Verified solution · Exam P

SOA Exam P Sample Question #195 Solution

This Exam P sample reference tests Conditional Probability. Inclusion–exclusion gives joint working probability 0.8+0.6-0.9=0.5. Dividing by the probability that A works gives P(B works given A works)=0.5/0.8=5/8, choice C.

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

Recover the joint working probability from the two marginals and their union.

\Pr(A\cap B)=\Pr(A)+\Pr(B)-\Pr(A\cup B)

Model

Insert the supplied probabilities and then form the conditional ratio.

\Pr(A\cap B)=0.8+0.6-0.9=0.5

\Pr(B\mid A)=\frac{\Pr(A\cap B)}{\Pr(A)}

Compute

Normalize the joint event by the conditioning marginal.

\Pr(B\mid A)=\frac{0.5}{0.8}=0.625=\frac58

Answer

The conditional probability is five eighths.

\boxed{\frac58\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 1/2 is the joint probability P(A and B). It is the numerator before conditioning on A.
BThe value 3/5 is the marginal probability P(B). It would equal P(B|A) only under independence, which is not stated and is inconsistent with the union.
DThe value 3/4 is 0.6/0.8, dividing the marginal of B by the marginal of A rather than using the joint probability.
EThe value 5/6 is 0.5/0.6=P(A|B), which reverses the conditioning direction.

Original practice · fully worked

Original variant: recover indicator covariance from an exactly-one probability

Events A and B satisfy P(A)=0.70, P(B)=0.50, and P(exactly one of A and B)=0.40. Let I_A and I_B be their event indicators. Calculate Cov(I_A,I_B).

A
-0.05
B
0
C
0.05
D
0.35
E
0.40

Variant answer in brief

If x is the intersection probability, exactly one has probability 0.70+0.50-2x. Setting this to 0.40 gives x=0.40, so covariance is 0.40-(0.70)(0.50)=0.05, choice C.

Setup

Let x be the joint-event probability and express the exactly-one region through x.

\Pr(\text{exactly one})=\Pr(A)+\Pr(B)-2\Pr(A\cap B)

0.40=0.70+0.50-2x

Model

Solve for the joint probability and use the covariance identity for indicators.

x=0.40

\operatorname{Cov}(I_A,I_B)=\operatorname{E}[I_AI_B]-\operatorname{E}[I_A]\operatorname{E}[I_B]

Compute

Substitute the joint probability and two marginals.

\operatorname{Cov}(I_A,I_B)=0.40-(0.70)(0.50)=0.05

Answer

The indicator covariance is 0.05.

\boxed{0.05\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #195 — 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: recover indicator covariance from an exactly-one probability

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution