Worked solution · Exam P

SOA Exam P Sample Question #24 Solution

This Exam P sample reference tests Bayes’ Theorem. A positive test comes from disease with weight 0.0095 and without disease with weight 0.00495. Normalization gives 0.65744, choice B.

Independent analysis by ActuaryProof · reviewed October 6, 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

Compute the two joint weights that can produce a positive test: disease with a true positive and no disease with a false positive.

w_D=(0.01)(0.95)=0.0095

w_{D^c}=(0.99)(0.005)=0.00495

Model

Apply Bayes' formula by normalizing the disease-positive weight over both positive-test weights.

\Pr(D\mid +)=\frac{w_D}{w_D+w_{D^c}}

Compute

The weights are 0.0095 and 0.00495, so the posterior disease probability is 0.0095/0.01445=0.657439.

\Pr(D\mid +)=\frac{0.0095}{0.01445}=0.657439

Answer

Given a positive test, the disease probability rounds to 0.657, corresponding to choice B.

\boxed{0.657\quad\text{(B)}}

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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.

DThe value 0.950 is the test sensitivity Pr(positive given disease), which reverses the conditioning direction.
EThe value 0.995 is the specificity, the probability of a negative result when disease is absent, not the posterior after a positive result.

Original practice · fully worked

Original variant: posterior defect probability after two positive screens

A manufacturing defect occurs in 4% of items. Each of two conditionally independent screens is positive for 90% of defective items and for 10% of nondefective items. Given that both screens are positive, find the probability that the item is defective.

A
0.04000
B
0.77143
C
0.27273
D
0.90000
E
0.96000

Answer and calculation

The two-positive weights are 0.04(0.9²)=0.0324 and 0.96(0.1²)=0.0096. Their normalization gives 0.77143, choice B.

Setup

For each defect state, multiply its prior probability by the likelihood that both conditionally independent screens are positive.

w_D=0.04(0.90)^2=0.0324

w_{D^c}=0.96(0.10)^2=0.0096

Model

Normalize the defective two-positive weight over the defective and nondefective two-positive weights.

\Pr(D\mid ++)=\frac{0.0324}{0.0324+0.0096}

Compute

The weights are 0.0324 and 0.0096, so the posterior defect probability is 0.0324/0.0420=27/35.

\Pr(D\mid ++)=27/35=0.771429

Answer

Given two positive screens, the defect probability is 0.77143, which is choice B.

\boxed{0.77143\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: posterior defect probability after two positive screens

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability