SOA Exam P Sample Question #619. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #619 — Verified Solution

Independent solution

How to solve this Poisson Distribution question

Answer in brief

Taking the ratio of the Poisson probabilities cancels their common exponential factor and gives λ^4/4!=54. Thus λ=6 per minute, and linearity of expectation gives 60(6)=360 over one hour, selecting choice E.

Setup

Let the common mean count in one minute be the positive parameter lambda.

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

Model

Form the stated probability ratio. The exponential terms cancel.

\frac{\Pr(N=4)}{\Pr(N=0)}=\frac{\lambda^4}{4!}=54

\lambda^4=54(24)=1296

Compute

Use the positive fourth root for a Poisson mean, then scale the expectation to sixty minutes.

\lambda=1296^{1/4}=6

\operatorname{E}[N_{60}]=60\lambda=360

Answer

The expected count in the requested period is 360.

\boxed{360\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 corresponds to guessing a rate of 2 per minute instead of solving the fourth-power probability-ratio equation.
BOmitting the factor 4! gives the incorrect equation lambda^4=54; rounding its fourth root to 3 produces 180 over sixty minutes.
CUsing the observed count label 4 as the mean rather than solving for lambda gives 60(4)=240.
DThis corresponds to using 5 as the fourth root of 1296; checking powers shows that 5^4 is only 625 while 6^4 is 1296.

Original practice · fully worked

Original variant: bright-meteor thinning

Meteors reach an automated detector according to a Poisson process with mean rate 12 per hour. Each detected meteor is independently classified as bright with probability one-third and dim otherwise. During a thirty-minute interval, calculate the probability of exactly two bright meteors and at least one dim meteor.

A
0.045
B
0.266
C
0.271
D
0.982
E
2.000

Variant answer in brief

Poisson thinning produces independent bright and dim counts with means 2 and 4. Multiplying the probability of two bright meteors by the probability of at least one dim meteor gives 0.265713, so choice B is correct.

Setup

Convert the hourly rate to the mean count over one half-hour.

N\sim\operatorname{Poisson}\!\left(12\cdot\frac12\right)=\operatorname{Poisson}(6)

Model

Thin the process by classification. The bright and dim counts are independent Poisson variables with their respective split means.

B\sim\operatorname{Poisson}\!\left(6\cdot\frac13\right)=\operatorname{Poisson}(2)

D\sim\operatorname{Poisson}\!\left(6\cdot\frac23\right)=\operatorname{Poisson}(4)

B\ \text{and}\ D\ \text{are independent}

Compute

Multiply the exact bright-count probability by the dim-count complement.

\Pr(B=2,D\ge1)=\left(e^{-2}\frac{2^2}{2!}\right)\left(1-e^{-4}\right)

\Pr(B=2,D\ge1)=0.2657130621\ldots

Answer

The requested joint probability rounds to 0.266.

\boxed{0.266\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #619 — Verified Solution.” Verified 2026-07-30.

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