Verified solution · Exam P

SOA Exam P Sample Question #274 Solution

This Exam P sample reference tests Conditional Probability. The joint probability of multiple-car and sports-car coverage is 0.20 × 0.62, or 0.124. Inclusion-exclusion makes the probability outside both sets 1-(0.62+0.15-0.124)=0.354, which is 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 M denote coverage of multiple cars and S denote coverage of a sports car.

\Pr(M)=0.62,\qquad \Pr(S)=0.15,\qquad \Pr(S\mid M)=0.20

Model

Recover the intersection from the conditional probability.

\Pr(M\cap S)=\Pr(S\mid M)\Pr(M)=0.20(0.62)=0.124

Compute

Exactly one non-sports car corresponds to being outside both the multiple-car and sports-car sets.

\Pr(M^c\cap S^c)=1-\Pr(M\cup S)

=1-[0.62+0.15-0.124]=0.354

Answer

The required joint probability is 0.354.

\boxed{0.354\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 0.230 is 0.38-0.15. It subtracts every sports-car customer from the one-car group, including the 0.124 intersection that belongs to the multiple-car group.
BThe value 0.260 is the rounded result of 0.38-0.124=0.256. It subtracts multiple-car sports customers from the one-car marginal even though those customers are already outside that group.
CThe value 0.323 is 0.38(0.85). It assumes the one-car and non-sports events are independent, contrary to the supplied conditional information.
EThe value 0.380 is 1-0.62, the probability of exactly one car without imposing the additional non-sports requirement.

Original practice · fully worked

Original variant: encryption among archived files

In a digital archive, 0.28 of files use compression but not encryption, 0.17 use encryption but not compression, and 0.35 use neither feature. Given that a selected file uses encryption, calculate the probability that it also uses compression.

A
0.200
B
0.350
C
0.370
D
0.459
E
0.541

Variant answer in brief

The three supplied exclusive cells total 0.80, so the both-features cell is 0.20. Encryption occurs in 0.17+0.20=0.37 of files, and 0.20/0.37=0.540541, making choice E correct.

Setup

Complete the four-cell partition formed by compression C and encryption E.

\Pr(C\cap E)=1-0.28-0.17-0.35=0.20

Model

Combine the two disjoint cells in which encryption is present.

\Pr(E)=\Pr(C^c\cap E)+\Pr(C\cap E)=0.17+0.20=0.37

Compute

Normalize the both-features cell by the encryption marginal.

\Pr(C\mid E)=\frac{\Pr(C\cap E)}{\Pr(E)}

=\frac{0.20}{0.37}=\frac{20}{37}=0.5405405405\ldots

Answer

Among encrypted files, approximately 0.541 also use compression.

\boxed{0.541\quad\text{(E)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #274 — 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: encryption among archived files

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution