Worked solution · Exam P

SOA Exam P Sample Question #382 Solution

This Exam P sample reference tests Hypergeometric Distribution. This problem conditions a without-replacement sample on three selections from one category. The remaining selection is from a pool with high-category probability 15/20, so the conditional variance is 0.1875 and choice A.

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 Hypergeometric Distribution question

Setup

After fixing three moderate-content selections, exactly one sampled page remains to be classified.

X=3\quad\Longrightarrow\quad Y\in\{0,1\}

Model

The final page must come from the twenty pages outside the conditioned category. Fifteen of those twenty belong to the high-content category.

\Pr(Y=1\mid X=3)=\frac{15}{15+5}=\frac34

Y\mid(X=3)\sim\operatorname{Bernoulli}\!\left(\frac34\right)

Compute

Apply the Bernoulli variance formula to the conditional success probability.

\operatorname{Var}(Y\mid X=3)=\frac34\left(1-\frac34\right)

\operatorname{Var}(Y\mid X=3)=\frac3{16}=0.1875

Answer

The conditional variance is 0.1875.

\boxed{0.1875\quad\text{(A)}}

Test the method on unseen problems

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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.

BThis removes three selected pages from 30 but incorrectly leaves the seven unselected moderate pages eligible, giving p=15/27 and p(1-p)=20/81=0.2469.
CThis squares the conditional mean, (3/4)²=0.5625, instead of subtracting that square from the second moment.
DBecause the conditional variable is Bernoulli, E[Y²|X=3]=E[Y|X=3]=0.75. This reports that second moment rather than the variance.
EThis adds rather than subtracts the squared mean: E[Y²|X=3]+E[Y|X=3]²=0.75+0.5625=1.3125.

Original practice · fully worked

Original variant: parity count after an even-sum report

Two distinct cards are selected uniformly from cards labeled 1 through 6. The selected labels are reported to have an even sum. Let K be the number of selected cards carrying odd labels. Calculate the conditional variance of K.

A
0.25
B
0.50
C
1.00
D
2.00
E
4.00

Answer and calculation

An even sum requires the two labels to have the same parity. There are three odd-odd pairs and three even-even pairs, so K is equally likely to be two or zero and has conditional variance 1, making choice C correct.

Setup

Partition the labels into their two parity classes.

\{1,3,5\}\text{ are odd},\qquad \{2,4,6\}\text{ are even}

Model

A two-label sum is even exactly when both selected labels have the same parity.

\#\{\text{odd-odd pairs}\}=\binom32=3

\#\{\text{even-even pairs}\}=\binom32=3

\Pr(K=0\mid\text{even sum})=\Pr(K=2\mid\text{even sum})=\frac12

Compute

Calculate the first two conditional moments of the two-point count.

E[K\mid\text{even sum}]=0\left(\frac12\right)+2\left(\frac12\right)=1

E[K^2\mid\text{even sum}]=0^2\left(\frac12\right)+2^2\left(\frac12\right)=2

\operatorname{Var}(K\mid\text{even sum})=2-1^2=1

Answer

The conditional variance of the odd-label count is one.

\boxed{1.00\quad\text{(C)}}

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Next steps for Exam P Multivariate Random Variables

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How to solve this Hypergeometric Distribution question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: parity count after an even-sum report

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables