Worked solution · Exam P

SOA Exam P Sample Question #259 Solution

This Exam P sample reference tests Conditional Probability. This is a two-event complement calculation with one conditional probability. The joint multiple-car and sports-car share is 0.096, making the union 0.744 and the complementary target 0.256, so choice C.

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 M denote insuring multiple cars and let S denote insuring a sports car. Convert the conditional percentage into a joint probability.

\Pr(M)=0.64,\qquad \Pr(S)=0.20

\Pr(M\cap S)=\Pr(S\mid M)\Pr(M)=0.15(0.64)=0.096

Model

Exactly one non-sports car means neither M nor S occurs, which is the complement of their union.

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

Compute

Apply the two-event union formula and take its complement.

\Pr(M\cup S)=0.64+0.20-0.096=0.744

\Pr(M^c\cap S^c)=1-0.744=0.256

Answer

The requested probability rounds to 0.26.

\boxed{0.26\quad\text{(C)}}

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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 0.16 is 0.36-0.20. It subtracts the entire sports-car population from the one-car population, incorrectly treating all sports-car customers as one-car customers.
BThe value 0.19 comes from the incorrect chain 0.20(1-0.15)=0.17 for one-car sports customers, followed by 0.36-0.17. The 15% rate is conditional on multiple-car customers, so its complement cannot be applied to the sports-car marginal.
DThe value 0.29 is 0.36(0.80)=0.288 rounded. It assumes the number-of-cars event and sports-car event are independent despite the supplied conditional information.
EThe value 0.31 is 0.36(0.85)=0.306 rounded. It incorrectly applies the 15% sports-car rate among multiple-car customers to the one-car group.

Original practice · fully worked

Original variant: exactly one of two release reviews

A software release may need a security review S and a load review L. The probability of S is 0.30, P(L given S)=0.40, and P(S given L)=0.24. What is the chance that a release needs one review but not both?

A
0.12
B
0.18
C
0.38
D
0.56
E
0.68

Answer and calculation

The first conditional probability gives a joint review probability of 0.12, and the reverse conditional then gives P(L)=0.50. Removing the overlap from both one-review regions gives 0.56 and choice D.

Setup

Use the load-review rate within the security-review group to obtain the intersection.

\Pr(S\cap L)=\Pr(L\mid S)\Pr(S)=0.40(0.30)=0.12

Model

The reverse conditional probability recovers the marginal probability of a load review.

\Pr(S\mid L)=\frac{\Pr(S\cap L)}{\Pr(L)}=0.24

\Pr(L)=\frac{0.12}{0.24}=0.50

Compute

Add the two disjoint one-review regions after removing the shared intersection from each marginal.

\Pr(S\cap L^c)=0.30-0.12=0.18

\Pr(S^c\cap L)=0.50-0.12=0.38

\Pr(\text{exactly one})=0.18+0.38=0.56

Answer

The probability of exactly one review is 0.56.

\boxed{0.56\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: exactly one of two release reviews

Answer and calculation

Setup

Model

Compute

Answer

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Next steps for Exam P General Probability