Worked 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 July 31, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

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)}}

Test the method on unseen problems

General Probability Practice Workbook

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.

See the cover and sample pages

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

Answer and calculation

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)}}

Keep practicing

Next steps for Exam P General Probability

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P General Probability practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Conditional Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: encryption among archived files

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability