Worked solution · Exam P

SOA Exam P Sample Question #292 Solution

This Exam P sample reference tests Joint Distributions. This is a marginal-variance calculation from a uniform triangular joint support. Counting the admissible x-values gives P(Y=y)=(6-y)/21, from which E[Y]=5/3 and E[Y squared]=5, so Var(Y)=20/9=2.22 and choice B.

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 Distributions question

Setup

For a fixed y, the compatible x-values run from y through 5. There are 6-y such joint cells.

\Pr(Y=y)=\frac{6-y}{21},\qquad y=0,1,\ldots,5

Model

Use the marginal mass function to form the first two moments.

E[Y]=\frac1{21}\sum_{y=0}^{5}y(6-y)

E[Y^2]=\frac1{21}\sum_{y=0}^{5}y^2(6-y)

Compute

Evaluate both finite sums and subtract the squared mean.

E[Y]=\frac{35}{21}=\frac53

E[Y^2]=\frac{105}{21}=5

\operatorname{Var}(Y)=5-\left(\frac53\right)^2=\frac{20}{9}=2.222222\ldots

Answer

The variance rounds to 2.22.

\boxed{2.22\quad\text{(B)}}

Test the method on unseen problems

Probability Multivariate & Risk Models Practice Workbook

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.

AThe value 1.67 is the rounded mean E[Y]=5/3. A variance requires the centered second moment, not the first moment.
CThe value 3.33 results from subtracting E[Y] rather than its square: 5-5/3=10/3. The variance identity uses E[Y²]-(E[Y])².
DThe value 5.00 is the raw second moment E[Y²]. It still includes the squared location of the distribution and has not been centered.
EThe value 5.56 results from a misapplied total-variance formula: using the raw second moment 5 as the within-X term and adding Var(E[Y|X])=5/9 gives 50/9. The correct within-X term is E[Var(Y|X)]=5/3.

Original practice · fully worked

Original variant: active checkpoints under a random route class

A dispatch day selects a route class K with probabilities P(K=1)=0.20, P(K=3)=0.50, and P(K=5)=0.30. Given K, each of the K checkpoints is active independently with probability 0.40. Let Y be the number of active checkpoints. Calculate Var(Y).

A
0.3136
B
0.7680
C
1.0816
D
1.2800
E
1.9600

Answer and calculation

Conditional on K, the active count is binomial. The within-route variance component is 0.7680 and the between-route component is 0.3136, so total variance is 1.0816 and choice C.

Setup

Compute the first two moments of the random route size.

E[K]=1(0.20)+3(0.50)+5(0.30)=3.20

E[K^2]=1^2(0.20)+3^2(0.50)+5^2(0.30)=12.20

\operatorname{Var}(K)=12.20-3.20^2=1.96

Model

Given the route class, the active count is binomial with size K and probability 0.40.

E[Y\mid K]=0.40K

\operatorname{Var}(Y\mid K)=K(0.40)(0.60)=0.24K

Compute

Add the expected conditional variance and the variance of the conditional mean.

E[\operatorname{Var}(Y\mid K)]=0.24(3.20)=0.7680

\operatorname{Var}(E[Y\mid K])=0.40^2(1.96)=0.3136

\operatorname{Var}(Y)=0.7680+0.3136=1.0816

Answer

The active-count variance is 1.0816.

\boxed{1.0816\quad\text{(C)}}

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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: active checkpoints under a random route class

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables