SOA Exam P Sample Question #716. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #716 — Verified Solution

Independent solution

How to solve this Conditional Distributions question

Answer in brief

First average the conditional low-cost probabilities over the market states to obtain 0.32. The complementary high-cost probability is 0.68, so the expected cost is 5(0.32)+10(0.68)=8.4 million, choice D.

Setup

Use the market-state probabilities to remove the conditioning from the two-point cost distribution.

\Pr(C=5)=\sum_s \Pr(C=5\mid S=s)\Pr(S=s)

Model

Weight each conditional low-cost probability by its corresponding market-state probability.

\Pr(C=5)=0.05(0.10)+0.10(0.25)+0.25(0.30)+0.50(0.25)+0.90(0.10)

Compute

Complete the marginal cost distribution and take its expectation, in millions.

\Pr(C=5)=0.32,\qquad \Pr(C=10)=0.68

\mathbb{E}[C]=5(0.32)+10(0.68)=8.4

Answer

The expected recall cost is 8.4 million.

\boxed{\mathbb{E}[C]=8.4\text{ million}\quad\text{(D)}}

Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AA value of 6.4 million would require a high-cost probability of only 0.28, contradicting the weighted total of 0.68.
BThis is 10(0.68), so it counts the high-cost contribution but omits the low-cost contribution 5(0.32).
CThis is the unweighted midpoint of the two possible costs and ignores all conditional and market-state probabilities.
EThis would require a high-cost probability of 0.84, overstating the high-cost branch relative to the weighted table.

Original practice · fully worked

Original variant: field-station repair cost

A field station faces either a 2-thousand-dollar routine repair or an 8-thousand-dollar major repair. Weather is clear, rainy, or snowy with probabilities 0.50, 0.30, and 0.20. Conditional on those states, the probabilities of a major repair are 0.10, 0.40, and 0.80, respectively. Calculate the expected repair cost.

A
2.64 thousand dollars
B
3.02 thousand dollars
C
3.98 thousand dollars
D
4.60 thousand dollars
E
9.80 thousand dollars

Variant answer in brief

The total probability of a major repair is 0.50(0.10)+0.30(0.40)+0.20(0.80)=0.33. The expected cost is therefore 2(0.67)+8(0.33)=3.98 thousand dollars, choice C.

Setup

Average the major-repair probabilities over the weather distribution.

\Pr(M)=\sum_w\Pr(M\mid W=w)\Pr(W=w)

Model

Insert the three weather-state probabilities.

\Pr(M)=0.50(0.10)+0.30(0.40)+0.20(0.80)

Compute

Use the resulting two-point distribution for repair cost, measured in thousands.

\Pr(M)=0.33,\qquad \Pr(M^c)=0.67

\mathbb{E}[C]=8(0.33)+2(0.67)=3.98

Answer

The expected repair cost is 3.98 thousand dollars.

\boxed{3.98\text{ thousand dollars}\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #716 — Verified Solution.” Verified 2026-07-30.

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