SOA Exam P Sample Question #736. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #736 — Verified Solution

Independent solution

How to solve this Poisson Distribution question

Answer in brief

Use the Poisson second-moment identity to identify the rate, then evaluate the upper tail by complement. The verified value is approximately 0.323324, corresponding to choice C.

Setup

Write the Poisson mean and variance in terms of its rate parameter.

\mathbb{E}[X]=\lambda,\qquad \operatorname{Var}(X)=\lambda

Model

Connect the given second moment to the rate through the variance identity.

\mathbb{E}[X^2]=\operatorname{Var}(X)+\mathbb{E}[X]^2=\lambda+\lambda^2

\lambda^2+\lambda=6

Compute

Only the nonnegative root is admissible for a Poisson rate. Compute the requested tail by subtracting the first three masses.

(\lambda-2)(\lambda+3)=0\quad\Longrightarrow\quad\lambda=2

\Pr(X\ge 3)=1-\sum_{k=0}^{2}e^{-2}\frac{2^k}{k!}

\Pr(X\ge 3)=1-5e^{-2}=0.3233235838\ldots

Answer

The computed value rounds to the value listed under choice C.

\boxed{\Pr(X\ge 3)\approx 0.323324\quad\text{(C)}}

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 probability of four or more events, so it results from shifting the tail cutoff upward by one.
BThis keeps only the probability of exactly three events instead of adding the entire upper tail.
DThis is produced by using 3 as the Poisson rate before evaluating the same tail.
EThis is the probability of two or more events, so it results from shifting the cutoff downward by one.

Original practice · fully worked

Original variant: observatory alert count

An observatory models the number of false instrument alerts during a 12-hour shift with a Poisson random variable N. The probability of no alerts during a shift is e^{-1.6}. Calculate the probability that at most two alerts occur during a shift.

A
0.202
B
0.323
C
0.525
D
0.783
E
0.921

Variant answer in brief

The zero-count probability identifies the rate as 1.6. Summing the masses for zero, one, and two gives approximately 0.783.

Setup

For a Poisson count, the zero-count probability is the exponential of the negative rate.

\Pr(N=0)=e^{-\lambda}

Model

Match the supplied zero-count probability to the Poisson formula.

e^{-\lambda}=e^{-1.6}\quad\Longrightarrow\quad\lambda=1.6

Compute

Add the probabilities of zero, one, and two alerts.

\Pr(N\le 2)=e^{-1.6}\left(1+1.6+\frac{1.6^2}{2}\right)

\Pr(N\le 2)=3.88e^{-1.6}=0.783358\ldots

Answer

The closest listed value is choice D.

\boxed{\Pr(N\le 2)\approx 0.783\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #736 — Verified Solution.” Verified 2026-07-30.

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