Worked solution · Exam P

SOA Exam P Sample Question #501 Solution

This Exam P sample reference tests Conditional Probability. Conditioning on a total of three affected individuals makes every three-person subset equally likely. Counting subsets with two or three members of the eight-person group gives 112/120=0.933333, so choice E is correct.

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 Conditional Probability question

Setup

Given that exactly three members of the ten-person population are selected by identical independent event mechanisms, the selected set is uniform over all three-person subsets.

\#\{\text{possible selected sets}\}=\binom{10}{3}=120

Model

There are eight members in the distinguished group and two outside it. The favorable set contains either exactly two or exactly three distinguished members.

K\mid(T=3)\sim\operatorname{Hypergeometric}(N=10,K_0=8,n=3)

\Pr(K\ge2\mid T=3)=\frac{\binom82\binom21+\binom83\binom20}{\binom{10}{3}}

Compute

Count the two disjoint favorable cases and divide by the total number of possible selected sets.

\binom82\binom21=28(2)=56,\qquad \binom83\binom20=56

\Pr(K\ge2\mid T=3)=\frac{56+56}{120}=\frac{14}{15}=0.933333\ldots

Answer

The conditional probability rounds to 0.933.

\boxed{0.933\quad\text{(E)}}

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30 original general probability 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 0.384 is the with-replacement binomial probability C(3,2)(0.8)²(0.2) of exactly two distinguished members. It ignores both sampling without replacement and the all-three case.
BThe value 56/120=0.466667 counts exactly two distinguished members under the correct without-replacement model, but it omits the 56 subsets containing three distinguished members.
CCounting all 56 all-distinguished subsets but only C(8,2)=28 of the exactly-two subsets gives (56+28)/120=0.700. The omitted factor C(2,1)=2 chooses which outside member is present.
DTreating the three selected people as independent draws with distinguished probability 0.8 gives C(3,2)(0.8)²(0.2)+(0.8)³=0.896. A fixed finite population requires hypergeometric sampling instead.

Original practice · fully worked

Original variant: three-category laboratory kit

A storeroom contains five calibration cartridges, four assay cartridges, and three control cartridges. A technician chooses four of the twelve cartridges uniformly without replacement to build a kit. Calculate the probability that the kit contains at least one cartridge of every type.

A
0.121212
B
0.181818
C
0.242424
D
0.454545
E
0.545455

Answer and calculation

A four-cartridge kit representing all three types must have category counts 2,1,1. Summing the three possible doubled categories gives 270 favorable kits out of 495, or 6/11, so choice E is correct.

Setup

All four-cartridge subsets are equally likely. First count the complete sample space.

\#\{\text{kits}\}=\binom{12}{4}=495

Model

With four selections and three represented types, one type appears twice and each other type appears once.

(n_C,n_A,n_R)\in\{(2,1,1),(1,2,1),(1,1,2)\}

Compute

Count the kits for each possible doubled type and add the disjoint cases.

\binom52\binom41\binom31=120

\binom51\binom42\binom31=90

\binom51\binom41\binom32=60

\Pr(\text{all types})=\frac{120+90+60}{495}=\frac6{11}=0.545454\ldots

Answer

The probability that every cartridge type is represented is approximately 0.545455.

\boxed{0.545455\quad\text{(E)}}

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How to solve this Conditional Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: three-category laboratory kit

Answer and calculation

Setup

Model

Compute

Answer

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Next steps for Exam P General Probability