Worked solution · Exam P

SOA Exam P Sample Question #191 Solution

This Exam P sample reference tests Inclusion–Exclusion. Independence gives intersections 0.02 for A with B and 0.03 for B with C, while mutual exclusivity makes the A-C and triple intersections zero. Inclusion–exclusion gives union probability 0.55, choice D.

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 Inclusion–Exclusion question

Setup

Translate the independence and exclusion statements into intersection probabilities.

\Pr(A\cap B)=0.2(0.1)=0.02

\Pr(B\cap C)=0.1(0.3)=0.03

\Pr(A\cap C)=\Pr(A\cap B\cap C)=0

Model

Apply the three-event inclusion–exclusion identity.

\Pr(A\cup B\cup C)=\sum\Pr(\text{single})-\sum\Pr(\text{pair})+\Pr(A\cap B\cap C)

Compute

Substitute the three marginals and the valid intersection values.

\Pr(A\cup B\cup C)=0.2+0.1+0.3-0.02-0-0.03+0

\Pr(A\cup B\cup C)=0.55

Answer

The union probability is 0.550.

\boxed{0.550\quad\text{(D)}}

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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 0.496 is 1-(0.8)(0.9)(0.7), which assumes all three events are mutually independent. That contradicts the stated mutual exclusivity of A and C.
BThe value 0.540 subtracts an extra 0.010 as a spurious A-C overlap. Mutual exclusivity makes that overlap exactly zero.
CThe value 0.544 subtracts an invented triple product 0.2(0.1)(0.3)=0.006 after the valid pair corrections. The triple intersection is zero and would have a plus sign in inclusion–exclusion.
EThe value 0.600 simply adds the three marginal probabilities. It double-counts the A-B and B-C overlaps.

Original practice · fully worked

Original variant: probability that exactly two of three checks trigger

Three checks A, B, and C have pairwise intersection probabilities P(A and B)=0.25, P(A and C)=0.20, and P(B and C)=0.18. The probability that all three trigger is 0.12. Calculate the probability that exactly two checks trigger.

A
0.12
B
0.27
C
0.39
D
0.63
E
0.84

Answer and calculation

The sum of pairwise intersections counts each exactly-two outcome once and the all-three outcome three times. Subtracting three copies of 0.12 from 0.63 gives 0.27, choice B.

Setup

Separate each pairwise intersection into its pair-only region and the common triple region.

\Pr(A\cap B\text{ only})=0.25-0.12=0.13

\Pr(A\cap C\text{ only})=0.20-0.12=0.08

\Pr(B\cap C\text{ only})=0.18-0.12=0.06

Model

The three pair-only regions are mutually exclusive and exhaust the exactly-two event.

\Pr(\text{exactly two})=\Pr(AB\text{ only})+\Pr(AC\text{ only})+\Pr(BC\text{ only})

Compute

Add the three pair-only region probabilities.

\Pr(\text{exactly two})=0.13+0.08+0.06=0.27

Answer

Exactly two checks trigger with probability 0.27.

\boxed{0.27\quad\text{(B)}}

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How to solve this Inclusion–Exclusion question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: probability that exactly two of three checks trigger

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability