Worked solution · Exam P

SOA Exam P Sample Question #585 Solution

This Exam P sample reference tests Continuous Distributions. Intersecting the two interval events leaves 2<Y<3. Integrating the density gives probability 13/54 for that intersection and 13/27 for the conditioning interval 2<Y<4. Their ratio is 1/2, so choice D 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.

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Independent solution

How to solve this Continuous Distributions question

Setup

Replace the numerator by the intersection of the target interval and the conditioning interval.

(1<Y<3)\cap(2<Y<4)=\{2<Y<3\}

Model

Use an antiderivative of the density to evaluate both interval probabilities.

G(y)=\int\left(\frac{y}{6}-\frac{y^2}{36}\right)dy

G(y)=\frac{y^2}{12}-\frac{y^3}{108}

Compute

Calculate the intersection probability and the conditioning probability before taking their ratio.

\Pr(2<Y<3)=G(3)-G(2)=\frac{13}{54}

\Pr(2<Y<4)=G(4)-G(2)=\frac{13}{27}

\Pr(1<Y<3\mid2<Y<4)=\frac{13/54}{13/27}=\frac12

Answer

The conditional probability equals 0.500.

\boxed{0.500\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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Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis does not result from consistently intersecting the target interval with the condition and dividing by the full conditioning probability.
BThis is the unnormalized numerator probability P(2<Y<3)=13/54, so it omits division by P(2<Y<4).
CThis is the conditioning probability P(2<Y<4)=13/27 rather than the requested conditional ratio.
EThis uses P(1<Y<3) as the numerator without intersecting it with 2<Y<4; dividing that larger probability by the conditioning probability gives about 0.885.

Original practice · fully worked

Original variant: conditional load median

A normalized equipment-load index X has density f(x)=x/18 for 0<x<6 and zero otherwise. Given that the load index exceeds 2, determine the median of the resulting conditional distribution.

A
3.000
B
4.000
C
4.243
D
4.472
E
4.690

Answer and calculation

The CDF is x squared divided by 36. A conditional median m balances the probability mass on (2,m) and (m,6), giving m squared minus 4 equal to 36 minus m squared. Thus m=√(20)=4.472, choice D.

Setup

Integrate the density to obtain the distribution function on its support.

F(x)=\int_0^x\frac{t}{18}\,dt=\frac{x^2}{36},\qquad 0<x<6

Model

Let m be the conditional median above 2. Half of the conditional mass must lie between 2 and m.

\frac{F(m)-F(2)}{1-F(2)}=\frac12

Compute

Substitute the quadratic CDF and solve for the support value greater than 2.

\frac{m^2/36-4/36}{1-4/36}=\frac12

m^2-4=16

m=\sqrt{20}=4.472135955\ldots

Answer

The median after conditioning above 2 is approximately 4.472.

\boxed{4.472\quad\text{(D)}}

Keep practicing

Next steps for Exam P Univariate Random Variables

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More Exam P Univariate Random Variables practice problems Free Exam P formula sheet Exam P study manual PDF — $49

How to solve this Continuous Distributions question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: conditional load median

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables