Worked solution · Exam P

SOA Exam P Sample Question #335 Solution

This Exam P sample reference tests Hypergeometric Distribution. This is a two-case hypergeometric acceptance probability. An accepted sample contains either two good items and one defective item or three good items; the two cases each have 56 favorable samples, giving 112/120=14/15≈0.933 and choice E.

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

The finite box contains eight good items and two defective items, and three are sampled without replacement.

G=8,\qquad D=2,\qquad n=3

Model

Acceptance occurs for exactly two good items or exactly three good items.

\Pr(\text{accept})=\frac{\binom82\binom21+\binom83\binom20}{\binom{10}{3}}

Compute

Evaluate the two disjoint composition counts.

\binom82\binom21=56,\qquad \binom83\binom20=56

\Pr(\text{accept})=\frac{112}{120}=\frac{14}{15}=0.9333333333

Answer

The acceptance probability rounds to 0.93.

\boxed{0.93\quad\text{(E)}}

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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 rejection probability is 1/15≈0.0667. Rounding that probability to one decimal as 0.10 and reporting it without complementing reverses the event.
BThe value 0.47 is one of the two acceptance components, 56/120. It omits the equally large all-good sample component.
CThe value 0.70 results from counting the two-good component fully but counting only half of the equally numerous all-good component: (56+28)/120.
DThe value 0.90 is the with-replacement binomial approximation Pr(Binomial(3,0.8)≥2)=0.896. The actual sample is drawn without replacement.

Original practice · fully worked

Original variant: both supervisors on a represented panel

A project group contains two supervisors and seven analysts. Three people are selected uniformly without replacement for a review panel. Given that the panel contains at least one supervisor, calculate the probability that it contains both supervisors.

A
0.0833
B
0.1429
C
0.4167
D
0.5833
E
0.8571

Answer and calculation

Seven of the 84 possible panels contain both supervisors. Forty-nine panels contain at least one supervisor. Conditioning gives 7/49, or 1/7, approximately 0.1429. This is choice B.

Setup

Count all three-person panels and panels with no supervisor.

N_{\mathrm{all}}=\binom93=84,\qquad N_0=\binom73=35

Model

Use the complement to count panels represented by at least one supervisor.

N_{\ge1}=84-35=49

Compute

Count panels with both supervisors and divide by the conditioning count.

N_2=\binom22\binom71=7

\Pr(H=2\mid H\ge1)=\frac7{49}=\frac17=0.142857\ldots

Answer

Given supervisor representation, both supervisors serve with probability approximately 0.1429.

\boxed{0.1429\quad\text{(B)}}

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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: both supervisors on a represented panel

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability