Worked solution · Exam P

SOA Exam P Sample Question #584 Solution

This Exam P sample reference tests Independent Poisson Sum. A one-year zero-count probability of 0.90 implies a Poisson mean of -ln(0.90). Adding 15 independent yearly counts gives a Poisson total whose variance is 15[-ln(0.90)]=1.5804, so choice E is correct.

Independent analysis by ActuaryProof · reviewed July 30, 2026

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Independent solution

How to solve this Independent Poisson Sum question

Setup

Let λ be the mean count for one year and convert the positive-count probability to a zero-count probability.

\Pr(N=0)=1-0.10=0.90

Model

For a Poisson count, the zero probability determines its parameter. Independent Poisson counts add by summing their parameters.

e^{-\lambda}=0.90

\lambda=-\ln(0.90)

S=\sum_{j=1}^{15}N_j\sim\operatorname{Poisson}(15\lambda)

Compute

Use equality of the Poisson mean and variance for the aggregated count.

\operatorname{Var}(S)=15\lambda

\operatorname{Var}(S)=-15\ln(0.90)=1.5804077349\ldots

Answer

The variance of the 15-year total rounds to 1.58.

\boxed{1.58\quad\text{(E)}}

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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 is approximately the one-year Poisson mean -ln(0.90), so it omits aggregation over all 15 years.
BThis scales the one-year parameter by the square root of 15, mixing a standard-deviation scaling rule with the Poisson variance calculation.
CThis treats each year as a Bernoulli indicator with variance 0.10(0.90) and adds 15 such variances, discarding the possibility of multiple counts.
DThis replaces the Poisson mean -ln(0.90) by the positive-count probability 0.10 and reports 15(0.10).

Original practice · fully worked

Original variant: quiet monitoring window

Independent daily alert counts for a monitoring system follow Poisson distributions with a common mean. The variance of the total alert count over eight days is 12. Calculate the probability of receiving no alerts during a separate three-day period.

A
0.000006
B
0.011109
C
0.125000
D
0.223130
E
0.988891

Answer and calculation

The eight-day Poisson total has mean and variance 12, so the daily mean is 1.5. A three-day total therefore has mean 4.5, and its zero-count probability is exp(-4.5)=0.011109, which is choice B.

Setup

Let λ denote the common mean of one daily Poisson count.

N_d\sim\operatorname{Poisson}(\lambda)

Model

The eight-day total is Poisson with parameter 8lambda, and its variance is therefore 8lambda.

\operatorname{Var}\!\left(\sum_{d=1}^{8}N_d\right)=8\lambda=12

Compute

Recover the daily parameter, form the three-day parameter, and evaluate the Poisson zero probability.

\lambda=1.5

\sum_{d=1}^{3}N_d\sim\operatorname{Poisson}(4.5)

\Pr\!\left(\sum_{d=1}^{3}N_d=0\right)=e^{-4.5}=0.0111089965\ldots

Answer

The probability that the three-day period is quiet is approximately 0.011109.

\boxed{0.011109\quad\text{(B)}}

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How to solve this Independent Poisson Sum question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: quiet monitoring window

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables