Worked solution · Exam P

SOA Exam P Sample Question #159 Solution

This Exam P sample reference tests Poisson Distribution. Exceeding mean three means a count of at least four, while exceeding mean 1.5 means a count of at least two. The two Poisson tails differ by 0.08941, choice B.

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.

Read the solution Open official source

Independent solution

How to solve this Poisson Distribution question

Setup

Translate each strict inequality into the correct integer threshold.

A=\Pr(N_Q\ge4),\qquad N_Q\sim\operatorname{Poisson}(3)

B=\Pr(N_R\ge2),\qquad N_R\sim\operatorname{Poisson}(1.5)

Model

Write both tails as complements of their finite lower sums.

A=1-e^{-3}\left(1+3+\frac{3^2}{2}+\frac{3^3}{6}\right)

B=1-e^{-1.5}(1+1.5)

Compute

Evaluate the tails separately before subtracting.

A=0.3527681,\qquad B=0.4421746

B-A=0.4421746-0.3527681=0.0894065

Answer

The difference rounds to 0.09.

\boxed{0.09\quad\text{(B)}}

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 0.00 assumes that a Poisson count has the same exceed-mean probability for every mean. Discreteness makes the two thresholds behave differently.
CThe value 0.13 results after treating a count of three as exceeding the mean three and taking the absolute difference: |0.44217-0.57681|=0.13464.
DThe value 0.19 is inconsistent with the separately evaluated tails 0.44217 and 0.35277. Both strict integer thresholds must be set before subtraction.
EThe value 0.31 can arise by adding an extra Poisson point mass to the verified difference. The problem asks only for the difference of the two tails.

Original practice · fully worked

Original variant: locate the third inspection alert

Inspection alerts follow a homogeneous Poisson process averaging one alert every two minutes. Calculate the probability that the third alert arrives after minute four but no later than minute six.

A
0.42319
B
0.25349
C
0.57681
D
0.67668
E
0.74651

Answer and calculation

The third alert is later than time t exactly when the count by t is at most two. Subtracting the six-minute survival probability from the four-minute survival probability gives 5 exp(−2)-8.5 exp(−3)=0.25349, choice B.

Setup

Translate the two clock times into expected Poisson counts.

\lambda(4)=2,\qquad \lambda(6)=3

Model

The third-arrival time exceeds t if no more than two alerts have occurred by t.

\Pr(T_3>t)=\Pr(N(t)\le2)

Compute

Subtract the nested survival events at six and four minutes.

\Pr(T_3>4)=e^{-2}\left(1+2+\frac{2^2}{2}\right)=5e^{-2}

\Pr(T_3>6)=e^{-3}\left(1+3+\frac{3^2}{2}\right)=8.5e^{-3}

\Pr(4<T_3\le6)=5e^{-2}-8.5e^{-3}=0.2534863

Answer

The third alert falls in the specified two-minute interval with probability about 0.25349.

\boxed{0.25349\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: locate the third inspection alert

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables