Worked solution · Exam P

SOA Exam P Sample Question #76 Solution

This Exam P sample reference tests Joint Distributions. Sum the joint table over warranty count to get the marginal probabilities of X. Its two moments yield variance 83/144 = 0.5764, choice B.

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 Joint Distributions question

Setup

Obtain the marginal distribution of the shipment count by summing the joint probabilities across every warranty-count value for each fixed shipment count.

P(X=0)=\frac16

P(X=1)=\frac1{12}+\frac16=\frac14

P(X=2)=\frac7{12}

Model

Use the marginal probabilities to compute the first and second raw moments.

E[X]=\frac14+2\frac7{12}=\frac{17}{12}

E[X^2]=\frac14+4\frac7{12}=\frac{31}{12}

Compute

Subtracting the squared mean from the second raw moment gives variance approximately 0.576389.

\operatorname{Var}(X)=\frac{31}{12}-\left(\frac{17}{12}\right)^2=\frac{83}{144}=0.576389

Answer

The marginal shipment count has variance approximately 0.58, selecting choice B.

\boxed{0.58\quad\text{(B)}}

Test the method on unseen problems

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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 only one joint cell for each shipment-count value instead of summing the entire corresponding row or column.
CChoice C reports a probability-weighted quantity that does not subtract the squared mean.
DChoice D reports the mean of the shipment count rather than its variance.
EChoice E reports the second raw moment without centering it by the squared mean.

Original practice · fully worked

Original variant: marginal spread from a joint shipping table

A warehouse records packages shipped X and returns Y. Their joint probabilities are P(0,0)=0.20, P(1,0)=0.15, P(1,1)=0.25, P(2,0)=0.10, P(2,1)=0.20, and P(2,2)=0.10. Find Var(X).

A
0.36
C
0.49
B
0.56
D
0.64
E
0.80

Answer and calculation

The marginal probabilities for X are 0.20, 0.40, and 0.40. Thus E[X]=1.2, E[X²]=2.0, and Var(X)=0.56.

Setup

Sum the joint cells for each shipment count. The resulting marginal probabilities at zero, one, and two are 0.20, 0.40, and 0.40.

P_X(0)=0.20,\quad P_X(1)=0.40,\quad P_X(2)=0.40

Model

The marginal distribution gives mean 1.2 and second raw moment 2.0.

E[X]=0.4+0.8=1.2

E[X^2]=0.4+1.6=2.0

Compute

Subtracting the squared mean from the second raw moment gives variance 0.56.

\operatorname{Var}(X)=2.0-(1.2)^2=0.56

Answer

The shipment count has variance 0.56, selecting choice B.

\boxed{0.56\quad\text{(B)}}

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Next steps for Exam P Multivariate Random Variables

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How to solve this Joint Distributions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: marginal spread from a joint shipping table

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables