Verified solution · Exam P

SOA Exam P Sample Question #346 Solution

This Exam P sample reference tests Inclusion–Exclusion. Because event C cannot accompany either other event, exactly two events means the overlap of A and B, with probability 0.06. Adding that overlap to the 0.38 exactly-one probability gives 0.44 for the union of A and B. The remaining 0.46 belongs to C, choice B.

Independent analysis by ActuaryProof · reviewed August 4, 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 Inclusion–Exclusion question

Setup

Use the two mutual-exclusivity conditions to identify the only possible two-event overlap.

\Pr(A\cap B)=0.06

Model

Relate the exactly-one probability for A and B to their union.

\Pr(\text{exactly one of }A,B)=\Pr(A\cup B)-\Pr(A\cap B)

Compute

Recover the union of A and B, then use its disjoint union with C.

\Pr(A\cup B)=0.38+0.06=0.44

0.90=\Pr(A\cup B)+\Pr(C)

\Pr(C)=0.90-0.44=0.46

Answer

Event C has probability 0.46.

\boxed{0.46\quad\text{(B)}}

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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 0.32 subtracts the A-and-B overlap from the exactly-one probability. The overlap must instead be added to recover their union.
CThe value 0.52 subtracts the exactly-one probability directly from the three-event union. This leaves both C and the A-and-B overlap.
DThe value 0.56 is the complement of the A-or-B union in the full sample space. It includes the 0.10 probability that none of the three events occurs.
EThe value 0.58 first subtracts the A-and-B overlap from the exactly-one probability, then removes that incorrect quantity from the three-event union.

Original practice · fully worked

Original variant: exactly one divisibility rule

An integer is selected uniformly from 1 through 60. Calculate the probability that it is divisible by exactly one of 2, 3, and 5.

A
0.0333
B
0.3333
C
0.4667
D
0.7333
E
0.9667

Variant answer in brief

There are 16 integers divisible only by 2, eight divisible only by 3, and four divisible only by 5. Thus 28 of the 60 integers satisfy exactly one rule, giving 7/15, approximately 0.4667 and choice C.

Setup

Count the multiples and their intersections within the selected range.

|A|=30,\quad |B|=20,\quad |C|=12

|A\cap B|=10,\quad |A\cap C|=6,\quad |B\cap C|=4

|A\cap B\cap C|=2

Model

Remove the other two divisibility rules from each single-rule count and restore the triple intersection.

|A\text{ only}|=30-10-6+2=16

|B\text{ only}|=20-10-4+2=8

|C\text{ only}|=12-6-4+2=4

Compute

Add the disjoint exactly-one regions and divide by 60.

\Pr(\text{exactly one rule})=\frac{16+8+4}{60}

=\frac7{15}=0.466666\ldots

Answer

The selected integer satisfies exactly one rule with probability approximately 0.4667.

\boxed{0.4667\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #346 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Inclusion–Exclusion question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: exactly one divisibility rule

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution