Worked solution · Exam P

SOA Exam P Sample Question #157 Solution

This Exam P sample reference tests Sampling Without Replacement. A winning nine-number set must contain all four fixed selections and any five of the other eight numbers. The ratio C(8,5)/C(12,9) is 0.25455, choice B.

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 Sampling Without Replacement question

Setup

Treat the gambler’s four selections as a fixed subset of the twelve available numbers.

N_{\text{all draws}}=\binom{12}{9}

Model

After including the four required numbers, choose the remaining five drawn numbers from the other eight.

N_{\text{win}}=\binom85

Compute

Divide favorable draw sets by all equally likely draw sets.

\Pr(\text{win})=\frac{\binom85}{\binom{12}{9}}=\frac{56}{220}=\frac{14}{55}

\Pr(\text{win})=0.2545455

Answer

The jackpot probability rounds to 0.255.

\boxed{0.255\quad\text{(B)}}

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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 0.002 is approximately 1/C(12,4). It treats one four-number set as the entire draw and ignores that each nine-number draw contains C(9,4) different four-number subsets.
CThe value 0.296 is not supported by the fixed-subset count. Even the common independence shortcut gives (9/12)⁴ = 0.31641; the exact without-replacement ratio is required.
DThe value 0.573 has no valid favorable-set numerator. Draws containing exactly three fixed selections are losses and must not be added to the 56 winning sets.
EThe value 0.625 does not arise from the four required sequential inclusions. The exact product is (9/12)(8/11)(7/10)(6/9)=14/55.

Original practice · fully worked

Original variant: first report in a conditioned audit

An archive holds six priority reports and five routine reports. An auditor draws four reports one at a time without replacement. Given that the four-report sample contains at least one priority report, calculate the probability that the first report drawn was a priority report.

A
0.54545
B
0.55385
C
0.98485
D
0.44615
E
0.25000

Answer and calculation

A four-report sample contains a priority report with probability 65/66. A priority report on the first draw guarantees that condition and has probability 6/11, so the conditional probability is (6/11)/(65/66)=36/65=0.55385, choice B.

Setup

Let A denote a priority report on the first draw and B denote at least one priority report in the four-report sample.

A\subseteq B,\qquad \Pr(A)=\frac6{11}

Model

Use the complement consisting of four routine reports to calculate the conditioning probability.

\Pr(B)=1-\frac{\binom54}{\binom{11}{4}}

\Pr(B)=1-\frac5{330}=\frac{65}{66}

Compute

Because A already ensures B, normalize the first-draw probability by the probability of B.

\Pr(A\mid B)=\frac{\Pr(A)}{\Pr(B)}

\Pr(A\mid B)=\frac{6/11}{65/66}=\frac{36}{65}=0.5538462

Answer

After conditioning on at least one priority report, the first-draw probability is about 0.55385.

\boxed{0.55385\quad\text{(B)}}

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How to solve this Sampling Without Replacement question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: first report in a conditioned audit

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability