Worked solution · Exam P

SOA Exam P Sample Question #300 Solution

This Exam P sample reference tests Poisson Distribution. This is a lower-tail probability for an aggregate Poisson count. Adding three independent annual means gives a three-year mean of 0.864; the zero-count and one-count masses sum to e to the minus 0.864 × 1.864, or 0.7856 and choice E.

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 N be the claim count over the full three-year window and add the independent annual Poisson means.

N=N_1+N_2+N_3

N\sim\operatorname{Poisson}(3\cdot0.288)=\operatorname{Poisson}(0.864)

Model

Eligibility allows exactly zero or exactly one claim, two disjoint Poisson outcomes.

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

Compute

Evaluate the first two masses of the aggregate distribution.

\Pr(N\le1)=e^{-0.864}+0.864e^{-0.864}

=1.864e^{-0.864}=0.7856253267

Answer

The probability rounds to 0.79.

\boxed{0.79\quad\text{(E)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

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.

AThe value 0.01 is the three-year probability of exactly four claims, rounded from 0.009786. Four claims are outside the qualifying count range.
BThe value 0.36 is the probability of exactly one three-year claim, rounded from 0.364153. It omits qualifying policies with no claims.
CThe value 0.42 is the probability of no three-year claims, rounded from 0.421473. It omits the allowed one-claim outcome.
DThe value 0.54 cannot contain both permitted masses: those masses are 0.421473 and 0.364153 separately. Their disjoint sum, rather than a partial combination, is required.

Original practice · fully worked

Original variant: standing account in a random rebate split

The number N of unexpected support tickets in a service window is Poisson with mean 2. A rebate of 90 credits is divided equally among the N tickets and one standing account. Calculate the expected number of credits assigned to the standing account.

A
12.18 credits
B
30.00 credits
C
38.91 credits
D
45.00 credits
E
77.82 credits

Answer and calculation

Conditional on N, the standing account receives 90/(N+1). Summing this amount against the Poisson masses and shifting the factorial index gives 45(1-e to the minus 2)=38.9099 credits, so choice C.

Setup

Write the standing-account allocation as a function of the random ticket count.

N\sim\operatorname{Poisson}(2),\qquad R=\frac{90}{N+1}

Model

Average the conditional allocation over the Poisson mass function.

E[R]=90e^{-2}\sum_{n=0}^{\infty}\frac{2^n}{(n+1)n!}

Compute

Use (n+1)n!=(n+1)! and shift the exponential-series index.

\sum_{n=0}^{\infty}\frac{2^n}{(n+1)!}=\frac12\sum_{k=1}^{\infty}\frac{2^k}{k!}=\frac{e^2-1}{2}

E[R]=45e^{-2}(e^2-1)=45(1-e^{-2})=38.90991225

Answer

The standing account receives about 38.91 credits on average.

\boxed{38.91\text{ credits}\quad\text{(C)}}

Keep practicing

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: standing account in a random rebate split

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables