Worked solution · Exam P

SOA Exam P Sample Question #209 Solution

This Exam P sample reference tests Conditional Probability. The exponential cdf is F(x)=1-exp(-2x). Because the numerator event is contained in the conditioning event, the conditional probability is F(0.5)/F(1)=(1-exp(−1))/(1-exp(−2))=0.731059, 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.

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Independent solution

How to solve this Conditional Probability question

Setup

Integrate the density to obtain its cumulative distribution function.

F(x)=\int_0^x2e^{-2t}\,dt=1-e^{-2x}

Model

The smaller-threshold event is a subset of the conditioning event, so their intersection is the smaller event.

\Pr(X\le0.5\mid X\le1)=\frac{F(0.5)}{F(1)}

Compute

Evaluate the cdf at both thresholds and form the ratio.

F(0.5)=1-e^{-1}

F(1)=1-e^{-2}

\frac{1-e^{-1}}{1-e^{-2}}=0.7310585786

Answer

The conditional probability rounds to 0.731.

\boxed{0.731\quad\text{(D)}}

Test the method on unseen problems

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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 value 0.433 is below the unconditional numerator F(0.5)=0.632. Conditioning on X≤1 removes only larger outcomes, so such a decrease fails a basic direction check.
BThe value 0.547 is F(0.5)F(1)=0.546572 after rounding. Conditional probability requires dividing by the conditioning probability, not multiplying.
CThe value 0.632 is F(0.5)=1-exp(−1), the unconditioned numerator with the normalization by F(1) omitted.
EThe value 0.865 is F(1)=1-exp(−2), the probability of the conditioning event itself.

Original practice · fully worked

Original variant: conditional mean from weighted integer scores

A repair priority score K can be any integer from 1 through 6. Its probability mass is P(K=k)=k/21 for k=1,...,6. Given that a logged score is even, calculate the conditional expected score.

A
0.5714
B
2.6667
C
4.0000
D
4.6667
E
5.3333

Answer and calculation

The even scores have total probability (2+4+6)/21=4/7 and an unnormalized first moment (2²+4²+6²)/21=8/3. Dividing gives E[K | K even]=14/3=4.6667, choice D.

Setup

Collect the probability carried by the three even scores.

\Pr(K\text{ even})=\frac{2+4+6}{21}=\frac47

Model

Form the first-moment numerator using only outcomes in the conditioning event.

E[K\mathbf{1}_{\{K\text{ even}\}}]=\sum_{k\in\{2,4,6\}}k\frac{k}{21}

Compute

Evaluate the restricted first moment and normalize it by the even-score probability.

E[K\mathbf{1}_{\{K\text{ even}\}}]=\frac{4+16+36}{21}=\frac83

E[K\mid K\text{ even}]=\frac{8/3}{4/7}=\frac{14}{3}=4.6667

Answer

The conditional expected priority score is 4.6667.

\boxed{4.6667\quad\text{(D)}}

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Next steps for Exam P Univariate Random Variables

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How to solve this Conditional Probability question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: conditional mean from weighted integer scores

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables