Worked solution · Exam P

SOA Exam P Sample Question #651 Solution

This Exam P sample reference tests Marginal Distributions. Only 40% of the second-period loss remains unreimbursed, while the first-period loss is fully covered. Marginalizing the joint table gives a second-period mean of 0.90, so the expected unreimbursed amount is 0.36 and choice A.

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 Marginal Distributions question

Setup

Let Y be the loss in the partially covered period. Sum each row of the joint table to obtain the marginal distribution of Y.

\Pr(Y=0)=0.80,\quad \Pr(Y=2)=0.10

\Pr(Y=5)=0.06,\quad \Pr(Y=10)=0.04

Model

The fully covered period contributes nothing to the unreimbursed amount. The other period leaves 40% of its loss with the patient.

U=0.40Y

\operatorname{E}[U]=0.40\operatorname{E}[Y]

Compute

Calculate the marginal mean and apply the uncovered percentage.

\operatorname{E}[Y]=0(0.80)+2(0.10)+5(0.06)+10(0.04)=0.90

\operatorname{E}[U]=0.40(0.90)=0.36

Answer

The expected unreimbursed loss over the two periods is 0.36.

\boxed{\operatorname{E}[U]=0.36\quad\text{(A)}}

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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 0.76 is 40% of the first-period mean 1.90. That period is fully covered, so it should not create any unreimbursed loss.
CThe value 0.90 is the second-period expected loss before applying its 40% uncovered share.
DThis amount exceeds the entire second-period mean of 0.90, so it cannot be the expectation of only 40% of that nonnegative loss; it reflects a misread table margin or coverage rate.
EThe value 2.80 is the expected total loss from both periods before either reimbursement provision is applied.

Original practice · fully worked

Original variant: festival vendor credit

A theatre festival vendor may incur a booth charge B and an electrical surcharge E. A sponsor pays all of B and 70% of E. Joint probabilities for columns B=0, 4, 10 are: row E=0 has 0.40, 0.15, 0.05; row E=5 has 0.10, 0.08, 0.02; and row E=12 has 0.08, 0.07, 0.05. Calculate the vendor's expected amount not paid by the sponsor.

A
0.30
B
0.40
C
1.02
D
1.74
E
3.40

Answer and calculation

The surcharge marginal probabilities are 0.60, 0.20, and 0.20 at 0, 5, and 12. Its mean is 3.40, of which the vendor pays 30%, giving 1.02 and choice C.

Setup

Sum across booth-charge columns to obtain the marginal distribution of the electrical surcharge.

\Pr(E=0)=0.40+0.15+0.05=0.60

\Pr(E=5)=0.10+0.08+0.02=0.20

\Pr(E=12)=0.08+0.07+0.05=0.20

Model

The booth charge is fully sponsored. The vendor's unpaid amount is therefore 30% of the electrical surcharge.

V=0.30E

\operatorname{E}[V]=0.30\operatorname{E}[E]

Compute

Evaluate the surcharge mean and then its uncovered portion.

\operatorname{E}[E]=0(0.60)+5(0.20)+12(0.20)=3.40

\operatorname{E}[V]=0.30(3.40)=1.02

Answer

The vendor expects to pay 1.02 credits beyond the sponsor's contribution.

\boxed{\operatorname{E}[V]=1.02\quad\text{(C)}}

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How to solve this Marginal Distributions question

Setup

Model

Compute

Answer

Probability Multivariate & Risk Models Practice Workbook

Why the other letters tempt

Original variant: festival vendor credit

Answer and calculation

Setup

Model

Compute

Answer

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Next steps for Exam P Multivariate Random Variables