Worked solution · Exam P

SOA Exam P Sample Question #735 Solution

This Exam P sample reference tests Joint Distributions. Marginalizing the joint mass function over the theft coordinate gives fire-loss probabilities 17/60, 20/60, and 23/60, so E[X] = 2.1. The policyholder retains 44% of that loss, producing 0.924 and choice C.

Independent analysis by ActuaryProof · reviewed July 30, 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 Joint Distributions question

Setup

Only the fire coordinate enters the requested loss, so first remove the other coordinate by summation.

f_X(x)=\sum_{y=1}^{3}\frac{x+y^2}{60}

Model

Evaluate the marginal mass for each possible fire-loss value.

f_X(x)=\frac{3x+14}{60}

f_X(1)=\frac{17}{60},\quad f_X(2)=\frac{20}{60},\quad f_X(3)=\frac{23}{60}

Compute

Find the fire-loss expectation and multiply by the fraction not reimbursed.

\mathbb E[X]=\frac{1(17)+2(20)+3(23)}{60}=2.1

\mathbb E[\text{unreimbursed fire loss}]=(1-0.56)(2.1)=0.924

Answer

The expected retained amount is choice C.

\boxed{0.924\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.

AThis is consistent with weighting the marginal probabilities without the loss values, which understates the expected fire loss.
BThis results from an incomplete marginal total, typically omitting part of the highest-loss row or column.
DThis applies the 56% reimbursement rate as though it were the retained fraction.
EThis overweights the larger fire-loss values before applying the retained percentage.

Original practice · fully worked

Original variant: automated fault diagnosis

A repair center observes X electrical faults and Y cosmetic faults in a device, where each count can be 0, 1, or 2 and their joint probability mass is p(x,y) = (x + 2y + 1)/36. Each electrical fault normally requires 50 minutes of diagnosis, but an automated bench completes 60% of that work. Calculate the expected number of technician minutes spent diagnosing electrical faults.

A
10.000
B
20.000
C
23.333
D
35.000
E
58.333

Answer and calculation

Summing the joint mass over cosmetic faults gives f_X(x) = (x + 3)/12 and E[X] = 7/6. Technicians perform 40% of 50 minutes per electrical fault, so the expected manual time is 20(7/6) = 23.333 minutes.

Setup

Marginalize over the cosmetic-fault count because the labor calculation depends only on X.

f_X(x)=\sum_{y=0}^{2}\frac{x+2y+1}{36}

Model

Simplify the marginal and compute the expected electrical-fault count.

f_X(x)=\frac{3x+9}{36}=\frac{x+3}{12}

\mathbb E[X]=0\left(\frac{3}{12}\right)+1\left(\frac{4}{12}\right)+2\left(\frac{5}{12}\right)=\frac{7}{6}

Compute

The technician supplies the 40% of diagnosis time not handled by the bench.

\text{manual minutes per fault}=0.40(50)=20

\mathbb E[\text{manual minutes}]=20\left(\frac{7}{6}\right)=\frac{70}{3}

Answer

The expected technician time rounds to choice C.

\boxed{23.333\text{ minutes}\quad\text{(C)}}

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How to solve this Joint Distributions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: automated fault diagnosis

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Multivariate Random Variables