SOA Exam P Sample Question #735. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #735 — Verified Solution

Independent solution

How to solve this Joint Distributions question

Answer in brief

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.

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)}}

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

Variant answer in brief

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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ActuaryProof. “SOA Exam P Sample Question #735 — Verified Solution.” Verified 2026-07-30.

Source reference: official SOA sample-question PDF. Last reviewed 2026-07-30.