Worked solution · Exam P

SOA Exam P Sample Question #700 Solution

This Exam P sample reference tests Poisson Distribution. Adding the 25 independent Poisson counts gives a manuscript total with mean 5. Summing its masses from zero through three produces 0.265026, which rounds to choice E.

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 item j and let T be the count across all 25 items.

X_j\sim\operatorname{Poisson}(0.20),\qquad j=1,\ldots,25

T=\sum_{j=1}^{25}X_j

Model

A sum of independent Poisson variables is Poisson with mean equal to the sum of the component means.

T\sim\operatorname{Poisson}(25\times 0.20)=\operatorname{Poisson}(5)

Compute

Three or fewer includes each total from zero through three, so add all four corresponding masses.

\Pr(T\le 3)=e^{-5}\sum_{k=0}^{3}\frac{5^k}{k!}

\Pr(T\le 3)=e^{-5}\left(1+5+\frac{25}{2}+\frac{125}{6}\right)=0.2650259153\ldots

Answer

The cumulative probability rounds to 0.265.

\boxed{\Pr(T\le 3)\approx 0.265\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.

AUsing an aggregate mean of 10 instead of 5 gives P(Poisson(10) at most 3)=0.010336, reflecting a doubled-exposure error.
BThe value exp(−5)(1+5+25/2)=0.124652 is P(T at most 2), which drops the allowed boundary value 3.
CThe value 5³ exp(−5)/3!=0.140374 is P(T=3) only, rather than the cumulative probability through 3.
DThe value exp(−5)(5+25/2+125/6)=0.258288 is P(1 at most T at most 3), which incorrectly excludes the zero-count outcome.

Original practice · fully worked

Original variant: telescope calibration hits

A telescope records cosmic-ray hits during 12 separate calibration windows. Counts in different windows are independent, and each window's count is Poisson with mean 0.25. Calculate the probability that no more than two hits are recorded over the full calibration session.

A
0.050
B
0.199
C
0.224
D
0.373
E
0.423

Answer and calculation

The 12 independent counts combine to a Poisson total with mean 3. Adding the zero-, one-, and two-hit masses gives 0.423190, so choice E is correct.

Setup

Let H be the total number of hits in all calibration windows.

H=\sum_{j=1}^{12}H_j,\qquad H_j\sim\operatorname{Poisson}(0.25)

Model

Superposition preserves the Poisson form and adds the 12 means.

H\sim\operatorname{Poisson}(12\times 0.25)=\operatorname{Poisson}(3)

Compute

No more than two hits means a total of zero, one, or two.

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

\Pr(H\le 2)=0.4231900811\ldots

Answer

The full-session probability is approximately 0.423.

\boxed{\Pr(H\le 2)\approx 0.423\quad\text{(E)}}

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: telescope calibration hits

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables