Worked solution · Exam P

SOA Exam P Sample Question #604 Solution

This Exam P sample reference tests Poisson Distribution. A zero count across six equal independent periods has probability exp(-6 λ)=0.008. The zero-count probability across the final two periods is therefore the cube root of 0.008, or 0.20; the observation in a disjoint period does not alter it, so choice A is correct.

Independent analysis by ActuaryProof · reviewed July 30, 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

Let λ be the common Poisson mean for one active month. A six-month total then has mean 6 λ.

N_6\sim\operatorname{Poisson}(6\lambda)

\Pr(N_6=0)=e^{-6\lambda}=0.008

Model

The two target months form a disjoint two-month count with mean 2 λ. Their count is independent of what happened in the earlier observed month.

N_2\sim\operatorname{Poisson}(2\lambda)

\Pr(N_2=0\mid\text{June observation})=\Pr(N_2=0)=e^{-2\lambda}

Compute

Relate the two-month survival factor to the supplied six-month factor.

e^{-2\lambda}=\left(e^{-6\lambda}\right)^{1/3}

e^{-2\lambda}=(0.008)^{1/3}=0.20

Answer

The conditional probability is 0.20.

\boxed{0.20\quad\text{(A)}}

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.

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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.

BThis is a numerical cube-root error. Since 0.20 cubed is exactly 0.008, no approximation above 0.20 is needed.
CThis is close to the probability that both target months are nonempty, obtained from the one-month nonzero probability. The requested event has both counts equal to zero.
DThis is close to the probability that a single active month is nonempty. It changes both the event direction and the two-month exposure length.
EMultiplying the two-month probability 0.20 by the three two-month blocks in six months treats intersecting zero-count requirements as a disjoint union. Independent block probabilities multiply.

Original practice · fully worked

Original variant: event locations within a shift

Signals arrive according to a homogeneous Poisson process during an eight-hour research shift. Exactly three signals were observed over the full shift. Conditional on that total, calculate the probability that at least two signals arrived during the first two hours.

A
0.0156
B
0.1406
C
0.1563
D
0.4219
E
0.5781

Answer and calculation

Given the three-event total, each event independently falls in the first quarter of the shift with probability 1/4. A Binomial(3,1/4) count is at least two with probability 10/64=0.15625, so choice C is correct.

Setup

Let K count signals in the first two hours, given that the eight-hour total equals three.

\frac{\text{early exposure}}{\text{total exposure}}=\frac28=\frac14

Model

Conditional on a homogeneous Poisson total, event locations are independent and uniform across the observation window.

K\mid(N_8=3)\sim\operatorname{Binomial}\!\left(3,\frac14\right)

Compute

Add the probabilities of exactly two and exactly three early signals.

\Pr(K\ge2\mid N_8=3)=\binom32\left(\frac14\right)^2\left(\frac34\right)+\left(\frac14\right)^3

\Pr(K\ge2\mid N_8=3)=\frac9{64}+\frac1{64}=\frac5{32}=0.15625

Answer

The conditional probability rounds to 0.1563.

\boxed{0.1563\quad\text{(C)}}

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: event locations within a shift

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables