Worked solution · Exam P

SOA Exam P Sample Question #403 Solution

This Exam P sample reference tests Counting Methods. Count the admissible category composition with a multivariate hypergeometric numerator. There are 120 favorable five-item subsets among 3003 total subsets, giving 0.03996 and choice C.

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 Counting Methods question

Setup

All five-item subsets of the fifteen distinct records are equally likely.

N_{\mathrm{all}}=\binom{15}{5}=3003

Model

Choose the required number from each of the three categories.

N_{\mathrm{good}}=\binom51\binom43\binom61=5(4)(6)=120

Compute

Divide the favorable subset count by the total subset count.

\Pr(\text{required composition})=\frac{120}{3003}

=0.03996003996\ldots

Answer

The probability rounds to 0.040.

\boxed{0.040\quad\text{(C)}}

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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.

AA with-replacement calculation using only ten category arrangements gives 0.0253. It both ignores depletion and treats the two singleton categories as interchangeable.
BUsing only 18 category orders gives 0.03596. There are actually 20 valid orders of the one-three-one category pattern.
DThe requested event is contained within the event of selecting exactly three items from the four-item category. That broader event has probability 0.0733, so 0.150 is impossible.
EThe same upper bound 0.0733 rules out 0.213, which exceeds the probability of even the less restrictive three-from-four event.

Original practice · fully worked

Original variant: exactly two reports reach their owners

Five labeled reports are randomly placed into five labeled folders, one report per folder. Calculate the probability that exactly two reports are placed in their matching folders.

A
1/12
B
1/6
C
3/8
D
11/30
E
1/2

Answer and calculation

There are ten choices for the two correctly placed reports, followed by two derangements of the remaining three. Thus 20 of the 120 assignments qualify, giving probability 1/6 and choice B.

Setup

Every assignment of the five distinct reports to the five folders is equally likely.

N_{\mathrm{all}}=5!=120

Model

Choose the two matching report-folder pairs. To avoid any additional match, the remaining three reports must be deranged.

\binom52=10,\qquad D_3=2

Compute

Multiply the choice of matching pairs by the number of valid assignments of the other reports.

N_{\mathrm{good}}=\binom52D_3=10(2)=20

\Pr(\text{exactly two matches})=\frac{20}{120}=\frac16

Answer

Exactly two reports match their folders with probability 1/6.

\boxed{\frac16\quad\text{(B)}}

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How to solve this Counting Methods question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: exactly two reports reach their owners

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability