Worked solution · Exam P

SOA Exam P Sample Question #264 Solution

This Exam P sample reference tests Bayes' Theorem. Female contributors supply probability mass 0.45(0.20)=0.09, while male contributors supply 0.55(0.30)=0.165. Bayes' theorem gives 0.09/(0.09+0.165)=6/17=0.352941, so 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 Bayes' Theorem question

Setup

Let F identify a female employee and C identify plan contribution.

\Pr(F)=0.45,\qquad \Pr(F^c)=0.55

Model

Weight each category-specific contribution rate by its employee share.

\Pr(F\cap C)=0.45(0.20)=0.09

\Pr(F^c\cap C)=0.55(0.30)=0.165

Compute

Normalize the female-contributor mass over all contributors.

\Pr(F\mid C)=\frac{0.09}{0.09+0.165}

=\frac6{17}=0.3529411765

Answer

The conditional probability rounds to 0.35.

\boxed{0.35\quad\text{(C)}}

Test the method on unseen problems

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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 0.09 is the joint probability of being female and contributing. It is the Bayes numerator, not the conditional probability.
BThe value 0.23 is 0.09/(0.09+0.30) rounded. It uses the male contribution rate in the denominator without multiplying by the male employee share.
DThe value 0.45 is the unconditional female share. It ignores the different contribution rates.
EThe value 0.55 is the unconditional male share, so it is both the wrong category and unconditioned.

Original practice · fully worked

Original variant: tray source after unlike materials

A routing device selects tray A with probability 0.40 and tray B otherwise. Tray A contains four copper pieces and two glass pieces; tray B contains one copper piece and five glass pieces. Two pieces are drawn without replacement from the selected tray, and their materials are different. Calculate the conditional probability that tray A was selected.

A
0.3333
B
0.4000
C
0.5161
D
0.5333
E
0.6000

Answer and calculation

The unlike-material likelihoods are 8/15 for tray A and 1/3 for tray B. Prior weighting gives contributions 16/75 and 15/75, so the posterior tray-A probability is 16/31=0.5161 and choice C.

Setup

Let D denote the observation that the two materials differ. Compute its likelihood under each tray.

\Pr(D\mid A)=2\left(\frac46\right)\left(\frac25\right)=\frac8{15}

\Pr(D\mid B)=2\left(\frac16\right)\left(\frac55\right)=\frac13

Model

Weight the two unlike-material likelihoods by the tray-selection probabilities.

\Pr(A\cap D)=0.40\left(\frac8{15}\right)=\frac{16}{75}

\Pr(B\cap D)=0.60\left(\frac13\right)=\frac{15}{75}

Compute

Normalize the tray-A contribution over both possible sources of unlike materials.

\Pr(A\mid D)=\frac{16/75}{16/75+15/75}

=\frac{16}{31}=0.5161290323

Answer

The posterior tray-A probability rounds to 0.5161.

\boxed{0.5161\quad\text{(C)}}

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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: tray source after unlike materials

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability