Worked solution · Exam P

SOA Exam P Sample Question #376 Solution

This Exam P sample reference tests Hypergeometric Distribution. This is a hypergeometric upper-tail probability from a population containing five senior and ten junior members. The favorable sample compositions contain either three or four seniors, giving 105/1365=1/13, approximately 0.077, 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 Hypergeometric Distribution question

Setup

Let X count senior members in the four-person sample. Every four-member subset of the fifteen-person population is equally likely.

\#\{\text{all samples}\}=\binom{15}{4}=1365

X\sim\operatorname{Hypergeometric}(15,5,4)

Model

At least three senior members means either exactly three seniors and one junior or exactly four seniors.

\#\{X=3\}=\binom53\binom{10}{1}=100

\#\{X=4\}=\binom54\binom{10}{0}=5

Compute

Add the two disjoint favorable counts and divide by the total number of samples.

\Pr(X\ge3)=\frac{100+5}{1365}

\Pr(X\ge3)=\frac1{13}=0.0769230769

Answer

The probability rounds to 0.077.

\boxed{0.077\quad\text{(C)}}

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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.

ACounting from an incorrect sixteen-person population and retaining only samples with exactly three seniors gives 110/1820, or 0.06044, which rounds to 0.060. The actual population has fifteen members, and the all-senior sample must also be included.
BCounting only samples with exactly three seniors gives 100/1365, or 0.07326; this omits the five all-senior samples.
DA with-replacement binomial calculation for exactly three seniors gives 8/81, or 0.09877. Sampling is without replacement.
EThis is the with-replacement binomial probability of at least three seniors, 8/81+1/81=1/9=0.11111. The changing composition requires a hypergeometric model.

Original practice · fully worked

Original variant: nonlinear score from a sample audit

An archive contains 12 chips, of which 4 are flagged and 8 are plain. Three chips are selected uniformly without replacement. If X is the number of flagged chips selected, define the audit score as the product of X and three minus X. Calculate its expected value.

A
8/11
B
1
C
16/11
D
17/11
E
2

Answer and calculation

The score is 2 when the sample has one or two flagged chips, and zero when it has none or all three. Those middle compositions have probability 8/11, so the expected score is 16/11 and choice C.

Setup

The flagged-chip count has a hypergeometric distribution. List how the score changes with that count.

X\sim\operatorname{Hypergeometric}(12,4,3)

R=X(3-X)=\begin{cases}0,&X=0\text{ or }3,\\2,&X=1\text{ or }2.\end{cases}

Model

It is shortest to subtract the two zero-score compositions from one.

\Pr(X=0)=\frac{\binom83}{\binom{12}3}=\frac{56}{220}

\Pr(X=3)=\frac{\binom43}{\binom{12}3}=\frac4{220}

\Pr(X=1\text{ or }2)=1-\frac{60}{220}=\frac8{11}

Compute

Multiply each possible score by its probability.

E[R]=0\left(\frac3{11}\right)+2\left(\frac8{11}\right)

E[R]=\frac{16}{11}=1.454545\ldots

Answer

The expected audit score is 16/11.

\boxed{\frac{16}{11}\quad\text{(C)}}

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How to solve this Hypergeometric Distribution question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: nonlinear score from a sample audit

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability