Worked solution · Exam P

SOA Exam P Sample Question #533 Solution

This Exam P sample reference tests Conditional Probability. Conditioning on exactly three affected employees makes every three-person subset equally likely. Only 20 of the 120 subsets lie entirely in department A, so the complement is 5/6=0.833333 and 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

All employees have the same independent event probability. Once the total count is fixed at three, the common Bernoulli factors cancel across candidate subsets.

\Pr(S\mid |S|=3)=\frac1{\binom{10}{3}}

Model

Use the complement: no department-B employee is included precisely when all three affected employees come from the six-person department A.

\Pr(\text{none from B}\mid |S|=3)=\frac{\binom63}{\binom{10}3}

Compute

Complement the all-A subset probability.

\Pr(\text{at least one from B}\mid |S|=3)=1-\frac{20}{120}

\Pr(\text{target})=\frac56=0.8333333333

Answer

The conditional probability rounds to 0.833.

\boxed{0.833\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 is approximately Pr(exactly three events)=C(10,3)(0.2)³(0.8)⁷=0.2013. It repeats the event already given instead of conditioning on it.
BThis assigns an unsupported one-half split between departments. The relevant three-person subsets are not divided evenly by whether they include department B.
CThis is 1-(0.8)⁴=0.5904, the unconditional probability that at least one of four department-B employees has an event; it ignores the fixed total of three.
DThis is 1-(6/10)³=0.784. It samples department labels with replacement, allowing the same employee to be selected repeatedly.

Original practice · fully worked

Original variant: sculpture-heavy inspection

A museum shipment contains eight framed prints and six sculptures. An inspector selects five of the fourteen objects uniformly without replacement. Calculate the probability that the inspection contains at least three sculptures.

A
0.0030
B
0.0599
C
0.2797
D
0.3427
E
0.6573

Answer and calculation

The favorable samples contain three, four, or five sculptures. Their 686 combinations out of 2002 five-object samples give 0.342657, so choice D is correct.

Setup

All five-object samples are equally likely.

N_{\text{all}}=\binom{14}{5}=2002

Model

Sum the hypergeometric counts for three, four, and five sculptures.

N_{\text{fav}}=\sum_{k=3}^{5}\binom6k\binom8{5-k}

Compute

Evaluate the three disjoint sample counts.

N_{\text{fav}}=\binom63\binom82+\binom64\binom81+\binom65\binom80

N_{\text{fav}}=560+120+6=686

\Pr(Y\ge3)=\frac{686}{2002}=0.3426573427

Answer

The required probability rounds to 0.3427.

\boxed{0.3427\quad\text{(D)}}

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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: sculpture-heavy inspection

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability