Worked solution · Exam P

SOA Exam P Sample Question #576 Solution

This Exam P sample reference tests Uniform Distribution. Conditioning on survival past a restricts the uniform lifetime to (a,40). The portion that ends before 30 therefore has conditional probability (30-a)/(40-a). Setting this equal to 0.60 gives a=15, so choice C is correct.

Independent analysis by ActuaryProof · reviewed July 30, 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 Uniform Distribution question

Setup

Intersect the failure event with the information that the lifetime has already exceeded a.

\Pr(T<30\mid T>a)=\frac{\Pr(a<T<30)}{\Pr(T>a)}

Model

For a uniform lifetime, these probabilities are proportional to interval lengths.

\Pr(T<30\mid T>a)=\frac{30-a}{40-a}

Compute

Set the conditional interval ratio to the supplied probability and solve.

\frac{30-a}{40-a}=0.60

30-a=24-0.60a

a=15

Answer

The conditioning age is 15 months.

\boxed{15\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.

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.

AUsing the full 40-month support in the numerator equation, (30-a)/40=0.60, ignores conditioning and gives a=6.
BUsing 30 rather than 40 as the upper endpoint of the conditioned support gives (30-a)/30=0.60 and a=12.
DThis sets the unconditional probability P(T<a) equal to 0.40, giving a=0.40(40)=16, instead of forming the conditional event.
EThis simply takes 60% of 30 to get 18 and does not account for the remaining support after survival past a.

Original practice · fully worked

Original variant: remaining battery-life percentile

A battery lifetime T is uniformly distributed from 10 to 50 hours. The battery is known to have lasted beyond 26 hours. Let R=T-26 be its remaining lifetime. Calculate the 75th percentile of R under this condition.

A
6 hours
B
12 hours
C
18 hours
D
24 hours
E
40 hours

Answer and calculation

Conditioning on T>26 makes T uniform on (26,50), so the remaining life R is uniform on (0,24). Its 75th percentile is 0.75(24)=18 hours, making choice C correct.

Setup

Restrict the original uniform support using the survival information.

T\mid(T>26)\sim\operatorname{Uniform}(26,50)

Model

Subtracting the elapsed 26 hours shifts the conditioned support to a remaining-life interval starting at zero.

R\mid(T>26)\sim\operatorname{Uniform}(0,24)

Compute

Take 75% of the 24-hour remaining-life support.

q_{0.75}=0+0.75(24)=18

Answer

The conditional 75th percentile of remaining life is 18 hours.

\boxed{18\text{ hours}\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 Uniform Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: remaining battery-life percentile

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables