SOA Exam P Sample Question #728. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #728 — Verified Solution

Independent solution

How to solve this Poisson Distribution question

Answer in brief

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.

Setup

Let lambda 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)}}

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 e^{-lambda}(1+lambda), the fewer-than-two probability for one hour; it fails to aggregate both hours.
EThe value 0.919698 is e^{-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

Variant answer in brief

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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ActuaryProof. “SOA Exam P Sample Question #728 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.