Worked solution · Exam P

SOA Exam P Sample Question #198 Solution

This Exam P sample reference tests Conditional Probability. Split the callers into attendant, same-day return, and next-day return groups. Their joint application probabilities are 0.48, 0.18, and 0.04, so Bayes' theorem gives 0.48/0.70=24/35=0.6857, choice E.

Independent analysis by ActuaryProof · reviewed July 31, 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 A denote initial attendant contact, S a same-day return call, N a next-day return call, and M a mortgage application.

\Pr(A)=0.60

\Pr(S)=0.40(0.75)=0.30

\Pr(N)=0.40(0.25)=0.10

Model

Weight each caller group by its conditional application probability.

\Pr(A\cap M)=0.60(0.80)=0.48

\Pr(S\cap M)=0.30(0.60)=0.18

\Pr(N\cap M)=0.10(0.40)=0.04

Compute

First find the total application probability, then normalize the attendant branch.

\Pr(M)=0.48+0.18+0.04=0.70

\Pr(A\mid M)=\frac{0.48}{0.70}=\frac{24}{35}=0.6857143

Answer

The conditional probability rounds to 0.69.

\boxed{0.69\quad\text{(E)}}

Test the method on unseen problems

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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.

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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 0.06 is 0.10(0.60), the joint probability of a next-day return caller who does not apply. It is neither the requested branch nor conditioned on applying.
BThe value 0.26 is approximately 0.18/0.70, the posterior probability of the same-day return group among applicants.
CThe value 0.48 is the joint probability of attendant contact and application. It must still be divided by the total application probability.
DThe value 0.60 is the prior probability of attendant contact and ignores the information that the caller applied.

Original practice · fully worked

Original variant: source inference after split test results

A laboratory receives sealed samples from sources A, B, and C in proportions 0.40, 0.35, and 0.25. Conditional on the source, each of two screening checks passes independently. The pass probabilities per check are 0.80 for A, 0.50 for B, and 0.20 for C. A selected sample passes exactly one of the two checks. Calculate the probability that it came from source B.

A
0.320
B
0.350
C
0.383
D
0.457
E
0.500

Answer and calculation

For a source with per-check pass probability p, exactly one pass has probability 2p(1-p). The prior-weighted likelihoods are 0.128, 0.175, and 0.080, so the posterior for source B is 0.175/0.383=0.4569, choice D.

Setup

Let E be the event that exactly one screening check passes.

\Pr(E\mid p)=2p(1-p)

Model

Compute the likelihood of the observed split result for each possible source.

\Pr(E\mid A)=2(0.80)(0.20)=0.32

\Pr(E\mid B)=2(0.50)(0.50)=0.50

\Pr(E\mid C)=2(0.20)(0.80)=0.32

Compute

Multiply by the source proportions and normalize the weight for source B.

\Pr(A\cap E)=0.40(0.32)=0.128

\Pr(B\cap E)=0.35(0.50)=0.175

\Pr(C\cap E)=0.25(0.32)=0.080

\Pr(B\mid E)=\frac{0.175}{0.128+0.175+0.080}=0.4569191

Answer

Among samples with one passing check, about 45.7% came from source B.

\boxed{0.457\quad\text{(D)}}

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How to solve this Conditional Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: source inference after split test results

Answer and calculation

Setup

Model

Compute

Answer

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Next steps for Exam P General Probability