Worked solution · Exam P

SOA Exam P Sample Question #698 Solution

This Exam P sample reference tests Poisson Superposition. The stated ratio of the one-count and zero-count masses fixes each Poisson mean at 2. Adding three independent counts gives a Poisson mean of 6, whose probability of at least two events is 0.982649, matching 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 Superposition question

Setup

Let λ be the common mean for one location and translate the supplied relationship between its first two Poisson masses.

\Pr(X=0)=e^{-\lambda},\qquad \Pr(X=1)=\lambda e^{-\lambda}

\lambda e^{-\lambda}=2e^{-\lambda}

Model

Cancel the positive exponential factor to find the individual mean, then superpose the three independent Poisson counts.

\lambda=2

T=X_1+X_2+X_3\sim\operatorname{Poisson}(3\lambda)=\operatorname{Poisson}(6)

Compute

Use the complement of total counts zero and one.

\Pr(T\ge 2)=1-\Pr(T=0)-\Pr(T=1)

\Pr(T\ge 2)=1-e^{-6}(1+6)=0.9826487348\ldots

Answer

Rounding the computed probability to three decimals gives the final listed value.

\boxed{\Pr(T\ge 2)\approx 0.983\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.

AThe value 18 exp(−6)=0.044618 is the probability that the combined count equals exactly two, not that it is at least two.
BThe value exp(−6)(1+6+18)=0.061969 is the probability that the combined count is at most two, which reverses the tail and includes the boundary.
CThe value 1-exp(−2)(1+2)=0.593994 is the at-least-two probability for one location and fails to combine all three counts.
DThe value 1-exp(−6)(1+6+18)=0.938031 is the probability of at least three events, an off-by-one threshold error.

Original practice · fully worked

Original variant: satellite receiver alerts

Four satellite receivers record independent Poisson numbers of false alerts during a diagnostic interval, all with the same positive mean. At one receiver, the probability of two alerts is one-half the probability of one alert. Calculate the probability of at least three alerts across all four receivers.

A
0.080
B
0.195
C
0.238
D
0.762
E
0.908

Answer and calculation

The within-receiver mass ratio gives a mean of 1, so the four-receiver total is Poisson with mean 4. Its probability of at least three alerts is 0.761897, selecting choice D.

Setup

Write the one- and two-alert masses for a Poisson variable with positive mean λ.

\Pr(X=1)=\lambda e^{-\lambda},\qquad \Pr(X=2)=\frac{\lambda^2e^{-\lambda}}{2}

Model

Apply the stated mass ratio and then add the independent receiver counts.

\frac{\lambda^2e^{-\lambda}}{2}=\frac{1}{2}\lambda e^{-\lambda}\quad\Longrightarrow\quad \lambda=1

S\sim\operatorname{Poisson}(4)

Compute

Subtract the probabilities of zero, one, and two alerts from one.

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

\Pr(S\ge 3)=0.7618966944\ldots

Answer

The required four-receiver tail probability rounds to 0.762.

\boxed{\Pr(S\ge 3)\approx 0.762\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Poisson Superposition question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: satellite receiver alerts

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables