Worked solution · Exam P

SOA Exam P Sample Question #364 Solution

This Exam P sample reference tests Exponential Distribution. One part survives a year with probability about 0.8187. Both survive with probability 0.6703, while at least one survives with probability 0.9671. Conditioning the first event on the second gives approximately 0.693, choice D.

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

Use the exponential survival function for one year with mean lifetime five years.

q=\Pr(T>1)=e^{-1/5}=0.8187307531\ldots

Model

Independence gives the probability that both parts survive, and complements give the probability that the machine still functions.

\Pr(\text{both})=q^2

\Pr(\text{at least one})=1-(1-q)^2=2q-q^2

Compute

Form the requested conditional probability.

\Pr(\text{both}\mid\text{at least one})=\frac{q^2}{2q-q^2}

=\frac{q}{2-q}=0.6930941064\ldots

Answer

Given that the machine functions after one year, both parts survive with probability approximately 0.693.

\boxed{0.693\quad\text{(D)}}

Test the method on unseen problems

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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 0.003 is far smaller than the probability that both parts fail, which is about 0.033. It does not represent any required reliability event.
BThe value 0.409 is one half of the single-part survival probability. The functioning cases are not divided equally between one and two surviving parts.
CThe value 0.670 is the unconditional probability that both parts survive. It omits conditioning on the machine's functioning.
EThe value 0.819 is the survival probability of one specified part, not the probability that both survive given at least one.

Original practice · fully worked

Original variant: future status of two operating transmitters

A field station operates two independent transmitters whose lifetimes are exponential. The primary transmitter has mean lifetime 4 hours, and the secondary has mean lifetime 8 hours. Both are confirmed operating at hour 3. Calculate the probability that, during the next 2 hours, the primary fails while the secondary remains operating.

A
0.099
B
0.221
C
0.306
D
0.393
E
0.779

Answer and calculation

Memorylessness resets both lifetime clocks at the observation time. The primary fails within two more hours with probability about 0.3935, while the secondary survives that period with probability about 0.7788. Independence gives approximately 0.3064 and choice C.

Setup

By memorylessness, the residual lifetimes at hour three remain exponential with their original means.

X\sim\operatorname{Exp}(1/4),\qquad Y\sim\operatorname{Exp}(1/8)

Model

Calculate the primary failure probability and the secondary survival probability over the next two hours.

\Pr(X\le2)=1-e^{-2/4}=1-e^{-1/2}

\Pr(Y>2)=e^{-2/8}=e^{-1/4}

Compute

The residual lifetimes are independent, so multiply the two required probabilities.

\Pr(X\le2,Y>2)=(1-e^{-1/2})e^{-1/4}

=0.3064342303\ldots

Answer

The required future-status probability is approximately 0.306.

\boxed{0.306\quad\text{(C)}}

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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: future status of two operating transmitters

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables