Worked solution · Exam P

SOA Exam P Sample Question #262 Solution

This Exam P sample reference tests Binomial Distribution. Taking the ratio of the adjacent Binomial(100,p) masses cancels their large common factors. The condition becomes (98/3)p/(1-p)=2, giving p=3/52=0.0576923 and choice C.

Independent analysis by ActuaryProof · reviewed July 31, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Binomial Distribution question

Setup

Let X be the defective count in a shipment and p the one-unit defect probability.

X\sim\operatorname{Binomial}(100,p)

Model

Express the given comparison as a ratio of adjacent binomial masses.

\frac{\Pr(X=3)}{\Pr(X=2)}=\frac{\binom{100}{3}p^3(1-p)^{97}}{\binom{100}{2}p^2(1-p)^{98}}=2

Compute

Cancel the common factors and solve the remaining linear equation.

\frac{98}{3}\frac{p}{1-p}=2

98p=6(1-p),\qquad p=\frac6{104}=\frac3{52}=0.0576923077

Answer

The defect probability rounds to 0.058.

\boxed{0.058\quad\text{(C)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThe value 0.040 results from setting the expected count to four by combining the referenced count two with the factor of two, then dividing by 100. The statement compares probability masses, not the mean count.
BStarting from that erroneous 0.040 rate and then reporting its odds gives 0.040/0.960=0.04167, or 0.042. Neither step uses the adjacent-mass ratio.
DA Poisson shortcut gives P(X=3)/P(X=2)=λ/3=2, hence λ=6 and p=λ/100=0.060. That is only an approximation; the exact binomial finite-n factor gives 0.057692.
EAt p=0.072, the exact adjacent-mass ratio is (98/3)(0.072/0.928)=2.534, not 2. This choice reflects an incorrect binomial-coefficient or trial-count factor.

Original practice · fully worked

Original variant: count spread from a conditional audit

A review batch contains eight independent records, each accepted with the same unknown probability p. Let X be the accepted-record count. Conditional on X being either 2 or 3, an audit reports probability 0.60 that X equals 3. Calculate the standard deviation of X.

A
0.4286
B
0.6000
C
1.3997
D
1.9592
E
3.4286

Answer and calculation

The conditional audit implies adjacent-mass ratio P(X=3)/P(X=2)=0.60/0.40=3/2. For Binomial(8,p) that ratio is 2p/(1-p), giving p=3/7 and standard deviation √(96/49)=1.3997, choice C.

Setup

Model the accepted-record count as binomial with eight trials.

X\sim\operatorname{Binomial}(8,p)

Model

Convert the conditional audit percentage into odds between the two adjacent counts.

\frac{\Pr(X=3)}{\Pr(X=2)}=\frac{0.60}{0.40}=\frac32

\frac{\Pr(X=3)}{\Pr(X=2)}=\frac{\binom83p^3(1-p)^5}{\binom82p^2(1-p)^6}=2\frac{p}{1-p}

Compute

Solve for the acceptance probability, then compute and square-root the binomial variance.

2\frac{p}{1-p}=\frac32,\qquad p=\frac37

\operatorname{Var}(X)=8\left(\frac37\right)\left(\frac47\right)=\frac{96}{49}

\operatorname{SD}(X)=\frac{4\sqrt6}{7}=1.3997084244

Answer

The accepted-record count has standard deviation about 1.3997.

\boxed{1.3997\quad\text{(C)}}

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How to solve this Binomial Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: count spread from a conditional audit

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables