Worked solution · Exam P

SOA Exam P Sample Question #156 Solution

This Exam P sample reference tests Hypergeometric Distribution. The all-uninsured probability forces the block to contain exactly three uninsured houses. Counting damage sets with zero or one insured house gives 22/120 = 11/60, 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

There are 120 equally likely three-house damage sets. Express the all-uninsured event using k.

\binom{10}{3}=120

\frac{\binom{k}{3}}{\binom{10}{3}}=\frac1{120}

Model

The probability equation determines the number of uninsured and insured houses.

\binom{k}{3}=1\quad\Longrightarrow\quad k=3

10-k=7

Compute

At most one insured house means either zero insured and three uninsured, or one insured and two uninsured.

\Pr(I\le1)=\frac{\binom70\binom33+\binom71\binom32}{\binom{10}{3}}

\Pr(I\le1)=\frac{1+21}{120}=\frac{11}{60}

Answer

The probability is 11/60.

\boxed{\frac{11}{60}\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.

AThe value 1/5 equals 24/120 and does not correspond to the hypergeometric count. The valid numerator is 1+21=22.
BThe value 7/40 equals 21/120, counting exactly one insured house but omitting the one all-uninsured damage set.
DThe value 49/60 is the probability of at least two insured houses, the complement of the requested event.
EThe value 119/120 is the probability that at least one damaged house is insured. The target also restricts the insured count to at most one.

Original practice · fully worked

Original variant: two-stage quality audit

A shipment contains five tagged units and eight untagged units. An inspector first draws three units and sets them aside, then draws four more from the remaining ten. Calculate the probability that the first group contains exactly two tagged units and the second group contains no tagged units.

A
0.02739
B
0.16667
C
0.04662
D
0.27972
E
0.95338

Answer and calculation

The first group has exactly two tagged units with probability 40/143. It then leaves seven untagged units among ten, so the second group is all untagged with probability 1/6. The product is 20/429 = 0.04662, choice C.

Setup

Count the tagged and untagged compositions of the first three-unit group.

\Pr(K_1=2)=\frac{\binom52\binom81}{\binom{13}{3}}=\frac{40}{143}

Model

After that event, three tagged and seven untagged units remain. The second group must use four of the seven untagged units.

\Pr(K_2=0\mid K_1=2)=\frac{\binom74}{\binom{10}{4}}=\frac16

Compute

Multiply the first-stage probability by the conditional second-stage probability.

\Pr(K_1=2,K_2=0)=\frac{40}{143}\frac16

\Pr(K_1=2,K_2=0)=\frac{20}{429}=0.0466200

Answer

The two-stage audit pattern occurs with probability about 0.04662.

\boxed{0.04662\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: two-stage quality audit

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability