Worked solution · Exam P

SOA Exam P Sample Question #149 Solution

This Exam P sample reference tests Sampling Without Replacement. For the fourth damaged item to appear in position 12, exactly three of the first eleven positions and three of the last eight positions must be damaged. The resulting combination ratio is 0.11920, choice B.

Independent analysis by ActuaryProof · reviewed October 6, 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

View the seven damaged-position locations as a uniformly selected subset of the twenty positions.

N_{\mathrm{all}}=\binom{20}{7}

Model

Fix position 12 as damaged, place three earlier damaged items, and place the remaining three after it.

N_{\mathrm{target}}=\binom{11}{3}\binom{8}{3}

Compute

Divide the favorable damaged-position sets by all possible damaged-position sets.

\Pr(T_4=12)=\frac{\binom{11}{3}\binom{8}{3}}{\binom{20}{7}}

\Pr(T_4=12)=0.1191950

Answer

The probability rounds to 0.119.

\boxed{0.119\quad\text{(B)}}

Test the method on unseen problems

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30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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 0.079 is the with-replacement negative-binomial calculation C(11,3)(7/20)⁴(13/20)⁸. The damaged fraction changes under sampling without replacement.
CThe value 0.237 is approximately the with-replacement binomial probability of four damaged items among the first twelve. It neither respects depletion nor requires the fourth one at position 12.
DThe value 0.243 uses an incorrect remaining-composition factor. After exactly three damaged items in the first eleven, the next-item damaged probability is 4/9.
EThe value 0.358 is the hypergeometric probability of exactly four damaged items among the first twelve. It allows the fourth damaged item at any of those twelve positions.

Original practice · fully worked

Original variant: conditional marked-token count

A box contains 15 tokens, of which 5 are marked. Six tokens are selected uniformly without replacement. Given that the selection contains at least one marked token, determine the probability that it contains exactly two marked tokens.

A
0.04196
B
0.41958
C
0.43796
D
0.56204
E
0.95804

Answer and calculation

The unconditional two-mark probability is C(5,2)C(10,4)/C(15,6), while the conditioning probability excludes the all-unmarked sample. Their ratio is 0.43796, choice C.

Setup

Let X count marked tokens in the six-token sample.

X\sim\operatorname{Hypergeometric}(15,5,6)

Model

Write the target mass and the positive-count conditioning probability.

\Pr(X=2)=\frac{\binom52\binom{10}{4}}{\binom{15}{6}}

\Pr(X\ge1)=1-\frac{\binom{10}{6}}{\binom{15}{6}}

Compute

Cancel the common sample-space denominator before evaluating.

\Pr(X=2\mid X\ge1)=\frac{\binom52\binom{10}{4}}{\binom{15}{6}-\binom{10}{6}}

\Pr(X=2\mid X\ge1)=\frac{2100}{4795}=0.4379562

Answer

The conditional probability is approximately 0.43796.

\boxed{0.43796\quad\text{(C)}}

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Next steps for Exam P Univariate Random Variables

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How to solve this Sampling Without Replacement question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: conditional marked-token count

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables