Worked solution · Exam P

SOA Exam P Sample Question #162 Solution

This Exam P sample reference tests Poisson Distribution. The zero-count probability gives a Poisson mean of -log(0.60). The expected count beyond the first event is λ minus the probability of at least one event, so the expected annual payment is about 554, choice A.

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 event count and identify its mean from the probability of zero.

N\sim\operatorname{Poisson}(\lambda)

e^{-\lambda}=0.60\quad\Longrightarrow\quad \lambda=-\log(0.60)=0.5108256

Model

Only events after the first produce payments, so the reimbursed count is the positive part of N-1.

R=(N-1)_+

(N-1)_+=N-\mathbf1_{\{N\ge1\}}

Compute

Take expectations and multiply the expected reimbursed count by the payment per event.

E[R]=\lambda-\Pr(N\ge1)=0.5108256-(1-0.60)=0.1108256

5000E[R]=554.1281

Answer

The expected annual payment rounds to 554.

\boxed{554\quad\text{(A)}}

Test the method on unseen problems

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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.

BThe value 872 implies 0.1744 reimbursed events. Direct summation of (n-1)P(N=n) gives only 0.11083, so this choice overstates the covered count.
CThe value 1022 is approximately 5000 × [λ-P(N=1)]. That retains all n events in years with at least two events instead of removing the first event in every nonempty year.
DThe value 1354 corresponds to 0.2708 covered events, more than twice the exact stop-loss mean. It results from applying the payment amount before correctly truncating the count.
EThe value 1612 corresponds to 0.3224 covered events and likewise cannot equal E[(N-1)+]. It treats too much of the ordinary Poisson mean as reimbursable.

Original practice · fully worked

Original variant: Poisson reimbursement beyond a two-alert retention

A monitoring center receives a Poisson number N of urgent alerts per day, with a positive daily mean. The probability of exactly one alert equals the probability of exactly two alerts. A service contract reimburses 300 dollars for each alert after the second alert of the day. Calculate the expected daily reimbursement.

A
$97.00
B
$162.40
C
$178.20
D
$300.00
E
$600.00

Answer and calculation

Equality of the one- and two-alert masses gives λ=2. The expected count beyond two is 4 exp(−2), so the expected reimbursement is $162.40, choice B.

Setup

Use the two equal Poisson masses to calibrate the daily mean.

e^{-\lambda}\lambda=e^{-\lambda}\frac{\lambda^2}{2}

\lambda>0\quad\Longrightarrow\quad \lambda=2

Model

Represent the count beyond two by subtracting the first two survival indicators.

(N-2)_+=N-\mathbf1_{\{N\ge1\}}-\mathbf1_{\{N\ge2\}}

Compute

Evaluate the expected excess count at λ=2 and apply the reimbursement rate.

E[(N-2)_+]=2-(1-e^{-2})-(1-3e^{-2})=4e^{-2}

300E[(N-2)_+]=1200e^{-2}=162.4023

Answer

The expected daily reimbursement is approximately $162.40.

\boxed{\$162.40\quad\text{(B)}}

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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: Poisson reimbursement beyond a two-alert retention

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables