SOA Exam P Sample Question #698. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #698 — Verified Solution

Independent solution

How to solve this Poisson Distribution question

Answer in brief

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.

Setup

Let lambda 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)}}

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 18e^{-6}=0.044618 is the probability that the combined count equals exactly two, not that it is at least two.
BThe value e^{-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-e^{-2}(1+2)=0.593994 is the at-least-two probability for one location and fails to combine all three counts.
DThe value 1-e^{-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

Variant answer in brief

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 lambda.

\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)}}

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ActuaryProof. “SOA Exam P Sample Question #698 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.