Worked solution · Exam P

SOA Exam P Sample Question #572 Solution

This Exam P sample reference tests Bayes' Theorem. The zero-count likelihoods are exp(-0.5) and exp(-1) for the two intersection classes. Bayes' theorem gives 0.4exp(-0.5)/[0.4exp(-0.5)+0.6exp(-1)]=0.5236, so choice D is correct.

Independent analysis by ActuaryProof · reviewed July 30, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

Independent solution

How to solve this Bayes' Theorem question

Setup

Let S identify the lower-rate class and let N be the observed annual count. Record the prior probabilities and the zero-count likelihoods.

\Pr(S)=0.4,\qquad \Pr(S^c)=0.6

\Pr(N=0\mid S)=e^{-0.5},\qquad \Pr(N=0\mid S^c)=e^{-1}

Model

Apply Bayes' theorem using the zero observation as the evidence.

\Pr(S\mid N=0)=\frac{\Pr(N=0\mid S)\Pr(S)}{\Pr(N=0\mid S)\Pr(S)+\Pr(N=0\mid S^c)\Pr(S^c)}

Compute

Substitute the two Poisson likelihoods and prior class weights.

\Pr(S\mid N=0)=\frac{0.4e^{-0.5}}{0.4e^{-0.5}+0.6e^{-1}}

\Pr(S\mid N=0)=0.5236161378\ldots

Answer

The posterior probability rounds to 0.52.

\boxed{0.52\quad\text{(D)}}

Test the method on unseen problems

General Probability Practice Workbook

30 original general probability problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

See the cover and sample pages

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 approximately the joint probability 0.4 exp(−0.5)=0.2426 and omits normalization by the total probability of zero events.
BThis repeats the prior probability 0.40 and ignores the evidence supplied by the observed zero count.
CThis is approximately the evidence probability 0.4 exp(−0.5)+0.6 exp(−1)=0.4633, not the posterior class probability.
EUsing 0.4 for both prior weights gives exp(−0.5)/[exp(−0.5)+exp(−1)]=0.6225 and erases the 60% prior weight of the higher-rate class.

Original practice · fully worked

Original variant: reconstructing a shipment prior

A shipment arrives from either North Dock or South Dock. A scan flags 20% of North Dock shipments and 50% of South Dock shipments. Among flagged shipments, the probability of North Dock origin is 1/3. Calculate the prior probability that a shipment comes from North Dock.

A
0.1667
B
0.3333
C
0.5000
D
0.5556
E
0.7143

Answer and calculation

Writing the unknown prior as p and applying Bayes gives 0.2p/[0.2p+0.5(1-p)]=1/3. Solving yields p=5/9, so choice D is correct.

Setup

Let p be the unknown prior probability of North Dock origin and let F denote a scan flag.

\Pr(N)=p,\qquad \Pr(F\mid N)=0.2,\qquad \Pr(F\mid N^c)=0.5

Model

Express the supplied posterior probability in terms of the unknown prior.

\frac{0.2p}{0.2p+0.5(1-p)}=\frac13

Compute

Cross-multiply and solve the resulting linear equation.

0.6p=0.2p+0.5(1-p)

0.9p=0.5

p=\frac59

Answer

The North Dock prior probability is five-ninths.

\boxed{\frac59\quad\text{(D)}}

Keep practicing

Next steps for Exam P General Probability

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P General Probability practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Bayes' Theorem question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: reconstructing a shipment prior

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability