Worked solution · Exam P

SOA Exam P Sample Question #291 Solution

This Exam P sample reference tests Joint Probability Functions. This is the correlation of two binary variables from a four-cell joint distribution. Their covariance is 1/16 and the product of their standard deviations is √(15)/16, so the correlation is 1/√(15)=0.2582 and choice C.

Independent analysis by ActuaryProof · reviewed July 31, 2026

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Independent solution

How to solve this Joint Probability Functions question

Setup

For binary variables, each first moment is its probability of taking value one, while XY equals one only in the joint one-one cell.

E[X]=0.250+0.375=0.625=\frac58

E[Y]=0.125+0.375=0.500=\frac12

E[XY]=0.375=\frac38

Model

Use Bernoulli variances for the marginals and center the product moment for covariance.

\operatorname{Var}(X)=\frac58\left(1-\frac58\right)=\frac{15}{64}

\operatorname{Var}(Y)=\frac12\left(1-\frac12\right)=\frac14

\operatorname{Cov}(X,Y)=\frac38-\frac58\frac12=\frac1{16}

Compute

Normalize the covariance by the two marginal standard deviations.

\operatorname{Corr}(X,Y)=\frac{1/16}{\sqrt{(15/64)(1/4)}}

=\frac{1/16}{\sqrt{15}/16}=\frac1{\sqrt{15}}=0.2581989

Answer

The correlation 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 0.06 is the rounded covariance 0.0625. Correlation must also divide by the product of the two marginal standard deviations.
BThe value 0.23 is the rounded marginal variance Var(X)=0.234375. A single-variable variance is not the normalized joint association.
DThe value 0.38 is the rounded product moment E[XY]=0.375, which is just the probability of the one-one cell. The product of the means must be removed before normalization.
EThe value 0.63 is the rounded marginal mean E[X]=0.625. It measures the frequency of X=1, not dependence between X and Y.

Original practice · fully worked

Original variant: correlation from conditional completion rates

Let A indicate that a technician has advanced certification and let B indicate that the technician completes a repair within the target time. In a workforce, P(A=1)=0.40, P(B=1 given A=1)=0.70, and P(B=1 given A=0)=0.20. Calculate the correlation coefficient of A and B.

A
0.1200
B
0.2800
C
0.4000
D
0.5000
E
0.7000

Answer and calculation

The conditional rates give P(B=1)=0.40 and P(A=1,B=1)=0.28. Thus the covariance is 0.12, both marginal variances are 0.24, and the correlation is 0.12/0.24=0.5000, so choice D.

Setup

Convert the conditional completion rates into a joint success probability and the marginal probability for B.

\Pr(A=1,B=1)=0.40(0.70)=0.28

\Pr(B=1)=0.40(0.70)+0.60(0.20)=0.40

Model

Because A and B are indicators, their expectations equal their success probabilities.

E[A]=E[B]=0.40

\operatorname{Var}(A)=\operatorname{Var}(B)=0.40(0.60)=0.24

Compute

Center the product moment, then divide by the marginal standard deviations.

\operatorname{Cov}(A,B)=0.28-0.40(0.40)=0.12

\operatorname{Corr}(A,B)=\frac{0.12}{\sqrt{0.24(0.24)}}=\frac{0.12}{0.24}=0.5000

Answer

The indicator correlation is 0.5000.

\boxed{0.5000\quad\text{(D)}}

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How to solve this Joint Probability Functions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: correlation from conditional completion rates

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables