Verified 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 August 4, 2026

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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)}}

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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 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

Variant answer in brief

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)}}

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Cite this solution

ActuaryProof. “SOA Exam P Sample Question #159 — 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: locate the third inspection alert

Variant answer in brief

Setup

Model

Compute

Answer

Related Exam P solutions

Turn this solution into a complete study sequence

Cite this solution