Worked solution · Exam P

SOA Exam P Sample Question #426 Solution

This Exam P sample reference tests Continuous Random Variables. Normalizing the linear density gives F(t)=t²⁄²⁵⁰⁰ on its support. The relevant conditional ratio is (F(25)-F(20))/(1-F(20))=3/28=0.1071, so choice B.

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

Normalize the density over its finite support.

1=\int_0^{50}kt\,dt=1250k

k=\frac1{1250}

Model

Write the conditional event as an interval inside the surviving population.

\Pr(T\le25\mid T>20)=\frac{\Pr(20<T\le25)}{\Pr(T>20)}

Compute

Integrate the normalized density over the numerator and denominator ranges.

\Pr(20<T\le25)=\int_{20}^{25}\frac{t}{1250}\,dt=0.09

\Pr(T>20)=\int_{20}^{50}\frac{t}{1250}\,dt=0.84

\frac{0.09}{0.84}=\frac3{28}=0.1071428571\ldots

Answer

The conditional probability rounds to 0.11.

\boxed{0.11\quad\text{(B)}}

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 0.09 is the unconditional interval probability Pr(20<T≤25); it omits division by the surviving probability 0.84.
CThe value 0.16 is F(20)=20²⁄⁵⁰², the probability below the conditioning threshold.
DThe value 0.17 is the uniform shortcut (25-20)/(50-20)=1/6. The actual density increases with t rather than remaining constant.
EThe value 0.84 is Pr(T>20), the denominator of the conditional probability rather than the requested ratio.

Original practice · fully worked

Original variant: conditional position of a print mark

The position X of a printed mark, measured in centimeters from the left edge of an eight-centimeter strip, has density f(x)=(8-x)/32 for 0<x<8. Given that the mark lies in the first four centimeters, calculate its expected position.

A
1.333
B
1.778
C
2.000
D
2.667
E
4.000

Answer and calculation

The first-four-centimeter event has probability 0.75, and its truncated first moment is 4/3. Their ratio is 16/9=1.7778 centimeters, so choice B.

Setup

Find the probability of the conditioning region.

\Pr(X<4)=\int_0^4\frac{8-x}{32}\,dx=\frac34

Model

Compute the first moment restricted to that region.

E[X\mathbf1_{\{X<4\}}]=\int_0^4x\frac{8-x}{32}\,dx=\frac43

Compute

Normalize the truncated first moment by the conditioning probability.

E[X\mid X<4]=\frac{4/3}{3/4}=\frac{16}{9}=1.777777\ldots

Answer

The conditional expected position is about 1.778 centimeters.

\boxed{1.778\quad\text{(B)}}

Keep practicing

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: conditional position of a print mark

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables