Worked solution · Exam P

SOA Exam P Sample Question #333 Solution

This Exam P sample reference tests Hypergeometric Distribution. Choose three of the ten target-group members and three of the other seven. This gives 4,200 favorable samples out of 12,376, so the probability is 75/221, approximately 0.339 and choice D.

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

Treat the two smoking categories together as the seven people outside the requested group.

10\text{ target members},\qquad 7\text{ other members},\qquad 6\text{ selections}

Model

An unordered favorable sample contains three people from each of the two groups.

\Pr(X=3)=\frac{\binom{10}{3}\binom{7}{3}}{\binom{17}{6}}

Compute

Evaluate the combination counts and reduce the ratio.

\begin{aligned}\binom{10}{3}\binom{7}{3}&=120(35)=4200,\qquad \binom{17}{6}=12376,\\\Pr(X=3)&=\frac{4200}{12376}=\frac{75}{221}=0.3393665158.\end{aligned}

Answer

The probability rounds to 0.339.

\boxed{0.339\quad\text{(D)}}

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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 value 0.176 is the probability that a three-person sample lies entirely in the target group. It ignores the other three selected people.
BThe value 0.284 is the binomial approximation obtained by treating the six selections as independent. The sample is drawn without replacement.
CThe value 0.300 is the ratio 3/10, comparing the desired count with the target-group size instead of counting six-person samples.
EThe value 0.588 is 10/17, the probability that one selected person belongs to the target group, not the probability that exactly three of six do.

Original practice · fully worked

Original variant: all three instrument types in a field kit

A field kit contains four thermal sensors, five optical sensors, and three pressure sensors. A technician draws four sensors at random without replacement. What is the probability that all three instrument types appear in the draw?

A
0.1212
B
0.1818
C
0.2424
D
0.4545
E
0.5455

Answer and calculation

With four selections and three instrument types, one type must appear twice and the other two once. The three possible doubled types contribute 90, 120, and 60 selections out of 495, giving probability 6/11, approximately 0.5455. This is choice E.

Setup

Count all four-sensor selections from the twelve instruments.

N_{\mathrm{all}}=\binom{12}{4}=495

Model

A favorable selection doubles exactly one type and includes one sensor from each remaining type.

N_T=\binom42\binom51\binom31=90

N_O=\binom41\binom52\binom31=120

N_P=\binom41\binom51\binom32=60

Compute

Add the three disjoint compositions and divide by all selections.

\Pr(\text{all types})=\frac{90+120+60}{495}

=\frac{270}{495}=\frac6{11}=0.5454545\ldots

Answer

The probability of including all three instrument types is approximately 0.5455.

\boxed{0.5455\quad\text{(E)}}

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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: all three instrument types in a field kit

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability