Worked solution · Exam P

SOA Exam P Sample Question #694 Solution

This Exam P sample reference tests Hypergeometric Distribution. At most two defective selections fails only when all three defectives are among the four selected items. That excluded event has probability 1/30, leaving 29/30 = 0.966667 and choice E.

Independent analysis by ActuaryProof · reviewed July 30, 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

Let X count defectives in a simple random selection of four from a population containing three defectives and seven usable items.

X\sim\operatorname{Hypergeometric}(N=10,K=3,n=4)

Model

Because only three defectives exist, the complement of at most two defectives is exactly the event that all three are selected.

\Pr(X\le2)=1-\Pr(X=3)

Compute

Count selections containing all three defectives and one of the seven usable items, then divide by all four-item selections.

\Pr(X=3)=\frac{\binom33\binom71}{\binom{10}4}=\frac7{210}=\frac1{30}

\Pr(X\le2)=1-\frac1{30}=\frac{29}{30}=0.966666\ldots

Answer

The exact without-replacement probability rounds to 0.97.

\boxed{\Pr(X\le2)\approx0.97\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.

AThis repeats the population defective fraction 3/10 = 0.30 instead of calculating a probability for the four-item selection.
BThe excluded event has probability 1/30 = 0.0333, not 0.33; this choice reflects both taking the complement event and losing a decimal place.
CUsing the ratio 2/3 because two of three defectives are allowed ignores the random selection mechanism and is not a hypergeometric probability.
DTreating the four draws as independent Bernoulli trials with defect probability 0.30 gives P(X ≤ 2) = 0.9163, about 0.92; sampling is actually without replacement.

Original practice · fully worked

Original variant: firmware audit sample

A warehouse tray contains 12 control modules, four of which require a firmware update. A technician selects five modules uniformly without replacement for an audit. Calculate the probability that at most three of the selected modules require the update.

A
0.333
B
0.010
C
0.667
D
0.955
E
0.990

Answer and calculation

The only way to exceed three update-required modules is to select all four of them. That event has probability 1/99, so the requested probability is 98/99 = 0.989899 and choice E.

Setup

Let Y be the number of update-required modules among five selected from four update-required and eight current modules.

Y\sim\operatorname{Hypergeometric}(N=12,K=4,n=5)

Model

The complement of at most three is selecting all four update-required modules, together with one current module.

\Pr(Y\le3)=1-\Pr(Y=4)

Compute

Evaluate the single complement count and subtract it from one.

\Pr(Y=4)=\frac{\binom44\binom81}{\binom{12}5}=\frac8{792}=\frac1{99}

\Pr(Y\le3)=\frac{98}{99}=0.989898\ldots

Answer

The audit contains at most three modules needing an update with probability approximately 0.990.

\boxed{0.990\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: firmware audit sample

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability