Verified solution · Exam P

SOA Exam P Sample Question #288 Solution

This Exam P sample reference tests Conditional Probability. This is a Bayes-theorem update after a positive test outcome. True positives contribute 0.18 of the tested population and false positives contribute 0.16, so the posterior is 0.18/0.34=0.5294 and choice D.

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 Conditional Probability question

Setup

Let P denote pregnancy and + denote a positive test outcome. Convert the false-negative rate to sensitivity.

\Pr(P)=0.20,\qquad \Pr(P^c)=0.80

\Pr(+\mid P)=1-0.10=0.90,\qquad \Pr(+\mid P^c)=0.20

Model

Bayes' theorem compares the positive mass from the target group with all positive results.

\Pr(P\mid +)=\frac{\Pr(+\mid P)\Pr(P)}{\Pr(+\mid P)\Pr(P)+\Pr(+\mid P^c)\Pr(P^c)}

Compute

Weight both test error rates by their population shares.

\Pr(P\cap +)=0.90(0.20)=0.18

\Pr(+)=0.18+0.20(0.80)=0.34

\Pr(P\mid +)=\frac{0.18}{0.34}=\frac9{17}=0.5294118\ldots

Answer

The posterior probability rounds to 0.53.

\boxed{0.53\quad\text{(D)}}

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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 0.10 is the false-negative probability. It concerns a negative outcome within the pregnant group, not pregnancy within the positive-result group.
BThe value 0.20 is the prior pregnancy share and also the false-positive rate here. Reporting either number without population weighting skips the Bayes update.
CThe value 0.50 comes from normalizing the prior 0.20 against the false-positive rate 0.20 as though they were comparable positive masses. The likelihood and the non-pregnant base rate are both missing.
EThe value 0.90 is the sensitivity P(+|P). It reverses the conditioning direction and is not P(P|+).

Original practice · fully worked

Original variant: two diagnostic flags from prior odds

A service incident comes from configuration R or configuration S, with prior odds R:S of 1:4. Two diagnostic flags are conditionally independent given the configuration. The first flag has likelihood ratio 3 in favor of R, and the second has likelihood ratio 2 in favor of R. Both flags appear. Calculate the posterior probability that configuration R caused the incident.

A
0.2000
B
0.3333
C
0.4286
D
0.6000
E
1.5000

Variant answer in brief

Conditional independence allows the likelihood ratios to multiply. The prior odds 1:4 become posterior odds 6:4=3:2, which convert to probability 3/5=0.6000 and choice D.

Setup

Represent the prior information as odds for R against S.

O_{\mathrm{prior}}(R:S)=\frac14

Model

For conditionally independent evidence, the combined likelihood ratio is the product of the two individual ratios.

LR_{\mathrm{both}}=3(2)=6

O_{\mathrm{post}}(R:S)=O_{\mathrm{prior}}LR_{\mathrm{both}}

Compute

Update the odds and then convert odds to a probability.

O_{\mathrm{post}}(R:S)=\frac14(6)=\frac32

\Pr(R\mid\text{both flags})=\frac{3}{3+2}=\frac35=0.6000

Answer

The posterior probability of configuration R is 0.6000.

\boxed{0.6000\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #288 — Verified Solution.” Verified 2026-07-31.

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

How to solve this Conditional Probability question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: two diagnostic flags from prior odds

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution