Worked solution · Exam P

SOA Exam P Sample Question #217 Solution

This Exam P sample reference tests Poisson Distribution. If N is the Poisson death count, the company's retained benefit is 100 min(N,4). Summing the first four Poisson tail probabilities gives an expected retained count of 1.924859, hence cost 192.49 and choice C.

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

Let N be the annual count and express the company's cost after the insurance layer.

N\sim\operatorname{Poisson}(2)

C=100\min(N,4)

Model

For a nonnegative integer count, the expected capped count is the sum of its first four survival probabilities.

\operatorname{E}[\min(N,4)]=\sum_{j=1}^{4}\Pr(N\ge j)

Compute

Evaluate the Poisson tails through the attachment count.

\operatorname{E}[C]=100\left[\Pr(N\ge1)+\Pr(N\ge2)+\Pr(N\ge3)+\Pr(N\ge4)\right]

\operatorname{E}[C]=100\left[4-\frac{46}{3}e^{-2}\right]

\operatorname{E}[C]=400-\frac{4600}{3}e^{-2}=192.4858990

Answer

The expected retained annual cost rounds to 192.

\boxed{192\quad\text{(C)}}

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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.

ASumming payments only for N=1 through N=4 gives 171.4247. That omits all outcomes with five or more deaths, even though the company still pays the capped amount 400 in those years.
BRounding the four Poisson tail probabilities before adding can produce a retained count near 1.9 and a listed value near 189. Keeping the exact tails gives 192.4859.
DThe value 200 is 100 E[N], the expected gross benefit before insurance. It does not subtract the insurer's excess-layer payment.
EThe insurer's expected excess payment is 200-192.4859=7.5141. Adding that amount to the gross mean instead of subtracting it gives 207.5141, nearest 208.

Original practice · fully worked

Original variant: probability an annual warranty cap binds

The annual number N of covered device failures is Poisson with mean 2. A warranty pays 75 credits per failure, subject to a maximum annual payment of 225 credits. Given that the warranty makes a positive payment during a year, calculate the probability that it pays the annual maximum.

A
0.1353
B
0.3233
C
0.3739
D
0.6870
E
0.8647

Answer and calculation

The warranty reaches 225 credits exactly when N is at least 3, while a positive payment means N is at least 1. The conditional probability is (1-5 exp(−2))/(1-exp(−2))=0.3739, choice C.

Setup

Translate the payment conditions into count events.

N\sim\operatorname{Poisson}(2)

Y=\min(75N,225)

\{Y>0\}=\{N\ge1\},\qquad \{Y=225\}=\{N\ge3\}

Model

Write the requested probability as one Poisson tail divided by another.

\Pr(Y=225\mid Y>0)=\frac{\Pr(N\ge3)}{\Pr(N\ge1)}

Compute

Evaluate the two count tails at mean 2.

\Pr(N\ge3)=1-e^{-2}\left(1+2+\frac{2^2}{2}\right)=1-5e^{-2}

\Pr(N\ge1)=1-e^{-2}

\Pr(Y=225\mid Y>0)=\frac{1-5e^{-2}}{1-e^{-2}}=0.3739294

Answer

Conditional on a positive payment, the cap binds with probability about 0.3739.

\boxed{0.3739\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: probability an annual warranty cap binds

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables