Worked solution · Exam P

SOA Exam P Sample Question #265 Solution

This Exam P sample reference tests Law of Total Probability. Let x be the territory-X share. The overall no-claim rate gives 0.20=0.15x+0.40(1-x), so x=0.80. The no-claim mass from X is then 0.15(0.80)=0.12, and 0.12/0.20=0.60 gives choice D.

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 Law of Total Probability question

Setup

Let x be the prior probability that a policyholder is from territory X, and let N denote no claims.

\Pr(X)=x,\qquad \Pr(X^c)=1-x

Model

Use total probability to recover the unknown territory mix.

0.20=0.15x+0.40(1-x)

Compute

Solve for the prior, then apply Bayes' theorem.

0.25x=0.20,\qquad x=0.80

\Pr(X\mid N)=\frac{0.15(0.80)}{0.20}=0.60

Answer

The posterior territory-X probability is 0.60.

\boxed{0.60\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.

AAfter solving Pr(X)=0.80, the territory-Y no-claim contribution is 0.20(0.40)=0.08, near this small choice. It is a joint mass for the wrong territory and is not normalized.
BThe value 0.27 is 0.15/(0.15+0.40)=0.2727. It assumes the two territories have equal prior shares instead of solving the supplied overall no-claim rate.
CThe value 0.50 assumes equal territory shares and ignores both the different no-claim rates and the overall rate.
EThe value 0.80 is the recovered prior probability Pr(X). It stops before conditioning on no claims.

Original practice · fully worked

Original variant: format of a uniformly sampled archive byte

In a large digital archive, 60% of files use format S and the remaining 40% use format L. The average file sizes are 2 megabytes for format S and 5 megabytes for format L. One stored byte is selected uniformly from all bytes in the archive. Calculate the probability that its containing file uses format L.

A
0.3750
B
0.4000
C
0.5000
D
0.6250
E
0.7143

Answer and calculation

Uniform byte sampling weights each file format by both its file share and its average size. The L and S byte-volume weights are 0.40(5)=2.0 and 0.60(2)=1.2, so the L share is 2.0/3.2=0.6250 and choice D.

Setup

Form a byte-volume weight for each format by multiplying file prevalence by average file size.

w_S=0.60(2)=1.20

w_L=0.40(5)=2.00

Model

A uniformly sampled byte belongs to a format in proportion to that format's total byte volume.

\Pr(L\mid\text{sampled byte})=\frac{w_L}{w_S+w_L}

Compute

Normalize the L byte-volume weight across both formats.

\Pr(L\mid\text{sampled byte})=\frac{2.00}{1.20+2.00}

=\frac58=0.6250

Answer

The sampled byte comes from format L with probability 0.6250.

\boxed{0.6250\quad\text{(D)}}

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How to solve this Law of Total Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: format of a uniformly sampled archive byte

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability