Worked solution · Exam P

SOA Exam P Sample Question #267 Solution

This Exam P sample reference tests Exponential Distribution. Exponential memorylessness resets the lifetime clock at the known survival time. The remaining interval is 0.30, and a mean of 0.50 gives rate 2, so the conditional survival probability is exp(-0.60)=0.548812 and 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.

Read the solution Open official source

Independent solution

How to solve this Exponential Distribution question

Setup

Convert the supplied exponential mean to a rate.

\lambda=\frac{1}{0.50}=2

Model

Use memorylessness to replace the conditional lifetime with survival through only the additional interval.

\Pr(T>0.70\mid T>0.40)=\Pr(T>0.70-0.40)

=\Pr(T>0.30)

Compute

Evaluate the exponential survival function at the remaining duration.

\Pr(T>0.30)=e^{-2(0.30)}=e^{-0.60}

=0.5488116361\ldots

Answer

The conditional probability rounds to 0.549.

\boxed{0.549\quad\text{(D)}}

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.

See the cover and sample pages

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 0.203 is approximately the unconditional interval probability P(0.40<T<0.70)=exp(-0.8)-exp(-1.4)=0.2027. It omits division by P(T>0.40).
BThe value 0.247 is the unconditional upper tail exp(-2(0.70))=0.2466. It does not use the information that the device has already survived to 0.40.
CThe value 0.449 is exp(-0.80)=0.4493, the unconditional probability of surviving past the already-observed time 0.40. It reports the conditioning event itself instead of the requested conditional probability.
EThe value 0.861 is exp[-0.50(0.30)]=0.8607. It substitutes the mean 0.50 for the exponential rate, which is its reciprocal.

Original practice · fully worked

Original variant: conditional charge for a delayed sterilization cycle

A sterilization cycle has an exponential duration with mean 5 hours. At hour 4 the cycle is still running. A maintenance agreement charges 20 credits if the cycle finishes by hour 7 and 80 credits if it finishes after hour 7. Calculate the conditional expected charge.

A
9.024
B
34.796
C
43.905
D
47.071
E
52.929

Answer and calculation

Memorylessness makes the chance of running beyond hour 7 equal to exp(-3/5)=0.548812 after the observation at hour 4. Weighting the 20- and 80-credit charges gives 52.928698, so choice E.

Setup

Let R be the residual duration after the cycle is observed running at hour four.

R=T-4\mid T>4

Model

By memorylessness, the residual is exponential with the original five-hour mean.

\Pr(R>3)=e^{-3/5}=0.5488116361\ldots

\Pr(R\le3)=1-e^{-3/5}=0.4511883639\ldots

Compute

Weight the two possible agreement charges by the conditional completion cases.

E[C\mid T>4]=20(1-e^{-3/5})+80e^{-3/5}

=52.92869817\ldots

Answer

The conditional expected charge is approximately 52.929 credits.

\boxed{52.929\quad\text{(E)}}

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: conditional charge for a delayed sterilization cycle

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables