Worked solution · Exam P

SOA Exam P Sample Question #82 Solution

This Exam P sample reference tests Conditional Distributions. Given disease is present, the positive-test probability is 0.125/0.175=5/7. The conditional Bernoulli variance is (5/7)(2/7)=10/49, choice C.

Independent analysis by ActuaryProof · reviewed October 6, 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 Distributions question

Setup

Within the conditioning class, the two displayed joint cells sum to probability 0.175.

P(X=1)=0.05+0.125=0.175

Model

Normalize the positive-test joint cell by that class probability to obtain Bernoulli success probability five sevenths.

P(Y=1\mid X=1)=\frac{0.125}{0.175}=\frac57

Compute

A Bernoulli variable with that success probability has conditional variance ten forty-ninths, approximately 0.204082.

\operatorname{Var}(Y\mid X=1)=\frac57\left(1-\frac57\right)=\frac{10}{49}=0.204082

Answer

The conditional Bernoulli variance is approximately 0.20, selecting choice C.

\boxed{0.20\quad\text{(C)}}

Test the method on unseen problems

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30 original multivariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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.

AChoice A uses a raw joint probability as the Bernoulli success parameter without conditional normalization.
BChoice B uses an incomplete denominator for the conditioning class.
DChoice D reports the conditional success probability rather than its Bernoulli variance.
EChoice E reports the complementary conditional probability rather than the variance.

Original practice · fully worked

Original variant: variance of an alert flag within a risk class

A monitoring table gives P(R=1,A=1)=0.18 and P(R=1,A=0)=0.12, where R indicates a high-risk device and A an alert. Find Var(A | R=1).

A
0.18
B
0.21
C
0.24
D
0.30
E
0.60

Answer and calculation

Within the high-risk class, the alert probability is 0.18/(0.18+0.12)=0.6. Its Bernoulli variance is 0.6(0.4)=0.24.

Setup

The high-risk class consists of the alert and no-alert joint cells, whose probabilities sum to 0.30.

P(R=1)=0.18+0.12=0.30

Model

Normalizing the alert cell within that class gives conditional alert probability 0.60.

P(A=1\mid R=1)=\frac{0.18}{0.30}=0.6

Compute

The alert indicator is Bernoulli within the class, so its conditional variance is 0.24.

\operatorname{Var}(A\mid R=1)=0.6(0.4)=0.24

Answer

The conditional alert variance is 0.24, selecting choice C.

\boxed{0.24\quad\text{(C)}}

Keep practicing

Next steps for Exam P Multivariate Random Variables

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How to solve this Conditional Distributions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: variance of an alert flag within a risk class

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables