Verified solution · Exam P

SOA Exam P Sample Question #130 Solution

This Exam P sample reference tests Inclusion-Exclusion. The two event probabilities are 0.30 and 0.15, while their union has probability 0.39. Inclusion-exclusion therefore gives intersection probability 0.30+0.15-0.39=0.060, 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

Represent the two types of utilization by events E and H and complement the given zero-use probabilities.

\Pr(E)=1-0.70=0.30

\Pr(H)=1-0.85=0.15

Model

The probability of at least one of the two utilization types is the complement of having neither type.

\Pr(E\cup H)=1-0.61=0.39

Compute

Rearrange the two-event inclusion-exclusion identity to isolate the overlap.

\Pr(E\cap H)=\Pr(E)+\Pr(H)-\Pr(E\cup H)

\Pr(E\cap H)=0.30+0.15-0.39=0.060

Answer

The probability of experiencing both utilization types is 0.060.

\boxed{0.060\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 0.045 is 0.30(0.15), which assumes the two utilization events are independent. The supplied probability of neither event shows that this assumption is unwarranted.
CThe value 0.390 is the probability of at least one utilization type. It is the union used in the calculation, not the intersection requested.
DThe value 0.667 is too large even to be an intersection: any overlap must be no greater than the smaller marginal probability, 0.15.
EThe value 0.840 exceeds both event probabilities and therefore cannot be their intersection. It confuses a high complement probability with joint utilization.

Original practice · fully worked

Original variant: inspections that trigger neither monitor

During equipment inspections, 42% trigger an acoustic monitor and 30% trigger a thermal monitor. Among inspections that trigger the acoustic monitor, 25% also trigger the thermal monitor. Determine the probability that a randomly chosen inspection triggers neither monitor.

A
0.105
B
0.280
C
0.385
D
0.615
E
0.720

Variant answer in brief

The joint trigger probability is 0.42(0.25)=0.105. Thus the probability of at least one trigger is 0.42+0.30-0.105=0.615, leaving 0.385 for neither, choice C.

Setup

Let A and T denote acoustic and thermal triggers.

\Pr(A)=0.42,\qquad \Pr(T)=0.30,\qquad \Pr(T\mid A)=0.25

Model

First recover the joint trigger probability from the conditional probability.

\Pr(A\cap T)=\Pr(A)\Pr(T\mid A)=0.42(0.25)=0.105

Compute

Find the union and then take its complement.

\Pr(A\cup T)=0.42+0.30-0.105=0.615

\Pr(A^{c}\cap T^{c})=1-0.615=0.385

Answer

The probability that neither monitor triggers is 0.385.

\boxed{0.385\quad\text{(C)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #130 — 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: inspections that trigger neither monitor

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution