SOA Exam P Sample Question #665. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #665

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.

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

Independent solution

How to solve this Poisson Distribution question

Answer in brief

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.

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)}}

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

Variant answer in brief

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)}}

Continue by topic

Build a complete proof notebook

The Probability Proof Manual reorganizes verified solutions by syllabus skill and adds formula proofs, error patterns, and original practice.

Explore study products Get the free formula sheet

Cite this solution

ActuaryProof. “SOA Exam P Sample Question #665 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.