Worked solution · Exam P

SOA Exam P Sample Question #517 Solution

This Exam P sample reference tests Joint Distributions. The cost information determines one missing table cell, and normalization determines the other. The resulting marginal moments of Y give variance 0.5675, so choice B is correct.

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

Convert the expected Type I cost into the mean count of Type I claims, then use the row for X=1 to determine its missing cell.

\operatorname{E}[X]=\frac{345}{750}=0.46

\Pr(X=1)=0.21+0.13+q=0.34+q

q=0.12

Model

The known table entries total 0.70, so normalization gives p+q=0.30 and hence p=0.18. Sum columns to obtain the marginal law of Y.

\Pr(Y=0)=0.52,\qquad \Pr(Y=1)=0.31,\qquad \Pr(Y=2)=0.17

Compute

Calculate the first two moments and apply the variance identity.

\operatorname{E}[Y]=0.31+2(0.17)=0.65

\operatorname{E}[Y^2]=0.31+4(0.17)=0.99

\operatorname{Var}(Y)=0.99-(0.65)^2=0.5675

Answer

The variance rounds to 0.57.

\boxed{0.57\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.

AThis treats Y as the indicator of the zero-count event, giving 0.52(1-0.52)=0.2496. Y can also equal 2, so a Bernoulli variance formula is invalid.
CThis results if the solved values p=0.18 and q=0.12 are interchanged. The resulting incorrect marginal has variance 0.6659, which rounds to 0.67.
DThis omits the factor 2 in E[Y], using 0.31+0.17=0.48; then 0.99-0.48²=0.7596, near 0.75.
EThis subtracts P(Y=2)=0.17 from E[Y²]=0.99 as though that cell probability were (E[Y])², producing 0.82.

Original practice · fully worked

Original variant: variance of conditional scan flags

A routing controller selects operating mode A with probability 0.40 and mode B otherwise. During the next two independent checks, conditional on the selected mode, each check raises a flag with probability 0.50 in mode A and 0.25 in mode B. Let Y be the total number of flags. Calculate Var(Y).

A
0.060
B
0.425
C
0.485
D
0.700
E
0.975

Answer and calculation

The expected conditional variance is 0.425 and the variance of the conditional means is 0.060. Their sum is 0.485, so choice C is correct.

Setup

Within either fixed mode, the two-check flag count is binomial.

\operatorname{E}[Y\mid A]=1,\quad \operatorname{Var}(Y\mid A)=0.5

\operatorname{E}[Y\mid B]=0.5,\quad \operatorname{Var}(Y\mid B)=0.375

Model

Use the law of total variance to combine within-mode variation and between-mode variation.

\operatorname{Var}(Y)=\operatorname{E}[\operatorname{Var}(Y\mid M)]+\operatorname{Var}(\operatorname{E}[Y\mid M])

Compute

First find the overall mean, then evaluate the two variance components.

\operatorname{E}[Y]=0.4(1)+0.6(0.5)=0.7

\operatorname{E}[\operatorname{Var}(Y\mid M)]=0.4(0.5)+0.6(0.375)=0.425

\operatorname{Var}(\operatorname{E}[Y\mid M])=0.4(1-0.7)^2+0.6(0.5-0.7)^2=0.060

\operatorname{Var}(Y)=0.425+0.060=0.485

Answer

The total flag count has variance 0.485.

\boxed{0.485\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: variance of conditional scan flags

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables