Worked 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 October 6, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

Read the solution Open official source

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)}}

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.

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

Answer and calculation

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)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: one-count probability from equal Poisson masses

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables