Worked solution · Exam P

SOA Exam P Sample Question #728 Solution

This Exam P sample reference tests Poisson Distribution. The zero-count probability determines the one-hour Poisson mean, and independent hourly counts add their means over the requested interval. The resulting probability is approximately 0.735758, matching choice A.

Independent analysis by ActuaryProof · reviewed July 30, 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 λ denote the common Poisson mean for one hour and recover it from the probability of a zero count.

e^{-\lambda}=0.60653

\lambda=-\log(0.60653)=0.5000010877\ldots

Model

The sum across two independent hours is Poisson with mean equal to the sum of the hourly means.

T\sim\operatorname{Poisson}(2\lambda)

\mu_T=2\lambda=1.0000021754\ldots

Compute

Fewer than two arrivals means that the two-hour total is either zero or one.

\Pr(T<2)=\Pr(T=0)+\Pr(T=1)

\Pr(T<2)=e^{-\mu_T}(1+\mu_T)=0.7357580821\ldots

Answer

The computed probability rounds to the first listed value.

\boxed{\Pr(T<2)\approx 0.736\quad\text{(A)}}

Test the method on unseen problems

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30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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.

BThe value 0.78694 is 2(1-0.60653); it doubles the chance of a nonempty hour and treats that expected number of nonempty hours as the requested count probability.
CThe value 0.803265 comes from 0.60653+0.5(1-0.60653), which mixes a zero-count probability with an invalid proxy for a one-count probability.
DThe value 0.909796 is exp(−λ)(1+λ), the fewer-than-two probability for one hour; it fails to aggregate both hours.
EThe value 0.919698 is exp(−1)(1+1+1/2), which includes totals of two even though the event is strictly fewer than two.

Original practice · fully worked

Original variant: backup security alerts

During an automated overnight backup, security alerts in each 15-minute block follow independent Poisson distributions with mean 0.4. A monitoring window contains three consecutive blocks. Calculate the probability that the window contains at least two alerts.

A
0.301
B
0.337
C
0.361
D
0.663
E
0.699

Answer and calculation

Three independent blocks produce a Poisson total with mean 1.2. Subtracting the zero- and one-alert probabilities from one gives approximately 0.337373, so choice B is correct.

Setup

Identify the mean in each block and the number of independent blocks in the monitoring window.

\lambda_{\text{block}}=0.4,\qquad m=3

Model

Add the independent Poisson counts to obtain the distribution of the window total.

S\sim\operatorname{Poisson}(3\times 0.4)=\operatorname{Poisson}(1.2)

Compute

Use the complement of zero or one alert.

\Pr(S\ge 2)=1-e^{-1.2}(1+1.2)

\Pr(S\ge 2)=0.3373727338\ldots

Answer

The closest listed probability is choice B.

\boxed{\Pr(S\ge 2)\approx 0.337\quad\text{(B)}}

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Next steps for Exam P Univariate Random Variables

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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: backup security alerts

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables