Worked solution · Exam P

SOA Exam P Sample Question #205 Solution

This Exam P sample reference tests Exponential Distribution. The stated timing event places the first-arrival time between days 365 and 730. The exponential survival difference is exp(-365/200)-exp(-730/200)=0.135227, giving choice B.

Independent analysis by ActuaryProof · reviewed July 31, 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 Exponential Distribution question

Setup

Let T be the waiting time to the first event. Its rate is the reciprocal of the mean.

T\sim\operatorname{Exponential}(1/200)

S(t)=e^{-t/200}

Model

No event in the first year followed by an event in the next year is equivalent to the first event falling in the second year.

\{365<T\le730\}

Compute

Take the difference of the survival probabilities at the two endpoints.

\Pr(365<T\le730)=S(365)-S(730)

=e^{-365/200}-e^{-730/200}=0.1352265154

Answer

Rounding to three decimals gives 0.135.

\boxed{0.135\quad\text{(B)}}

Test the method on unseen problems

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30 original univariate random variables problems with complete solutions and wrong-choice analysis. Written for Exam P, not copied from the SOA set. One-time $19 PDF.

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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.

AThe value 0.026 is exp(−730/200)=0.025991, the probability that the first event occurs after both one-year periods.
CThe value 0.161 is exp(−365/200)=0.161218, the probability of surviving the first year, without requiring an event in the second.
DThe value 0.704 is (1-exp(−365/200))²=0.703555. It describes at least one event in each year under an additional Poisson-process assumption.
EThe value 0.839 is 1-exp(−365/200)=0.838782, the probability that an event occurs during the first year.

Original practice · fully worked

Original variant: delayed exponential race

Two independent processing clocks A and B have exponential ringing times with means 4 hours and 6 hours, respectively. Calculate the probability that clock A rings before clock B and that neither clock rings during the first hour.

A
0.2637
B
0.3408
C
0.3955
D
0.6000
E
0.6592

Answer and calculation

Both clocks survive one hour with probability exp(-(1/4+1/6))=exp(-5/12). After that, memorylessness gives A a 3/5 chance to ring first, so the answer is 0.6exp(-5/12)=0.395544, choice C.

Setup

Write the two rates and the density for A's ringing time.

\lambda_A=\frac14

\lambda_B=\frac16

Model

If A rings at time t>1 and wins the race, B must still be silent at t.

\Pr(A<B,\,A>1)=\int_1^\infty \lambda_Ae^{-\lambda_A t}e^{-\lambda_B t}\,dt

Compute

Evaluate the exponential integral.

\int_1^\infty \frac14 e^{-(5/12)t}\,dt=\frac{1/4}{5/12}e^{-5/12}

=0.6e^{-5/12}=0.3955443781

Answer

The delayed-win probability is 0.3955.

\boxed{0.3955\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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How to solve this Exponential Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: delayed exponential race

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables