https://actuaryproof.inshihub.com/exam-p/sample-690/) — 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.
- [SOA Exam P Sample Question #692](https://actuaryproof.inshihub.com/exam-p/sample-692/) — 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.
- [SOA Exam P Sample Question #702](https://actuaryproof.inshihub.com/exam-p/sample-702/) — 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.
- [SOA Exam P Sample Question #716](https://actuaryproof.inshihub.com/exam-p/sample-716/) — First average the conditional low-cost probabilities over the market states to obtain 0.32. The complementary high-cost probability is 0.68, so the expected cost is 5(0.32)+10(0.68)=8.4 million, choice D.
- [SOA Exam P Sample Question #717](https://actuaryproof.inshihub.com/exam-p/sample-717/) — Restricting the joint distribution to the stated event gives conditional masses 0.125, 0.625, and 0.250 at values 0, 25, and 50, respectively. The resulting conditional variance is 224.609375 in millions of squared monetary units, which rounds to choice A.
- [SOA Exam P Sample Question #721](https://actuaryproof.inshihub.com/exam-p/sample-721/) — 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.
- [SOA Exam P Sample Question #724](https://actuaryproof.inshihub.com/exam-p/sample-724/) — 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.
- [SOA Exam P Sample Question #726](https://actuaryproof.inshihub.com/exam-p/sample-726/) — 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.
- [SOA Exam P Sample Question #735](https://actuaryproof.inshihub.com/exam-p/sample-735/) — 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.

## Dataset status

- Verified problem pages live: 160
- Rights mode: official stems withheld

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.