Worked solution · Exam P

SOA Exam P Sample Question #534 Solution

This Exam P sample reference tests Poisson Distribution. After the first two accident-free years are conditioned away, the first accident must be avoided in three further years and then occur in the next year. The resulting probability is 0.070498, so choice C 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

A year with a Poisson mean of 0.10 has a zero-count probability obtained directly from the Poisson mass function.

q=\Pr(N=0)=e^{-0.10}

Model

The stated conditioning fixes the first two years. For the first accident to arrive in year six, years three through five must have no accidents and year six must have at least one.

\Pr(\text{target}\mid\text{first two years empty})=q^3(1-q)

Compute

Independence across years permits multiplication of the four yearly factors.

q^3(1-q)=e^{-0.30}(1-e^{-0.10})

e^{-0.30}(1-e^{-0.10})=0.07049817465

Answer

The nearest listed probability is 0.0705.

\boxed{0.0705\quad\text{(C)}}

Test the method on unseen problems

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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.

AThis cubes the yearly positive-count probability, (1-exp(−0.1))³=0.000862, as though accidents rather than empty years were required in the intervening period.
BThis ignores the conditioning and requires all five years before year six to be empty: exp(−0.5)(1-exp(−0.1))=0.05772.
DThis treats the Poisson mean 0.10 as a Bernoulli accident probability, giving 0.9³(0.1)=0.0729.
EThis accumulates the probability of at least one accident over four years, 1-exp(−0.4)=0.3297, instead of requiring the first nonempty year to be the last one.

Original practice · fully worked

Original variant: first week with service requests

Requests arrive at a remote support desk according to an independent Poisson count with mean 0.25 per week. No requests arrived during the first four weeks of a monitoring period. Calculate the probability that week eight is the first later week containing at least one request.

A
0.0489
B
0.0814
C
0.1045
D
0.1723
E
0.5276

Answer and calculation

Weeks five through seven must be empty and week eight nonempty. Independence gives exp(−0.75)(1-exp(−0.25))=0.104487, so choice C is correct.

Setup

Let q be the probability of an empty week.

q=e^{-0.25}

Model

The earlier observation is already conditioned upon. Three additional empty weeks must precede one nonempty week.

\Pr(\text{first later request in week 8})=q^3(1-q)

Compute

Substitute the weekly zero-count probability.

e^{-0.75}(1-e^{-0.25})=0.1044871116

Answer

The requested probability rounds to 0.1045.

\boxed{0.1045\quad\text{(C)}}

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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: first week with service requests

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables