Worked solution · Exam P

SOA Exam P Sample Question #547 Solution

This Exam P sample reference tests Bayes' Theorem. A received one has mass 0.70(0.20)=0.14 from sent zeros and mass 0.30(0.80)=0.24 from sent ones. The posterior is 0.14/(0.14+0.24)=7/19=0.3684, so choice C 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.

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Independent solution

How to solve this Bayes' Theorem question

Setup

Let S_i denote the transmitted symbol and R_i the received symbol. A zero produces a received one only through a transmission error.

\Pr(S_0)=0.70,\qquad \Pr(S_1)=0.30

\Pr(R_1\mid S_0)=0.20,\qquad \Pr(R_1\mid S_1)=0.80

Model

Partition the evidence R_1 by the two mutually exclusive transmitted symbols.

\Pr(S_0\mid R_1)=\frac{\Pr(S_0)\Pr(R_1\mid S_0)}{\Pr(S_0)\Pr(R_1\mid S_0)+\Pr(S_1)\Pr(R_1\mid S_1)}

Compute

Evaluate the error-generated and correctly generated received-one masses.

\Pr(S_0\cap R_1)=0.70(0.20)=0.14

\Pr(S_1\cap R_1)=0.30(0.80)=0.24

\Pr(S_0\mid R_1)=\frac{0.14}{0.14+0.24}=\frac{7}{19}=0.3684210526

Answer

The posterior probability rounds to 0.37.

\boxed{0.37\quad\text{(C)}}

Test the method on unseen problems

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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 converts the joint error mass 0.14 into the odds 0.14/(1-0.14)=0.1628 and reports those odds as a posterior probability.
BA value of 0.34 would require an evidence probability 0.14/0.34=0.4118, but the disjoint evidence contributions total only 0.14+0.24=0.38.
DThis is the complementary posterior Pr(S_1|R_1)=0.24/0.38=0.6316, answering which symbol was correctly received instead.
EThis conditions on receiving zero rather than one: Pr(S_0|R_0)=0.56/(0.56+0.06)=0.9032, which rounds to 0.90.

Original practice · fully worked

Original variant: infer a monitoring mode

A security monitor uses strict mode on 30% of sessions and relaxed mode on 70%. Conditional on strict mode, each of two independent rules triggers with probability 0.80; under relaxed mode, each triggers with probability 0.40. Exactly one rule triggers during a session. Calculate the probability that strict mode was used.

A
0.0960
B
0.2222
C
0.3000
D
0.3360
E
0.7778

Answer and calculation

The exactly-one-trigger likelihood is 0.32 in strict mode and 0.48 in relaxed mode. Weighting by the mode priors gives equal-scale masses 0.096 and 0.336, so the strict posterior is 2/9=0.2222 and choice B.

Setup

Let E be the observation that exactly one of the two rules triggers.

\Pr(S)=0.30,\qquad \Pr(R)=0.70

Model

For two independent equal-probability rules, exactly one trigger has probability 2p(1-p) under either mode.

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

\Pr(E\mid R)=2(0.40)(0.60)=0.48

Compute

Weight the likelihoods by their mode priors and normalize the strict-mode mass.

\Pr(S\cap E)=0.30(0.32)=0.096

\Pr(R\cap E)=0.70(0.48)=0.336

\Pr(S\mid E)=\frac{0.096}{0.096+0.336}=\frac{2}{9}=0.2222222222

Answer

The posterior probability of strict mode is about 0.2222.

\boxed{0.2222\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: infer a monitoring mode

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability