Worked solution · Exam P

SOA Exam P Sample Question #518 Solution

This Exam P sample reference tests Poisson Distribution. For a Poisson variable, the ratio of its second and first masses is one half of its mean. The recovered mean is 0.86, and its tail from three upward is 0.056433, so choice D is correct.

Independent analysis by ActuaryProof · reviewed July 31, 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 ratio of the two adjacent Poisson probabilities so that the common exponential factor cancels.

\frac{\Pr(N=2)}{\Pr(N=1)}=\frac{e^{-\lambda}\lambda^2/2}{e^{-\lambda}\lambda}=\frac{\lambda}{2}

Model

Equate the ratio to the supplied multiplier to recover the Poisson mean.

\frac{\lambda}{2}=0.43

\lambda=0.86

Compute

Use the complement of the probabilities for zero, one, and two events.

\Pr(N\ge3)=1-e^{-0.86}\left(1+0.86+\frac{0.86^2}{2}\right)

\Pr(N\ge3)=0.05643318885

Answer

The tail probability rounds to 0.056.

\boxed{0.056\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.

AThis treats 0.43 as the Poisson mean instead of recognizing that the adjacent-mass ratio is λ/2. With λ=0.43, the tail is 0.00963, which rounds to 0.010.
BThis computes P(N at least 4) rather than P(N at least 3). With λ=0.86, removing the N=3 mass leaves 0.01157, which rounds to 0.012.
CThis reports only P(N=3)=exp(−0.86)(0.86)³⁄⁶=0.04486, which rounds to 0.045 and omits counts above three.
EThis rounds the given ratio 0.43 to 0.5 before solving, producing λ=1; the resulting tail 1-exp(−1)(1+1+1/2)=0.0803 rounds to 0.080.

Original practice · fully worked

Original variant: parity of a photon count

The number N of photons recorded in a fixed exposure follows a Poisson distribution. The probability of recording exactly four photons is one third of the probability of recording exactly two photons. Calculate the probability that N is odd.

A
0.0183
B
0.1353
C
0.2707
D
0.4908
E
0.8647

Answer and calculation

The ratio P(N=4)/P(N=2)=λ²⁄¹² determines λ=2. The Poisson odd-count identity then gives (1-exp(−4))/2=0.490842, so choice D is correct.

Setup

Form the ratio of the two nonadjacent Poisson masses.

\frac{\Pr(N=4)}{\Pr(N=2)}=\frac{\lambda^4/4!}{\lambda^2/2!}=\frac{\lambda^2}{12}

Model

Use the stated ratio and the nonnegative Poisson parameter to identify the mean.

\frac{\lambda^2}{12}=\frac13

\lambda=2

Compute

Separate the exponential series into even and odd powers.

\Pr(N\text{ odd})=e^{-\lambda}\sinh(\lambda)=\frac{1-e^{-2\lambda}}{2}

\Pr(N\text{ odd})=\frac{1-e^{-4}}{2}=0.4908421806

Answer

The probability of an odd photon count rounds to 0.4908.

\boxed{0.4908\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Poisson Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: parity of a photon count

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables