Worked solution · Exam P

SOA Exam P Sample Question #550 Solution

This Exam P sample reference tests Combinatorial Probability. Choose two people from the five in the specified category and three from the remaining four. The product C(5,2)C(4,3)=40 gives choice C.

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

A qualifying five-person selection consists of two people from the first category and three from its four-person complement.

k_1=2,\qquad k_2=3

Model

Choose the members independently within the two disjoint categories, then multiply the counts.

N=\binom{5}{2}\binom{4}{3}

Compute

Evaluate each combination factor.

\binom{5}{2}=10,\qquad \binom{4}{3}=4

N=10(4)=40

Answer

There are 40 qualifying combinations.

\boxed{40\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.

AThis counts only the choice of two people from the five-person category, C(5,2)=10, and omits the remaining three selections.
BThis adds the category counts, C(5,2)+C(4,3)=10+4=14, even though each first-category choice can be paired with every complementary choice.
DThis incorrectly treats both categories as having five members and computes C(5,2)C(5,3)=10(10)=100.
EThis counts all unrestricted five-person selections, C(9,5)=126, without enforcing exactly two members from the specified category.

Original practice · fully worked

Original variant: balanced museum exhibit

A museum store has 6 bronze objects, 5 glass objects, and 4 textile objects, all distinct. A seven-piece exhibit must contain exactly 3 bronze objects, 2 glass objects, and 2 textile objects. Calculate the number of possible exhibits.

A
36
B
120
C
200
D
800
E
1200

Answer and calculation

The three category choices are independent, so the exhibit count is C(6,3)C(5,2)C(4,2)=20(10)(6)=1200 and choice E.

Setup

Separate the seven positions by the three required material categories.

k_B=3,\qquad k_G=2,\qquad k_T=2

Model

Choose the distinct objects within each disjoint category and multiply the resulting counts.

N=\binom{6}{3}\binom{5}{2}\binom{4}{2}

Compute

Evaluate the three combination factors.

N=20\cdot 10\cdot 6=1200

Answer

There are 1200 exhibits satisfying all three material requirements.

\boxed{1200\quad\text{(E)}}

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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: balanced museum exhibit

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability