SOA Exam P Sample Question #685. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #685 — Verified Solution

Independent solution

How to solve this Conditional Probability question

Answer in brief

The continuation event occurs unless both spouses die during the period, so its probability is 1-(0.15)(0.05)=0.9925. Dividing the husband's 0.85 survival probability by 0.9925 gives 0.856423, choice B.

Setup

Let H and W be the events that the husband and wife, respectively, are alive after ten years.

\Pr(H)=0.85,\qquad \Pr(W)=0.95

Model

Payments continue on H union W. Independence makes the probability that both have died the product of their death probabilities.

\Pr(H\cup W)=1-\Pr(H^c\cap W^c)=1-(0.15)(0.05)

Compute

Because H is contained in the continuation event, its joint probability with that event is simply Pr(H).

\Pr(H\cup W)=0.9925

\Pr(H\mid H\cup W)=\frac{\Pr(H)}{\Pr(H\cup W)}

\frac{0.85}{0.9925}=0.8564231738\ldots

Answer

Conditional on continued payments, the husband's survival probability is approximately 0.856.

\boxed{0.856\quad\text{(B)}}

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 0.850 is the husband's unconditional survival probability and does not use the information that payments continue.
CThe ratio 0.95/0.9925=0.9572 instead calculates the wife's conditional survival probability under the same continuation event.
DThe value 0.9925 is the probability that payments continue, so it is the conditioning denominator rather than the requested conditional probability.
EContinued payments establish only that at least one spouse is alive; treating this as proof that the husband is alive ignores the wife-only state.

Original practice · fully worked

Original variant: redundant coolant pumps

A research freezer is kept cold by two redundant coolant pumps. Over the next service interval, pump A has failure probability 0.20 and pump B has failure probability 0.10, independently. At inspection the freezer is still being cooled, which means at least one pump is operating. Calculate the conditional probability that pump A is operating.

A
0.800
B
0.816
C
0.900
D
0.918
E
0.980

Variant answer in brief

Cooling continues with probability 1-(0.20)(0.10)=0.98. Pump A operates with probability 0.80 and that event guarantees cooling, so the conditional probability is 0.80/0.98=0.816327, choice B.

Setup

Let A and B denote the events that the corresponding pumps operate through the service interval.

\Pr(A)=0.80,\qquad \Pr(B)=0.90

Model

Cooling is the union of the two operating events. Its complement requires both independent failures.

\Pr(A\cup B)=1-(0.20)(0.10)=0.98

Compute

Since an operating pump A necessarily implies cooling, normalize Pr(A) over the cooling event.

\Pr(A\mid A\cup B)=\frac{0.80}{0.98}=0.8163265306\ldots

Answer

The conditional probability that pump A operates is approximately 0.816.

\boxed{0.816\quad\text{(B)}}

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ActuaryProof. “SOA Exam P Sample Question #685 — Verified Solution.” Verified 2026-07-30.

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