Worked solution · Exam P

SOA Exam P Sample Question #529 Solution

This Exam P sample reference tests Poisson Distribution. Each count has Poisson mean 0.25, and independence makes their aggregate Poisson with mean 0.50. Complementing the zero- and one-count masses gives 0.090204, so choice B is correct.

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.

Read the solution Open official source

Independent solution

How to solve this Poisson Distribution question

Setup

For a Poisson variable, the variance equals the mean. The supplied variance therefore identifies the rate for one item.

X_1,X_2\sim\operatorname{Poisson}(0.25)

Model

Independent Poisson variables add to another Poisson variable whose mean is the sum of the component means.

T=X_1+X_2\sim\operatorname{Poisson}(0.25+0.25)=\operatorname{Poisson}(0.50)

Compute

The required upper tail is most directly evaluated by removing the masses at zero and one.

\Pr(T>1)=1-\Pr(T=0)-\Pr(T=1)

\Pr(T>1)=1-e^{-0.5}-0.5e^{-0.5}=0.09020401043

Answer

The result rounds to the second listed probability.

\boxed{0.09\quad\text{(B)}}

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.

See the cover and sample pages

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 the one-item tail, 1-exp(−0.25)(1+0.25)=0.0265, rounded to 0.03; it never combines the two independent counts.
CThis is the probability of exactly one aggregate event, 0.5 exp(−0.5)=0.3033, rather than the probability of exceeding one.
DThis is the aggregate zero-count mass, exp(−0.5)=0.6065, so it selects the complement's first term instead of the upper tail.
EThis is Pr(T≤1)=1.5 exp(−0.5)=0.9098, the complement of the requested event.

Original practice · fully worked

Original variant: pooled sensor alerts

Three sensor clusters generate false alerts as independent Poisson processes with rates 0.15, 0.35, and 0.20 per hour. During a two-hour diagnostic window, calculate the probability that the monitoring service records at least three false alerts.

A
0.1665
B
0.2417
C
0.2466
D
0.3452
E
0.8335

Answer and calculation

Pooling the three streams over two hours gives a Poisson mean of 1.4. Complementing the masses for zero, one, and two alerts produces 0.166502, so choice A is correct.

Setup

Add the three hourly rates before applying the two-hour exposure.

\lambda_{\text{hour}}=0.15+0.35+0.20=0.70

Model

Superposition and time scaling give the diagnostic-window count.

N\sim\operatorname{Poisson}(2\cdot0.70)=\operatorname{Poisson}(1.40)

Compute

Remove all outcomes below three.

\Pr(N\ge3)=1-e^{-1.4}\left(1+1.4+\frac{1.4^2}{2}\right)

\Pr(N\ge3)=0.1665022619

Answer

The pooled upper tail rounds to 0.1665.

\boxed{0.1665\quad\text{(A)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: pooled sensor alerts

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables