Worked solution · Exam P

SOA Exam P Sample Question #340 Solution

This Exam P sample reference tests Continuous Random Variables. Integrating the inverse-cube density gives survival probability equal to the reciprocal of the squared threshold. Conditional on reaching 3, the chance of remaining below 4 is therefore 7/16, or 0.4375. The listed answer is 0.44, choice E.

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 Continuous Random Variables question

Setup

Integrate the density above a threshold to obtain its survival function.

S(x)=\int_x^{\infty}\frac{2}{t^3}\,dt=\frac1{x^2},\qquad x>1

Model

Express the conditional interval probability through survival values at three and four.

\Pr(X<4\mid X\ge3)=\frac{S(3)-S(4)}{S(3)}

Compute

Substitute the two survival probabilities and simplify.

\frac{1/9-1/16}{1/9}=1-\frac9{16}=\frac7{16}=0.4375

Answer

The conditional probability rounds to 0.44.

\boxed{0.44\quad\text{(E)}}

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.

AA conditional answer of 0.04 would imply unconditional interval mass below 0.005 after multiplying by survival to 3. The exact interval mass is about 0.0486.
BThe value 0.05 is the unconditional probability of lying between 3 and 4, rounded from 0.0486. It omits conditional normalization.
CThe value 0.06 is the probability of reaching 4, rounded from 0.0625. It is a survival probability rather than the requested conditional interval.
DThe value 0.11 is the probability of reaching 3, which is the conditioning denominator rather than the requested ratio.

Original practice · fully worked

Original variant: remaining life of a heavy-tailed component

A component lifetime T, measured in years, has survival probability beyond t years equal to the reciprocal of t cubed for t at least one. The component is still operating at age 2. Calculate its expected additional lifetime.

A
0.333
B
0.667
C
1.000
D
2.000
E
3.000

Answer and calculation

Conditional survival for x additional years is the cube of 2 divided by 2 plus x. Integrating this residual survival function gives an expected additional lifetime of 1 year and choice C.

Setup

Let R be the remaining lifetime after the component reaches age two.

R=T-2\mid T>2

Model

Form the residual survival function by conditioning the original tail.

\Pr(R>x)=\frac{S(2+x)}{S(2)}=\left(\frac{2}{2+x}\right)^3

Compute

Integrate residual survival to obtain the expected additional lifetime.

\operatorname{E}[R]=\int_0^\infty\left(\frac{2}{2+x}\right)^3dx=1

Answer

The component's expected additional lifetime is one year.

\boxed{1.000\quad\text{(C)}}

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Next steps for Exam P Univariate Random Variables

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How to solve this Continuous Random Variables question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: remaining life of a heavy-tailed component

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables