Worked solution · Exam P

SOA Exam P Sample Question #255 Solution

This Exam P sample reference tests Discrete Random Variables. This is an expected-payment calculation using linearity of expectation. The two benefit components have expectations 32.50 and 70.00, for a total of 102.50 and choice B.

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 Discrete Random Variables question

Setup

Let F and R be the two annual treatment counts. First calculate their expected values from the supplied mass functions.

E[F]=0(0.60)+1(0.20)+2(0.15)+3(0.05)=0.65

E[R]=0(0.80)+1(0.20)=0.20

Model

The annual benefit is the sum of the fully covered lower-cost treatment and the partially covered higher-cost treatment.

C=50F+0.70(500)R=50F+350R

Compute

Apply linearity; no assumption about dependence between the two counts is needed.

E[C]=50E[F]+350E[R]

=50(0.65)+350(0.20)=32.50+70.00=102.50

Answer

The expected annual payment is 102.50.

\boxed{102.50\quad\text{(B)}}

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.

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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 90.00 replaces the expected filling count by the probability of at least one filling: 50(0.40)+350(0.20)=90. It loses the extra payments from two or three fillings.
CThe value 132.50 is 50(0.65)+500(0.20). It reimburses the second treatment at 100% instead of the stated 70%.
DThe value 250.00 is 3(50)+0.20(500). It treats the maximum of three fillings as certain and also ignores the partial-coverage percentage on the second treatment.
EThe value 400.00 is 50+350, the payment for exactly one of each treatment. It treats that particular outcome as certain instead of averaging over the count distributions.

Original practice · fully worked

Original variant: reimbursement by operating class

A cooperative classifies leased equipment as light-use with probability 0.75 and heavy-use with probability 0.25. The conditional expected numbers of replacement units are 1.2 and 3.6, respectively. A contract reimburses 40 credits per replacement unit and adds a 90-credit calibration allowance for heavy-use equipment. Calculate the expected reimbursement.

A
22.50 credits
B
72.00 credits
C
90.00 credits
D
94.50 credits
E
162.00 credits

Answer and calculation

Averaging the two conditional replacement counts gives 1.8 units and 72 credits of unit reimbursement. The heavy-use allowance contributes 22.5 more, for 94.5 credits and choice D.

Setup

Let K be the replacement-unit count and H indicate the heavy-use class. Average the two conditional count means.

E[K]=0.75(1.2)+0.25(3.6)=1.8

E[H]=\Pr(H=1)=0.25

Model

Write the contract payment as the unit reimbursement plus the class-specific allowance.

C=40K+90H

E[C]=40E[K]+90E[H]

Compute

Evaluate the two expected payment components.

40E[K]=40(1.8)=72.00

90E[H]=90(0.25)=22.50

E[C]=72.00+22.50=94.50

Answer

The expected reimbursement is 94.50 credits.

\boxed{94.50\text{ credits}\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Discrete Random Variables question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: reimbursement by operating class

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables