SOA Exam P Sample Question #660. Official problem wording is withheld; follow the source link to read it.

Verified solution · Exam P

SOA Exam P Sample Question #660

An exponential variable with mean 2.5 has rate 0.4. Its probability of exceeding 2.5 is exp[-(0.4)(2.5)]=exp(-1), so the correct response is choice B.

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 Exponential Distribution question

Answer in brief

An exponential variable with mean 2.5 has rate 0.4. Its probability of exceeding 2.5 is exp[-(0.4)(2.5)]=exp(-1), so the correct response is choice B.

Setup

Convert the exponential mean into the corresponding rate parameter.

\lambda=\frac{1}{2.50}=0.40

Model

The question asks for an upper-tail probability, so use the exponential survival function rather than its complement.

S(t)=\Pr(X\ge t)=e^{-\lambda t}

Compute

At a time equal to the distribution's mean, the rate-time product equals one.

\Pr(X\ge2.50)=e^{-(0.40)(2.50)}=e^{-1}

e^{-1}=0.3678794412\ldots

Answer

The required survival probability is the exponential constant e to the power negative one.

\boxed{e^{-1}\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 expression 1-exp(-0.4) both omits the time factor 2.5 from the exponent and takes the cumulative probability instead of the survival probability.
CThe expression 1-exp(-1) is the probability that the waiting time ends by 2.5, the complement of the requested at-least probability.
DThe expression exp(-0.4) has the correct survival form but uses the rate alone as the exponent, omitting multiplication by 2.5.
EThe expression 1-exp(-6.25) multiplies the time by the mean rather than by the reciprocal mean, and it also selects the lower tail.

Original practice · fully worked

Original variant: filtration cartridge life

The operating life of a filtration cartridge is exponentially distributed with mean 8 weeks. A cartridge has already operated for 5 weeks without failing. Calculate the conditional probability that it continues operating for at least 6 additional weeks.

A
0.252840
B
0.527633
C
0.472367
D
0.535261
E
0.747160

Variant answer in brief

Exponential memorylessness makes the remaining lifetime exponential with the original mean. The six-week residual survival probability is exp(-6/8)=0.472367, which selects choice C.

Setup

Let T be total cartridge life. Translate the wording into a conditional survival probability.

\Pr(T\ge11\mid T\ge5)

Model

The exponential distribution is memoryless, so only the six additional weeks enter the residual survival calculation.

\Pr(T\ge5+6\mid T\ge5)=\Pr(T\ge6)

\lambda=\frac{1}{8}

Compute

Evaluate the exponential survival function for the required additional duration.

\Pr(T\ge6)=e^{-(1/8)6}=e^{-3/4}

e^{-3/4}=0.4723665527\ldots

Answer

The conditional probability of at least six more operating weeks is approximately 0.472367.

\boxed{0.472367\quad\text{(C)}}

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ActuaryProof. “SOA Exam P Sample Question #660 — Verified Solution.” Verified 2026-07-30.

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