Worked solution · Exam P

SOA Exam P Sample Question #725 Solution

This Exam P sample reference tests Poisson Tail Probability. 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.

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 Tail Probability question

Setup

Let M be the Poisson count with unknown rate λ 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)}}

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.

See the cover and sample pages

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

Answer and calculation

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 λ.

\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)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Poisson Tail Probability question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: telescope calibration alerts

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables