Worked solution · Exam P

SOA Exam P Sample Question #302 Solution

This Exam P sample reference tests Poisson Distribution. This is a lower tail of an aggregate Poisson count. Three independent daily counts with mean 4 combine to a Poisson mean of 12; adding the zero-count and one-count masses gives (1+12)e to the minus 12, so choice A.

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 S be the total count across the three independent days and add their Poisson means.

S=X_1+X_2+X_3

S\sim\operatorname{Poisson}(4+4+4)=\operatorname{Poisson}(12)

Model

At most one event consists of the two disjoint count outcomes zero and one.

\Pr(S\le1)=\Pr(S=0)+\Pr(S=1)

Compute

Evaluate and combine the first two Poisson masses.

\Pr(S\le1)=e^{-12}+12e^{-12}=13e^{-12}

13e^{-12}=0.00007987476059

Answer

The exact listed expression is 13e to the minus 12.

\boxed{13e^{-12}\quad\text{(A)}}

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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 expression 72e to the minus 12 is the probability of exactly two events, since 12 squared divided by 2 factorial is 72. The requested count is at most one.
CThe expression 85e to the minus 12 adds the exactly-two mass 72e to the minus 12 to the correct zero-and-one total. It calculates at most two events.
DThe expression 5e to the minus 4 is the zero-and-one probability for a single Poisson mean of 4. It fails to aggregate all three days.
EThe expression 13e to the minus 4 mixes the aggregate coefficient 13 with the one-day exponential factor. The full three-day mean is 12 in both parts of the Poisson mass.

Original practice · fully worked

Original variant: unresolved alerts after limited service

Two independent radio channels generate Poisson alert counts during a shift, with means 0.8 and 1.2. A duty operator can resolve one alert during the shift; any additional alerts remain unresolved. Calculate the expected number of unresolved alerts.

A
0.1353
B
0.8647
C
1.0000
D
1.1353
E
2.0000

Answer and calculation

The combined alert count is Poisson with mean 2. The number resolved is one whenever at least one alert arrives, so its expectation is 1-exp(−2). Subtracting this from the expected total gives 2-(1-exp(−2))=1+exp(−2)=1.1353, choice D.

Setup

Let X and Y be the independent channel counts and N their combined total.

N=X+Y\sim\operatorname{Poisson}(0.8+1.2)=\operatorname{Poisson}(2)

Model

The operator resolves one alert if N is positive and resolves none if N is zero.

U=(N-1)_+=N-\mathbf{1}_{\{N\ge1\}}

Compute

Take expectations and use the zero-count probability of a Poisson variable.

\operatorname{E}[U]=\operatorname{E}[N]-\Pr(N\ge1)

=2-\left(1-e^{-2}\right)=1+e^{-2}=1.1353352832

Answer

The expected unresolved count rounds to 1.1353.

\boxed{1.1353\quad\text{(D)}}

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How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: unresolved alerts after limited service

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables