Worked solution · Exam P

SOA Exam P Sample Question #309 Solution

This Exam P sample reference tests Bayes' Theorem. Zone A contributes 0.40(0.015)=0.006 to the fire-loss probability. The three zones contribute 0.01185 in total, so Bayes' theorem gives 0.006/0.01185=0.506329 and choice E.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Bayes' Theorem question

Setup

Multiply each zone's policy share by its annual fire-loss rate.

w_A=0.40(0.015)=0.006

w_B=0.35(0.011)=0.00385,\qquad w_C=0.25(0.008)=0.002

Model

The total fire-loss probability is the sum of the three disjoint zone contributions.

\Pr(F)=w_A+w_B+w_C=0.01185

Compute

Normalize Zone A's contribution by the probability of the observed fire loss.

\Pr(A\mid F)=\frac{w_A}{\Pr(F)}=\frac{0.006}{0.01185}

=\frac{40}{79}=0.5063291139\ldots

Answer

The posterior zone probability rounds to 0.506.

\boxed{0.506\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 0.349 is essentially the 0.35 prior share of policies in Zone B. It uses the wrong zone and does not condition on a fire loss.
BThe value 0.400 is Zone A's prior policy share. It ignores the differing zone-specific fire rates.
CThe value 0.441 is 0.015/(0.015+0.011+0.008). It normalizes the three fire rates as though the zones had equal policy shares.
DWith the verified numerator 0.006, a posterior of 0.465 would require total fire probability about 0.012903. The prior-weighted zone contributions instead sum to 0.01185, so this choice reflects incompatible weights.

Original practice · fully worked

Original variant: infer whether paired outputs are linked

A test unit operates in linked mode with probability 0.30 and independent mode with probability 0.70. It emits two binary outputs. In linked mode the outputs are always identical, with 0 and 1 equally likely. In independent mode the outputs are independent fair bits. The two observed outputs match. Calculate the conditional probability that the unit was in linked mode.

A
0.3000
B
0.3500
C
0.4615
D
0.5000
E
0.5385

Answer and calculation

A match is certain in linked mode and has probability 1/2 in independent mode. The prior-weighted match probabilities are 0.30 and 0.35, so the linked-mode posterior is 0.30/0.65=6/13, approximately 0.4615 and choice C.

Setup

Let M denote the event that the two outputs match.

\Pr(M\mid L)=1

Model

In independent mode, the matching patterns 00 and 11 are two of four equally likely patterns.

\Pr(M\mid I)=\frac24=\frac12

\Pr(L\cap M)=0.30,\qquad \Pr(I\cap M)=0.70\left(\frac12\right)=0.35

Compute

Normalize the linked-mode contribution over all matching outputs.

\Pr(L\mid M)=\frac{0.30}{0.30+0.35}=\frac6{13}=0.46153846\ldots

Answer

The conditional linked-mode probability rounds to 0.4615.

\boxed{0.4615\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: infer whether paired outputs are linked

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability