SOA Exam P Sample Question #725. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #725

Take the ratio of the adjacent Poisson masses to identify the rate as 4. The complement through count 2 then gives an upper-tail probability of approximately 0.761897, corresponding to choice E.

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Independent solution

How to solve this Poisson Distribution question

Answer in brief

Take the ratio of the adjacent Poisson masses to identify the rate as 4. The complement through count 2 then gives an upper-tail probability of approximately 0.761897, corresponding to choice E.

Setup

Let M be the Poisson count with unknown rate lambda and write its probability mass function.

\Pr(M=k)=e^{-\lambda}\frac{\lambda^k}{k!}

Model

For adjacent Poisson masses, their ratio removes the common exponential factor and most powers.

\frac{\Pr(M=4)}{\Pr(M=3)}=\frac{\lambda}{4}

\Pr(M=3)=\Pr(M=4)\quad\Longrightarrow\quad\frac{\lambda}{4}=1

Compute

Substitute the identified rate and calculate the requested upper tail by complement.

\lambda=4

\Pr(M\ge3)=1-\Pr(M\le2)

\Pr(M\ge3)=1-e^{-4}\left(1+4+\frac{4^2}{2}\right)=1-13e^{-4}=0.7618966944\ldots

Answer

The tail probability rounds to the value shown in choice E.

\boxed{\Pr(M\ge3)\approx0.762\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.

AThis is the single mass P(M=3)=0.195367..., not the probability of the whole upper tail.
BThis is the complement P(M <= 2)=0.238103..., so it reverses the requested event.
CThis is P(M <= 3)=0.433470..., obtained by adding the mass at 3 to choice B rather than taking the upper tail.
DThis is P(M >= 4)=0.566530..., caused by shifting the inclusive cutoff from 3 to 4.

Original practice · fully worked

Original variant: telescope calibration alerts

An autonomous telescope models its number of calibration alerts in one night by a Poisson distribution. The probability of exactly two alerts equals the probability of exactly three alerts. Calculate the probability that no more than one alert occurs during a night.

A
0.0498
B
0.1991
C
0.2240
D
0.4232
E
0.8009

Variant answer in brief

Equality of the masses at 2 and 3 forces the Poisson rate to equal 3. Summing the zero- and one-alert probabilities gives approximately 0.1991, so the answer is B.

Setup

Let N be the nightly alert count with Poisson rate lambda.

\Pr(N=k)=e^{-\lambda}\frac{\lambda^k}{k!}

Model

Use the ratio of the two adjacent masses to solve for the rate.

\frac{\Pr(N=3)}{\Pr(N=2)}=\frac{\lambda}{3}=1

\lambda=3

Compute

No more than one alert includes exactly zero and exactly one.

\Pr(N\le1)=\Pr(N=0)+\Pr(N=1)

\Pr(N\le1)=e^{-3}+3e^{-3}=4e^{-3}=0.1991482735\ldots

Answer

The probability rounds to choice B.

\boxed{\Pr(N\le1)\approx0.1991\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #725 — Verified Solution.” Verified 2026-07-30.

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