Worked solution · Exam P

SOA Exam P Sample Question #316 Solution

This Exam P sample reference tests Bayes' Theorem. Normalizing the three proportional category probabilities gives priors 0.5, 0.4, and 0.1. Their zero-count contributions are 0.05, 0.08, and 0.04, so the requested posterior is 0.04/0.17=0.235294 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 Bayes' Theorem question

Setup

Normalize the category probabilities from their stated ratios.

\Pr(C)=c,\quad \Pr(A)=5c,\quad \Pr(B)=4c

10c=1\implies (\Pr(A),\Pr(B),\Pr(C))=(0.5,0.4,0.1)

Model

Multiply each category prior by its conditional zero-count probability.

w_A=0.5(0.1)=0.05,\qquad w_B=0.4(0.2)=0.08

w_C=0.1(0.4)=0.04

Compute

Normalize category C's contribution by the total probability of zero claims.

\Pr(C\mid 0)=\frac{w_C}{w_A+w_B+w_C}

=\frac{0.04}{0.05+0.08+0.04}=\frac4{17}=0.2352941176\ldots

Answer

The conditional probability rounds to 0.235.

\boxed{0.235\quad\text{(C)}}

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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.

AThis is the joint contribution P(C and zero claims)=0.1(0.4)=0.04. It has not been divided by the total zero-claim probability.
BThis is the evidence probability P(zero claims)=0.05+0.08+0.04=0.17, which is the denominator rather than the requested posterior.
DThis is the posterior probability of category A, 0.05/0.17=0.294118. It normalizes the wrong category contribution.
EThis is the posterior probability of category B, 0.08/0.17=0.470588. It likewise uses the wrong numerator.

Original practice · fully worked

Original variant: mode inference from a continuous score

A signal is generated in mode R, S, or T with prior probabilities 0.45, 0.35, and 0.20. Conditional on the mode, its score Z lies between 0 and 1 with densities f_R(z)=2z, f_S(z)=2(1-z), and f_T(z)=1. A score of 0.80 is observed. Calculate the posterior probability that the signal used mode T.

A
0.132
B
0.189
C
0.200
D
0.450
E
0.679

Answer and calculation

At the observed score, the prior-weighted density contributions of modes R, S, and T are 0.72, 0.14, and 0.20. Normalizing T's contribution gives 0.20/1.06=0.188679, so choice B.

Setup

Evaluate each conditional density at the observed score.

f_R(0.8)=1.6,\qquad f_S(0.8)=0.4,\qquad f_T(0.8)=1

Model

For a continuous observation, Bayes' theorem uses likelihood densities. Multiply each density by its mode prior.

w_R=0.45(1.6)=0.72

w_S=0.35(0.4)=0.14,\qquad w_T=0.20(1)=0.20

Compute

Normalize mode T's contribution by the mixture density at the observation.

\Pr(T\mid Z=0.8)=\frac{0.20}{0.72+0.14+0.20}

=\frac{10}{53}=0.1886792453\ldots

Answer

The posterior probability rounds to 0.189.

\boxed{0.189\quad\text{(B)}}

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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: mode inference from a continuous score

Answer and calculation

Setup

Model

Compute

Answer

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Next steps for Exam P General Probability