Verified solution · Exam P

SOA Exam P Sample Question #29 Solution

This Exam P sample reference tests Poisson Distribution. The Poisson probability ratio reduces to λ²=4. The positive rate is λ=2, and a Poisson variable has variance λ, so choice D.

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

Write the stated relation between the Poisson probabilities at counts two and four.

\Pr(N=2)=3\Pr(N=4)

Model

Substitute the Poisson mass formula on both sides and cancel the common exponential factor.

e^{-\lambda}\frac{\lambda^2}{2!}=3e^{-\lambda}\frac{\lambda^4}{4!}

Compute

The remaining equation gives λ²=4. Since a Poisson rate is nonnegative, λ=2, and the Poisson variance equals λ.

\lambda^2=4,\qquad \lambda=2,\qquad \operatorname{Var}(N)=\lambda

Answer

The variance is 2, corresponding to official choice D.

\boxed{2\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.

AThe value 3 is the multiplier in the supplied probability relation, not the Poisson rate or variance.
CThis choice does not satisfy the displayed Poisson-mass relation as a separate rate. That relation forces λ²=4, and the only admissible Poisson rate is λ=2, whose variance is also 2.
EThe value 4 is λ², the intermediate result before taking the positive square root. A Poisson variance equals λ, not λ².

Original practice · fully worked

Original variant: one-count probability from equal Poisson masses

A Poisson count N has equal probabilities of being zero and being two. Determine P(N=1).

A
0.24312
B
0.34382
C
0.36788
D
0.41421
E
0.50000

Variant answer in brief

Equality of the zero and two masses gives λ=√(2). Substitution into the one-count mass gives 0.34382, choice B.

Setup

Set the Poisson zero-count and two-count masses equal as stated.

e^{-\lambda}=e^{-\lambda}\frac{\lambda^2}{2}

Model

Cancel the common exponential factor and solve the resulting equation λ²/2=1 for the positive rate.

\lambda^2=2,\qquad \lambda=\sqrt2

Compute

At the solved rate √2, the Poisson one-count probability is 0.343819.

\Pr(N=1)=\lambda e^{-\lambda}=\sqrt2e^{-\sqrt2}=0.343819

Answer

Therefore Pr(N=1)=0.34382, which is choice B.

\boxed{0.34382\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #29 — 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: one-count probability from equal Poisson masses

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution