Worked solution · Exam P

SOA Exam P Sample Question #158 Solution

This Exam P sample reference tests Poisson Distribution. The sum of three independent monthly Poisson counts is Poisson with mean three. Complementing counts zero through two gives probability 0.57681, 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 Poisson Distribution question

Setup

Let T be the total count across the three mutually independent months.

T=X_1+X_2+X_3

X_j\sim\operatorname{Poisson}(1)

Model

Independent Poisson means add, so the three-month total has mean three.

T\sim\operatorname{Poisson}(1+1+1)=\operatorname{Poisson}(3)

Compute

Use the complement of the first three Poisson masses.

\Pr(T>2)=1-e^{-3}\left(1+3+\frac{3^2}{2}\right)

\Pr(T>2)=1-8.5e^{-3}=0.5768099

Answer

The three-month upper-tail probability rounds to 0.577.

\boxed{0.577\quad\text{(D)}}

Test the method on unseen problems

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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 0.199 does not use the three-month mean of 3. Aggregating independent Poisson months requires adding their means before evaluating the tail.
BThe value 0.224 is P(T=2)=4.5 exp(−3), a single point probability. The event also includes counts three and above.
CThe value 0.423 is P(T≤2)=0.4231901, the complement of the requested upper tail.
EThe value 0.801 is too large for the verified complement. The exact lower-tail mass through two is 8.5 exp(−3), leaving 0.57681.

Original practice · fully worked

Original variant: largest count across two shifts

Two independent work shifts each produce a Poisson number of inspection alerts with mean one. Define M as the maximum of the two shift totals. Calculate the probability that M is exactly two.

A
0.18394
B
0.03383
C
0.30450
D
0.54134
E
0.84585

Answer and calculation

For one shift, the cumulative probabilities through one and two are 2/e and 2.5/e. Independence gives P(M=2)=(2.5/e)²-(2/e)²=2.25/exp(2)=0.30450, choice C.

Setup

Write the one-shift cumulative probabilities needed to locate an exact maximum.

F(1)=e^{-1}(1+1)=\frac2e

F(2)=e^{-1}\left(1+1+\frac12\right)=\frac{2.5}{e}

Model

The maximum equals two when both counts are at most two but they are not both at most one.

\Pr(M=2)=\Pr(M\le2)-\Pr(M\le1)

Compute

Use independence to square the two one-shift cumulative probabilities.

\Pr(M=2)=\left(\frac{2.5}{e}\right)^2-\left(\frac2e\right)^2

\Pr(M=2)=\frac{2.25}{e^2}=0.3045044

Answer

The larger shift count equals two with probability about 0.30450.

\boxed{0.30450\quad\text{(C)}}

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How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: largest count across two shifts

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables