Verified solution · Exam P

SOA Exam P Sample Question #318 Solution

This Exam P sample reference tests Bayes' Theorem. A positive result has probability 0.90 for the target group and 0.20 for its complement. Weighting by the 0.30 prevalence gives a target-group contribution of 0.27 out of 0.41 total positives, so the posterior is 0.658537 and choice C.

Independent analysis by ActuaryProof · reviewed August 4, 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

Translate the stated error rates into positive-result likelihoods for the two underlying states.

\Pr(+\mid P)=1-0.10=0.90

\Pr(+\mid P^c)=0.20

Model

Multiply each likelihood by its state prevalence to form the two positive-result contributions.

\Pr(P\cap +)=0.30(0.90)=0.27

\Pr(P^c\cap +)=0.70(0.20)=0.14

Compute

Normalize the target-state contribution by the total positive-result probability.

\Pr(P\mid +)=\frac{0.27}{0.27+0.14}

=\frac{27}{41}=0.6585365854\ldots

Answer

The posterior probability rounds to 0.66.

\boxed{0.66\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 multiplies the sensitivity and false-positive rate, 0.90(0.20)=0.18. Those likelihoods belong to mutually exclusive underlying states and should not be multiplied.
BThis repeats the 0.30 prevalence before observing the test result. A positive result changes the probability because the two positive likelihoods differ.
DThis is 1-0.90(0.20)=0.82, the complement of an irrelevant product of the two state-specific likelihoods.
EThis reports the sensitivity P(positive given the target state)=0.90. The requested conditioning direction is reversed.

Original practice · fully worked

Original variant: passenger-selected shuttle type

In a large fleet, 55% of the vehicles are minibuses with 18 seats and 45% are coaches with 42 seats. Every vehicle is carrying a full load. One passenger is selected uniformly from all passengers in the fleet. Calculate the probability that the selected passenger is riding in a coach.

A
0.344
B
0.450
C
0.550
D
0.656
E
0.700

Variant answer in brief

Passenger selection weights each vehicle type by its seat count. The coach contribution is 0.45(42)=18.9 seats per fleet vehicle out of 28.8 total, giving 21/32=0.65625 and choice D.

Setup

A passenger is more likely to come from a vehicle type that contributes more occupied seats. Form the occupied-seat weights.

w_M=0.55(18)=9.9

w_C=0.45(42)=18.9

Model

Conditioning on a uniformly selected passenger normalizes the vehicle-type weights by total occupied seats.

\Pr(C\mid \text{selected passenger})=\frac{w_C}{w_M+w_C}

Compute

Evaluate the normalized coach contribution.

\frac{18.9}{9.9+18.9}=\frac{18.9}{28.8}=\frac{21}{32}=0.65625

Answer

The probability rounds to 0.656.

\boxed{0.656\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #318 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Bayes' Theorem question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: passenger-selected shuttle type

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution