Worked solution · Exam P

SOA Exam P Sample Question #665 Solution

This Exam P sample reference tests Poisson Distribution. Independent Poisson counts add, so the full-document count is Poisson with mean 10. Complementing its cumulative probability through four gives 0.970747, which rounds to choice C.

Independent analysis by ActuaryProof · reviewed July 30, 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 X_j be the count on page j and let S be the count over the entire document.

X_j\sim\operatorname{Poisson}(0.10),\qquad S=\sum_{j=1}^{100}X_j

Model

The independent Poisson components combine into another Poisson variable whose mean is the sum of the page means.

S\sim\operatorname{Poisson}(100\cdot0.10)=\operatorname{Poisson}(10)

Compute

More than four is the complement of counts zero through four.

\Pr(S>4)=1-e^{-10}\sum_{k=0}^{4}\frac{10^k}{k!}

\Pr(S>4)=1-e^{-10}\left(1+10+50+\frac{1000}{6}+\frac{10000}{24}\right)

\Pr(S>4)=0.9707473119\ldots

Answer

The upper-tail probability rounds to 0.97.

\boxed{0.97\quad\text{(C)}}

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.

AThis is approximately the probability of four or fewer counts only through three, P(S≤3)=0.01034; it both uses the lower tail and drops the boundary value four.
BThis is the lower-tail probability P(S≤4)=0.02925. It is the complement of the requested result.
DThis is approximately 1-P(S=4)=0.98108, which excludes exactly four but incorrectly keeps counts zero through three.
EThis is approximately P(S≥4)=0.98966, an off-by-one tail that includes the four-count outcome.

Original practice · fully worked

Original variant: satellite mosaic false detections

A satellite image mosaic is divided into 40 independently processed zones. The number of false detections in each zone is Poisson with mean 0.075. Calculate the probability that the completed mosaic contains more than two false detections.

A
0.0498
B
0.2240
C
0.4232
D
0.5768
E
0.8009

Answer and calculation

The 40 independent zone counts produce a Poisson total with mean 3. Its probability above two is 0.576810, selecting choice D.

Setup

Let F be the total false-detection count across all zones.

F=\sum_{j=1}^{40}F_j,\qquad F_j\sim\operatorname{Poisson}(0.075)

Model

Poisson superposition adds the 40 zone means.

F\sim\operatorname{Poisson}(40\cdot0.075)=\operatorname{Poisson}(3)

Compute

Use the complement of the zero-, one-, and two-count masses.

\Pr(F>2)=1-e^{-3}\left(1+3+\frac{3^2}{2}\right)

\Pr(F>2)=0.5768099189\ldots

Answer

The probability of more than two false detections is approximately 0.5768.

\boxed{0.5768\quad\text{(D)}}

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: satellite mosaic false detections

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables