Worked solution · Exam P

SOA Exam P Sample Question #643 Solution

This Exam P sample reference tests Discrete Random Variables. This is a Poisson cumulative-probability calculation. Summing the masses from zero through four for a Poisson mean of two gives 7 times exp(-2), approximately 0.947347, which matches 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 Discrete Random Variables question

Setup

Let N be the count in the stated one-year interval. Its Poisson mean is two.

N\sim\operatorname{Poisson}(2),\qquad \Pr(N=k)=e^{-2}\frac{2^k}{k!}

Model

Not exceeding four means that the possible counts are zero, one, two, three, and four.

\Pr(N\le4)=\sum_{k=0}^{4}e^{-2}\frac{2^k}{k!}

Compute

Factor out the common exponential term and add the five finite-series coefficients.

\Pr(N\le4)=e^{-2}\left(1+2+2+\frac{4}{3}+\frac{2}{3}\right)

\Pr(N\le4)=7e^{-2}=0.9473469827

Answer

The cumulative probability through count four is 7 exp(-2).

\boxed{7e^{-2}\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.

AThis expression is only the single Poisson mass at count four. A cumulative probability must also include counts zero through three.
BThis takes the complement of the cumulative probability through three, producing the probability of four or more rather than no more than four.
CThis is the cumulative probability through count three. It is missing the nonzero mass at count four.
DThis is the probability of at least one occurrence, found by excluding only count zero; it does not impose the upper cutoff at four.

Original practice · fully worked

Original variant: meteor calibration window

During a twenty-minute calibration window, an observatory camera records a Poisson number of meteor streaks with mean 1.5. Calculate the probability that the camera records no more than three streaks during the window.

A
0.1255
B
0.2231
C
0.5578
D
0.9344
E
0.9814

Answer and calculation

Adding the Poisson masses for zero through three streaks gives 0.9343575. Rounded to four decimals, this is 0.9344 and selects choice D.

Setup

Let M denote the number of recorded streaks in the calibration window.

M\sim\operatorname{Poisson}(1.5)

Model

The requested event contains the four counts from zero through three.

\Pr(M\le3)=e^{-1.5}\sum_{k=0}^{3}\frac{1.5^k}{k!}

Compute

Evaluate the short cumulative sum and then apply the exponential factor.

\Pr(M\le3)=e^{-1.5}(1+1.5+1.125+0.5625)

\Pr(M\le3)=4.1875e^{-1.5}=0.9343575456

Answer

The four-decimal probability is 0.9344.

\boxed{0.9344\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Discrete Random Variables question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: meteor calibration window

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables