Worked solution · Exam P

SOA Exam P Sample Question #200 Solution

This Exam P sample reference tests Poisson Distribution. Independent marking thins the original Poisson count to a Poisson count with mean 1.25. Its zero-count probability is exp(-1.25) = 0.286505, which selects choice E.

Independent analysis by ActuaryProof · reviewed July 31, 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 U be the daily count in the marked category. The total count has mean 5, and an item enters U with probability 0.25.

\lambda=5

q=0.25

Model

Poisson thinning gives another Poisson variable whose mean is the original mean times the marking probability.

U\sim\operatorname{Poisson}(\lambda q)

\lambda_U=5(0.25)=1.25

Compute

Evaluate the probability mass at zero for the thinned count.

\Pr(U=0)=e^{-1.25}\frac{1.25^0}{0!}=e^{-1.25}

e^{-1.25}=0.2865047969

Answer

Rounding to the listed precision gives 0.287.

\boxed{0.287\quad\text{(E)}}

Test the method on unseen problems

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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.

AThe value 0.007 is exp(−5) = 0.006738, the probability that the total Poisson count is zero. It is unnecessarily restrictive.
BA zero-count probability of 0.010 would imply a Poisson mean of −ln(0.010) = 4.605, not the thinned mean 5(0.25) = 1.25.
CA zero-count probability of 0.124 corresponds to an unsupported mean −ln(0.124) = 2.087. The correct marked rate is 1.25.
DThe value 0.237 is 0.75⁵ = 0.237305. This incorrectly treats the random Poisson total as exactly five fixed trials.

Original practice · fully worked

Original variant: two-category Poisson splitting

An operations dashboard records a Poisson number of alerts with mean 6 during an evening. Each alert is independently labeled urgent with probability 0.20, routine with probability 0.50, or informational otherwise. Calculate the probability of exactly one urgent alert and no routine alerts.

A
0.0150
B
0.0180
C
0.0498
D
0.3010
E
0.3614

Answer and calculation

Poisson splitting makes the urgent and routine counts independent with means 1.2 and 3. Their joint event has probability 1.2 exp(−4.2) = 0.017995, choice B.

Setup

Denote the urgent and routine counts by U and R.

\lambda_U=6(0.20)=1.2

\lambda_R=6(0.50)=3

Model

Disjoint markings of a Poisson stream produce independent category counts.

U\sim\operatorname{Poisson}(1.2)

R\sim\operatorname{Poisson}(3)

U\perp R

Compute

Multiply the one-urgent and zero-routine probabilities.

\Pr(U=1,R=0)=(1.2e^{-1.2})(e^{-3})

=1.2e^{-4.2}=0.0179946922

Answer

The requested joint probability rounds to 0.0180.

\boxed{0.0180\quad\text{(B)}}

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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: two-category Poisson splitting

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables