Worked solution · Exam P

SOA Exam P Sample Question #119 Solution

This Exam P sample reference tests Conditional Probability from a Density. Conditioning on damage above 2 makes the normalizing constant cancel. The ratio of density integrals above 10 and above 2 is 0.2572, 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 from a Density question

Setup

Because the event above ten is contained in the event above two, the conditional probability is the ratio of the two corresponding density integrals.

G(x)=\frac{x^3}{3}-30x^2+800x

Model

An antiderivative of the unnormalized density is sufficient because the unknown normalizing constant cancels between numerator and denominator.

\Pr(X>10\mid X>2)=\frac{G(20)-G(10)}{G(20)-G(2)}

Compute

Evaluate the antiderivative over the intervals above ten and above two. Their ratio is approximately 0.257202.

\Pr(X>10\mid X>2)=0.257202

Answer

The conditional exceedance probability rounds to 0.26.

\boxed{0.26\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.

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Original practice · fully worked

Original variant: conditional tail of an increasing density

A measurement X has density f(x)=2x for 0<x<1. Given that the measurement exceeds 0.30, find the probability that it also exceeds 0.80.

A
0.2400
B
0.3600
C
0.3956
D
0.6400
E
0.9100

Answer and calculation

The survival probabilities at 0.80 and 0.30 are 0.36 and 0.91. Their ratio is 0.39560, choice C.

Setup

The event above 0.80 is contained in the conditioning event above 0.30, so the conditional probability is the ratio of the two survival probabilities.

F(x)=x^2,\quad 0<x<1

Model

The given density has cumulative distribution equal to the square of the measurement on its support.

\Pr(X>0.8\mid X>0.3)=\frac{1-F(0.8)}{1-F(0.3)}

Compute

The two survival probabilities are 0.36 and 0.91. Dividing gives approximately 0.395604.

\Pr(X>0.8\mid X>0.3)=\frac{0.36}{0.91}=0.395604

Answer

The conditional probability is about 0.3956.

\boxed{0.3956\quad\text{(C)}}

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How to solve this Conditional Probability from a Density question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Original variant: conditional tail of an increasing density

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables