Worked solution · Exam P

SOA Exam P Sample Question #618 Solution

This Exam P sample reference tests Uniform Distribution. The stated moments identify a uniform interval from 7 to 10. Conditioning on survival past 7.5 restricts the possible interval to length 2.5, while continuing past 9 leaves length 1, so the conditional probability is 1/2.5=0.40 and choice B 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 Uniform Distribution question

Setup

Use the uniform mean and variance to recover the two endpoints.

\frac{a+b}{2}=8.5

\frac{(b-a)^2}{12}=0.75

Model

The interval width is positive, so the moment equations give a width of 3 and an endpoint sum of 17.

b-a=\sqrt{12(0.75)}=3

a=7,\qquad b=10

Compute

Given that the duration exceeds 7.5, the favorable portion above 9 has length 1 out of a remaining length of 2.5.

\Pr(X>9\mid X>7.5)=\frac{10-9}{10-7.5}

\Pr(X>9\mid X>7.5)=0.40

Answer

The requested conditional probability is 0.40.

\boxed{0.40\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.

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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 is the unconditional probability of exceeding 9, namely the one-unit upper segment divided by the full width 3.
CThis uses the symmetry probability above the mean, 0.50, even though the cutoff is 9 and the sample is conditioned at 7.5.
DThis is the complement of the requested conditional tail, representing an endpoint at or below 9 after the condition is imposed.
EThis is the probability of the conditioning event itself, P(X>7.5)=2.5/3, rather than the conditional probability.

Original practice · fully worked

Original variant: dual-courier arrival order

Two couriers arrive independently at a loading bay, with each arrival time uniform over the same ten-minute receiving window. Given that at least one courier arrives during the first four minutes, calculate the probability that both couriers arrive during the first six minutes.

A
0.320
B
0.360
C
0.500
D
0.563
E
0.640

Answer and calculation

The conditioning event occupies probability 0.64. Both arrivals lie before minute six and at least one lies before minute four with probability 0.36-0.04=0.32, so the conditional probability is 0.32/0.64=0.50 and choice C is correct.

Setup

Let U and V be the two independent arrival times, each uniform from zero to ten. Express the condition and target using their minimum and maximum.

C=\{\min(U,V)<4\}

B=\{\max(U,V)<6\}

Model

Use the complement in which both couriers arrive at or after minute four to find the probability of the conditioning event.

\Pr(C)=1-\Pr(U\ge4,V\ge4)

\Pr(C)=1-\left(\frac6{10}\right)^2=0.64

Compute

Within the square where both arrivals are before minute six, remove the smaller square where both fall between minutes four and six.

\Pr(B\cap C)=\left(\frac6{10}\right)^2-\left(\frac2{10}\right)^2=0.32

\Pr(B\mid C)=\frac{0.32}{0.64}=0.50

Answer

The conditional probability is one half.

\boxed{0.500\quad\text{(C)}}

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Next steps for Exam P Univariate Random Variables

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How to solve this Uniform Distribution question

Setup

Model

Compute

Answer

Probability Univariate Distributions Practice Workbook

Why the other letters tempt

Original variant: dual-courier arrival order

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P Univariate Random Variables