SOA Exam P Sample Question #694. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #694 — Verified Solution

Independent solution

How to solve this Hypergeometric Distribution question

Answer in brief

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.

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)}}

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 \(\Pr(X\le2)=0.9163\), 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

Variant answer in brief

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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ActuaryProof. “SOA Exam P Sample Question #694 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.