Worked solution · Exam P

SOA Exam P Sample Question #561 Solution

This Exam P sample reference tests Combinatorial Probability. The desired selection contains one of the two distinguished objects and three of the six ordinary objects. The count ratio is C(2,1)C(6,3)/C(8,4)=4/7=0.5714, so choice E is correct.

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 Combinatorial Probability question

Setup

Among the eight objects, two have the distinguished property and six do not. The condition requires exactly one distinguished object in a sample of four.

K=2,\qquad N-K=6,\qquad n=4,\qquad X=1

Model

Count favorable unordered selections and divide by all unordered selections of four objects.

\Pr(X=1)=\frac{\binom{2}{1}\binom{6}{3}}{\binom{8}{4}}

Compute

Evaluate the combinations.

\binom{2}{1}\binom{6}{3}=2(20)=40

\binom{8}{4}=70

\Pr(X=1)=\frac{40}{70}=0.5714285714

Answer

The probability rounds to 0.57.

\boxed{0.57\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 uses only the two choices for the distinguished object, 2/C(8,4)=2/70=0.0286, and omits the three ordinary selections.
BThis treats the selections as independent with replacement and counts only one position for the distinguished object, giving (2/8)(6/8)³=0.1055.
CThis counts only one of the two distinguished objects, C(6,3)/C(8,4)=20/70=0.2857.
DThis treats the four selections as independent draws with replacement, giving C(4,1)(2/8)(6/8)³=0.421875.

Original practice · fully worked

Original variant: flagged cards after screening

An archive box contains 10 index cards, of which 4 are flagged and 6 are standard. Three cards are drawn uniformly without replacement. Given that the draw contains at least one flagged card, calculate the probability that it contains exactly one flagged card.

A
0.1667
B
0.5000
C
0.5510
D
0.6000
E
0.8333

Answer and calculation

There are 60 three-card selections with exactly one flag and 100 selections with at least one flag. The conditional probability is 60/100=0.60, so choice D is correct.

Setup

Count the exactly-one-flag selections and the selections admitted by the conditioning event.

N_{1}=\binom{4}{1}\binom{6}{2}=60

Model

Remove the no-flag selections from all three-card selections to obtain the conditional sample space.

N_{\geq 1}=\binom{10}{3}-\binom{6}{3}=120-20=100

Compute

Divide the favorable count by the conditioned count.

\Pr(X=1\mid X\geq 1)=\frac{60}{100}=0.60

Answer

The conditional probability is 0.6000.

\boxed{0.6000\quad\text{(D)}}

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How to solve this Combinatorial Probability question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: flagged cards after screening

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability