Verified 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 August 4, 2026

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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)}}

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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

Variant answer in brief

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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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #382 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Hypergeometric Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: parity count after an even-sum report

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution