SOA Exam P Sample Question #711. Official problem wording is withheld; follow the source link to read it.

SOA Exam P Sample Question #711 — Verified Solution

Independent solution

How to solve this Exponential Distribution question

Answer in brief

After conditioning on survival through year 2, exponential memorylessness restarts the waiting clock. The original years 5 and 8 become residual times 3 and 6, so the required probability is exp(-3/4) - exp(-6/4) = 0.249236, which rounds to choice C.

Setup

Let T be the waiting time and convert the stated mean to the exponential rate.

T\sim\operatorname{Exponential}(\lambda),\qquad \lambda=\frac{1}{4}

Model

Conditioning on no event before year 2 leaves an independent exponential residual wait with the same rate.

(T-2\mid T>2)\sim\operatorname{Exponential}\!\left(\frac14\right)

\Pr(5<T<8\mid T>2)=\Pr(3<T-2<6\mid T>2)

Compute

Subtract exponential survival probabilities at the two residual-time boundaries.

\Pr(5<T<8\mid T>2)=e^{-3/4}-e^{-6/4}

=0.4723665527-0.2231301601=0.2492363926

Answer

The conditional probability rounds to 0.25.

\boxed{0.25\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.

AUsing rate 1/2 instead of 1/4 gives exp(-5/2) - exp(-4) = 0.06377, near A. The two claim-free years are conditioning information, not a replacement for the stated mean.
BThe unconditional probability exp(-5/4) - exp(-8/4) = 0.15117 rounds to 0.15; it ignores the information that the waiting time has already exceeded two years.
DRounding exp(-3/4) to 0.5 and exp(-6/4) to 0.2 before subtracting gives 0.30. Such coarse intermediate rounding materially changes the result.
EThe value 0.75 is approximately the complement of the correct interval probability. The question asks for occurrence inside the interval, not outside it.

Original practice · fully worked

Original variant: remote pump repair

A remote irrigation pump requires an exponentially distributed amount of technician time with mean 10 hours. After 4 hours of work, the repair is still incomplete. Given that information, calculate the probability that the repair finishes between 7 and 12 hours after work began.

A
0.1954
B
0.2592
C
0.2915
D
0.4493
E
0.7085

Variant answer in brief

Memorylessness turns the requested original-time window into a residual-time window from 3 to 8 hours. Its probability is exp(-0.3) - exp(-0.8) = 0.291489, so choice C is correct.

Setup

Let R denote the total repair duration and record its rate and the elapsed-work condition.

R\sim\operatorname{Exponential}(0.1),\qquad R>4

Model

The residual repair time after hour 4 remains exponential with rate 0.1.

(R-4\mid R>4)\sim\operatorname{Exponential}(0.1)

7<R<12\quad\Longleftrightarrow\quad 3<R-4<8

Compute

Evaluate the probability between the two residual-time cutoffs.

\Pr(7<R<12\mid R>4)=e^{-0.1(3)}-e^{-0.1(8)}

=0.7408182207-0.4493289641=0.2914892566

Answer

The conditional completion probability is approximately 0.2915.

\boxed{0.2915\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #711 — Verified Solution.” Verified 2026-07-30.

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