Verified solution · Exam P

SOA Exam P Sample Question #512 Solution

This Exam P sample reference tests Poisson Distribution. A Poisson variable with mean 20 has second moment 20+20²=420. Adding its mean produces 440, so choice D is correct.

Independent analysis by ActuaryProof · reviewed August 4, 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 Distribution question

Setup

Identify the Poisson mean and variance.

\lambda=20

\operatorname{E}[X]=20,\qquad \operatorname{Var}(X)=20

Model

Recover the raw second moment from the variance identity.

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

Compute

Evaluate the second moment and then add the first moment.

\operatorname{E}[X^2]=20+20^2=420

\operatorname{E}[X^2]+\operatorname{E}[X]=420+20=440

Answer

The requested sum of moments is 440.

\boxed{440\quad\text{(D)}}

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Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis replaces E[X²] by Var(X)=20 and then adds E[X]=20, giving 40.
BThis writes E[X²] as Var(X)+E[X]=40 and then adds E[X] once more, giving 60; it omits the squared mean.
CThis correctly calculates E[X²]=420 but stops before adding the additional E[X]=20.
EThis doubles the second moment and then subtracts one mean, 2(420)-20=820, rather than adding one first moment to one second moment.

Original practice · fully worked

Original variant: urgent requests after Poisson splitting

A help desk receives a Poisson number of requests with mean 12 per hour. Each request is independently urgent with probability 1/4. During one hour exactly four nonurgent requests were observed. Calculate the conditional probability that at least two urgent requests arrived.

A
0.0498
B
0.1494
C
0.1991
D
0.8009
E
0.9502

Variant answer in brief

Poisson splitting makes urgent and nonurgent counts independent with respective means 3 and 9. The observed nonurgent count therefore does not change the urgent distribution, and P(U≥2)=1-exp(−3)(1+3)=0.800852, so choice D is correct.

Setup

Split the total Poisson stream by the independent urgent label.

U\sim\operatorname{Poisson}\left(12\cdot\frac14\right)=\operatorname{Poisson}(3)

R\sim\operatorname{Poisson}\left(12\cdot\frac34\right)=\operatorname{Poisson}(9)

Model

The split counts are independent, so conditioning on the nonurgent count leaves the urgent distribution unchanged.

\Pr(U\ge2\mid R=4)=\Pr(U\ge2)

Compute

Complement the zero- and one-urgent probabilities under a Poisson mean of three.

\Pr(U\ge2)=1-\Pr(U=0)-\Pr(U=1)

\Pr(U\ge2)=1-e^{-3}-3e^{-3}=1-4e^{-3}=0.8008517265

Answer

The conditional probability of at least two urgent requests is 0.8009.

\boxed{0.8009\quad\text{(D)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #512 — Verified Solution.” Verified 2026-07-31.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-31.

How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

All 718 Exam P sample solutions in syllabus order

Why the other letters tempt

Original variant: urgent requests after Poisson splitting

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution