Worked solution · Exam P

SOA Exam P Sample Question #492 Solution

This Exam P sample reference tests Exponential Distribution. For an exponential lifetime, S(x)=exp(−λ x). Since S(1)=0.80, S(x)=(0.80)ˣ and the CDF is F(x)=1-(0.80)ˣ, selecting choice C.

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

Write the exponential survival function in terms of its unknown rate.

S(x)=\Pr(X>x)=e^{-\lambda x}

S(1)=e^{-\lambda}=0.80

Model

Use the one-year survival probability as the exponential base for any nonnegative time.

S(x)=e^{-\lambda x}=\left(e^{-\lambda}\right)^x=(0.80)^x

Compute

Convert survival to cumulative probability by taking its complement.

F(x)=\Pr(X\le x)=1-S(x)

F(x)=1-(0.80)^x,\qquad x\ge0

Answer

The required cumulative distribution function is one minus 0.80 to the power x.

\boxed{F(x)=1-(0.80)^x\quad\text{(C)}}

Test the method on unseen problems

Probability Univariate Distributions Practice Workbook

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 expression (0.80)ˣ is the survival function S(x), not the cumulative distribution F(x). The CDF is its complement.
BThis density-like product is not a valid CDF: it does not approach 1 as x increases and uses 0.80 as though it were a rate.
DThis sets the exponential rate λ equal to 0.80. The supplied fact is exp(−λ)=0.80, so λ=-ln(0.80), not 0.80.
EThis both treats 0.80 as the rate and reports a decreasing survival expression instead of an increasing CDF.

Original practice · fully worked

Original variant: remaining decay time after inspection

A chemical cartridge has an exponential decay time T. By hour 6, 75% of cartridges have decayed. A cartridge is still active at hour 6. Calculate the probability that it remains active through hour 10.

A
0.0992
B
0.2500
C
0.3969
D
0.6300
E
0.7500

Answer and calculation

The six-hour survival probability is 0.25. Memorylessness reduces the conditional event to four additional hours, giving S(4)=(0.25)⁽⁴⁄⁶⁾=(0.25)⁽²⁄³⁾=0.3969 and choice C.

Setup

Translate the reported decay percentage into a six-hour survival probability.

S(6)=1-0.75=0.25

e^{-6\lambda}=0.25

Model

Conditioning on survival to hour six leaves an exponential remaining lifetime, so only the next four hours matter.

\Pr(T>10\mid T>6)=\Pr(T>4)=e^{-4\lambda}

Compute

Express the four-hour survival in terms of the known six-hour survival.

e^{-4\lambda}=\left(e^{-6\lambda}\right)^{4/6}

\Pr(T>10\mid T>6)=(0.25)^{2/3}=0.3968502630\ldots

Answer

The cartridge remains active through hour ten with conditional probability approximately 0.3969.

\boxed{0.3969\quad\text{(C)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

Work through the related free solutions for this topic, review the formulas you used, or keep every current Exam P solution in one searchable PDF.

More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Exponential Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: remaining decay time after inspection

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables