Worked solution · Exam P

SOA Exam P Sample Question #359 Solution

This Exam P sample reference tests Bayes' Theorem. This is a Bayes calculation using committee membership as the partition. Committee Y contributes 0.36 to the male probability, and the two committees contribute 0.48 in total. The posterior share from Y is therefore 0.75, choice E.

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 Bayes' Theorem question

Setup

Complement the supplied female rate within committee Y and the committee-Y marginal.

\Pr(M\mid Y)=1-0.40=0.60

\Pr(X)=1-0.60=0.40

Model

Compute the male contributions from the two committees.

\Pr(M\cap Y)=0.60(0.60)=0.36

\Pr(M\cap X)=0.30(0.40)=0.12

Compute

Normalize committee Y's contribution by the total male probability.

\Pr(M)=0.36+0.12=0.48

\Pr(Y\mid M)=\frac{0.36}{0.48}=0.75

Answer

Given that the person is male, the probability of committee Y is 0.75.

\boxed{0.75\quad\text{(E)}}

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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 0.36 is the joint probability of being male and in committee Y, the Bayes numerator before conditioning on all male members.
BThe value 0.48 is the total male probability, the denominator rather than the posterior ratio.
CThe value 0.60 is both the committee-Y proportion and the male rate within committee Y. Neither a prior nor a likelihood alone is the requested posterior.
DThe value 0.67 comes from normalizing the likelihoods 0.60 and 0.30 without their committee prior probabilities.

Original practice · fully worked

Original variant: source crate after two silver samples

A quality inspector selects crate A with probability 0.40 and crate B with probability 0.60. Crate A contains three silver and one black component; crate B contains two silver and two black components. Two components are drawn without replacement from the selected crate and both are silver. Calculate the conditional probability that crate A was selected.

A
0.100
B
0.200
C
0.400
D
0.600
E
0.667

Answer and calculation

The prior-weighted probabilities of drawing two silver components are 0.20 for crate A and 0.10 for crate B. Crate A therefore supplies two thirds of all such draws, choice E.

Setup

Calculate the probability of two silver components under each crate.

\Pr(SS\mid A)=\frac34\frac23=\frac12

\Pr(SS\mid B)=\frac24\frac13=\frac16

Model

Weight each likelihood by its crate-selection probability.

\Pr(A\cap SS)=0.40\left(\frac12\right)=0.20

\Pr(B\cap SS)=0.60\left(\frac16\right)=0.10

Compute

Normalize crate A's contribution by the total evidence probability.

\Pr(A\mid SS)=\frac{0.20}{0.20+0.10}=\frac23=0.6666666667

Answer

The posterior probability of crate A is two thirds.

\boxed{0.667\quad\text{(E)}}

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How to solve this Bayes' Theorem question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: source crate after two silver samples

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability