Worked solution · Exam P

SOA Exam P Sample Question #402 Solution

This Exam P sample reference tests Poisson Distribution. This problem recovers a one-period Poisson mean from the identity E[X²]=λ²+λ. Scaling that mean to fifteen periods and complementing counts zero through two gives 0.655362 and 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

For a Poisson variable, variance and mean both equal λ. Convert the supplied raw second moment into an equation for λ.

E[X^2]=\operatorname{Var}(X)+E[X]^2=\lambda+\lambda^2

\lambda^2+\lambda=0.2756

Model

Keep the positive quadratic root and add the independent one-period means across the longer interval.

\lambda=\frac{-1+\sqrt{1+4(0.2756)}}{2}=0.2249827584

\mu=15\lambda=3.3747413762

Compute

More than two events is the complement of the first three Poisson masses.

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

\Pr(Y>2)=0.6553624514

Answer

The upper-tail probability rounds to 0.655.

\boxed{0.655\quad\text{(A)}}

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.

BTreating the raw second moment 0.2756 as the one-period mean gives interval mean 4.134 and tail probability 0.780881, which rounds to 0.781.
CA tail probability of 0.805 would require interval mean about 4.319, or one-period mean about 0.288. That parameter has second moment λ+λ² about 0.371, not 0.2756.
DUsing the correct interval mean but complementing only counts zero and one gives Pr(Y>1)=0.850266. That calculates at least two events, not more than two.
EUsing 0.2756 as the one-period mean and also complementing only zero and one gives 0.917760, which rounds to 0.918. It combines the raw-moment error with the threshold error.

Original practice · fully worked

Original variant: covariance of overlapping event windows

Events in a system log follow a homogeneous Poisson process. For the count N in one hour, E[N²]=6. Let A count events from time 0 to 90 minutes, and let B count events from time 30 to 120 minutes. Calculate Cov(A,B).

A
0 covariance units
B
1 covariance unit
C
2 covariance units
D
3 covariance units
E
6 covariance units

Answer and calculation

The moment equation λ²+λ=6 gives rate 2 per hour. The two windows share one hour, and only the count in that shared interval contributes covariance, so Cov(A,B)=2 and choice C.

Setup

Recover the hourly Poisson rate from the raw second moment.

\lambda^2+\lambda=6

\lambda=2

Model

Split the two observation windows into disjoint half-hour, one-hour, and half-hour increments.

A=U+V,\qquad B=V+W

U,V,W\text{ are independent},\qquad V\sim\operatorname{Poisson}(2)

Compute

All cross-covariances vanish except the variance of the shared increment.

\operatorname{Cov}(A,B)=\operatorname{Cov}(U+V,V+W)

=\operatorname{Var}(V)=2

Answer

The overlapping counts have covariance 2.

\boxed{2\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: covariance of overlapping event windows

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables