Verified 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 August 4, 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)}}

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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

Variant answer in brief

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)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #402 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: covariance of overlapping event windows

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution