Worked solution · Exam P

SOA Exam P Sample Question #311 Solution

This Exam P sample reference tests Joint Probability Functions. Aggregating the six joint cells by Z=X-Y gives masses 2, 6, 14, and 9 over 31 at values 0, 1, 2, and 3. The resulting first two moments give variance 712/961=0.740895, 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 Joint Probability Functions question

Setup

Combine joint outcomes that produce the same value of the difference.

\Pr(Z=0)=\frac2{31},\quad \Pr(Z=1)=\frac6{31}

\Pr(Z=2)=\frac{14}{31},\quad \Pr(Z=3)=\frac9{31}

Model

Calculate the first and second moments of the induced one-dimensional distribution.

\operatorname{E}[Z]=\frac{6+2(14)+3(9)}{31}=\frac{61}{31}

\operatorname{E}[Z^2]=\frac{6+4(14)+9(9)}{31}=\frac{143}{31}

Compute

Subtract the squared mean from the second moment.

\operatorname{Var}(Z)=\frac{143}{31}-\left(\frac{61}{31}\right)^2

=\frac{712}{961}=0.7408949011\ldots

Answer

The variance rounds to 0.74.

\boxed{0.74\quad\text{(C)}}

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.

AThe value 0.20 is approximately P(Z=1)=6/31. It reports one probability mass rather than the variance of the full difference distribution.
BThe value 0.69 is Var(X)+Var(Y)=664/961, which assumes zero covariance. Here Cov(X,Y)=-24/961, so Var(X-Y) also contains -2Cov(X,Y).
DThe value 1.90 is near the mean E[Z]=61/31=1.9677. A mean measures location, not squared dispersion.
EThe value 2.65 is 143/31-61/31=82/31. It subtracts the mean from the second moment instead of subtracting the squared mean.

Original practice · fully worked

Original variant: variance after a failover adjustment

A controller's initial load level L is uniformly distributed over the integers 1 through 6. If L is 5 or 6, a failover routine subtracts 4 units from the recorded level; otherwise it makes no adjustment. Let S be the recorded level. Calculate Var(S).

A
0.6667
B
1.1389
C
2.1667
D
2.9167
E
5.8333

Answer and calculation

The six equally likely recorded levels are 1, 2, 3, 4, 1, and 2. Their mean is 13/6 and second moment is 35/6, giving variance 35/6-(13/6)²=41/36=1.1389, choice B.

Setup

Apply the failover rule to each possible initial level.

L:1,2,3,4,5,6

S:1,2,3,4,1,2

Model

Compute the first two moments from the six equally likely recorded values.

\operatorname{E}[S]=\frac{1+2+3+4+1+2}{6}=\frac{13}{6}

\operatorname{E}[S^2]=\frac{1+4+9+16+1+4}{6}=\frac{35}{6}

Compute

Subtract the squared mean from the second moment.

\operatorname{Var}(S)=\frac{35}{6}-\left(\frac{13}{6}\right)^2

=\frac{41}{36}=1.138888\ldots

Answer

The recorded-level variance rounds to 1.1389.

\boxed{1.1389\quad\text{(B)}}

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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: variance after a failover adjustment

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables