Worked solution · Exam P

SOA Exam P Sample Question #386 Solution

This Exam P sample reference tests Poisson Distribution. This problem expresses an increasing sequence of charges as a triangular-number function of a Poisson count. Using the first two Poisson moments gives an expected total of 12 and choice D.

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

Write the total charge incurred when the count equals n.

T(n)=1+2+\cdots+n=\frac{n(n+1)}2

T=\frac{N(N+1)}2

Model

Use the raw second moment of a Poisson variable with mean four.

E[N]=4,\qquad \operatorname{Var}(N)=4

E[N^2]=\operatorname{Var}(N)+E[N]^2=4+16=20

Compute

Apply linearity of expectation to the quadratic charge.

E[T]=\frac12\left(E[N^2]+E[N]\right)

E[T]=\frac12(20+4)=12

Answer

The expected total charge is 12.

\boxed{12\quad\text{(D)}}

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.

AThis evaluates the unordered-pair count at the mean of four, giving six. It uses the wrong charge function and moves the expectation through a nonlinear calculation.
BThis takes half the second factorial moment, giving eight unordered pairs, but omits the additional linear term in the triangular charge.
CThis inserts the mean of four into the triangular formula and gets ten. A nonlinear function evaluated at the mean does not generally equal its expectation.
EThis counts the linear contribution twice, combining the second moment 20 with twice the mean to obtain 14.

Original practice · fully worked

Original variant: expected review cost for alert triples

A monitoring system records a Poisson number N of alerts per shift with mean 3. After each shift, every unordered group of three distinct alerts is reviewed, at a cost of 2 credits per group. Calculate the expected review cost.

A
4.5 credits
B
9 credits
C
18 credits
D
27 credits
E
54 credits

Answer and calculation

The reviewed groups are the unordered triples of alerts. The third Poisson factorial moment gives an expected 4.5 groups, so the two-credit review cost has mean 9 and choice B.

Setup

Express the number of unordered three-alert groups as a function of the shift count.

G=\binom N3=\frac{N(N-1)(N-2)}{3!}

C=2G

Model

Use the third factorial moment of a Poisson variable.

E[N(N-1)(N-2)]=\lambda^3

\lambda=3

Compute

Convert ordered factorial triples to unordered groups and apply the cost per group.

E[G]=\frac{3^3}{3!}=\frac{27}{6}=4.5

E[C]=2(4.5)=9

Answer

The expected review cost is 9 credits per shift.

\boxed{9\ \text{credits}\quad\text{(B)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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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: expected review cost for alert triples

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables