Worked solution · Exam P

SOA Exam P Sample Question #368 Solution

This Exam P sample reference tests Inclusion–Exclusion. The smoker group contributes 17.5% of the population to the overlap. Inclusion–exclusion then gives a below-normal lung-function proportion of 32.5%, whose listed value is 33% and choice C.

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

Let S denote smoking and L denote below-normal lung function. First convert the conditional percentage into an intersection.

\Pr(S\cap L)=\Pr(L\mid S)\Pr(S)=(0.70)(0.25)=0.175

Model

Apply the two-event inclusion–exclusion identity to the reported union.

\Pr(S\cup L)=\Pr(S)+\Pr(L)-\Pr(S\cap L)

Compute

Isolate the lung-function probability and substitute the known proportions.

\Pr(L)=0.40-0.25+0.175=0.325

Answer

Thus 32.5% of the population has below-normal lung function; the corresponding listed percentage is 33%.

\boxed{33\%\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.

AFifteen percent is the below-normal group outside the smoker group: the union proportion 40% minus the smoker proportion 25%. It omits the 17.5% overlap.
BTwenty percent would leave only 5% in the overlap after inclusion–exclusion. The stated conditional rate instead fixes the overlap at 17.5%.
DFifty-five percent adds the 40% union and the 15% below-normal nonsmoker region. That region is already contained in the union.
ESixty percent is the complement of the reported 40% union. It describes people in neither group, not everyone with below-normal lung function.

Original practice · fully worked

Original variant: exactly one coordinate alarm

A controller generates a point (U,V) uniformly over the unit square. Alarm A activates when U<0.6, and alarm B activates when V<U. Calculate the probability that exactly one of the two alarms activates.

A
0.180
B
0.320
C
0.420
D
0.740
E
0.920

Answer and calculation

Alarm A has probability 0.6, alarm B covers half the square, and their intersection has area integral from zero to 0.6 of u du, or 0.18. Inclusion–exclusion for exactly one alarm gives 0.6+0.5-2(0.18)=0.74 and choice D.

Setup

Use unit-square areas for the two individual alarm events.

\Pr(A)=0.6,\qquad \Pr(B)=\Pr(V<U)=\frac12

Model

Within the strip U<0.6, the part below the diagonal has vertical length u.

\Pr(A\cap B)=\int_0^{0.6}u\,du=\frac{0.6^2}{2}=0.18

Compute

Add the two event probabilities and remove the overlap twice to retain only the exclusive regions.

\Pr(\text{exactly one})=\Pr(A)+\Pr(B)-2\Pr(A\cap B)

=0.6+0.5-2(0.18)=0.74

Answer

Exactly one coordinate alarm activates with probability 0.740.

\boxed{0.740\quad\text{(D)}}

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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: exactly one coordinate alarm

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability