Worked solution · Exam P

SOA Exam P Sample Question #237 Solution

This Exam P sample reference tests Continuous Uniform Distribution. A wait above three minutes occurs only for an arrival in the first two minutes of the five-minute closure, so each vehicle has probability 2/15. Independence gives 1-(13/15)²=56/225=0.2489, choice A.

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 Uniform Distribution question

Setup

Measure arrival time in minutes across the 15-minute observation window and identify the portion that creates more than three minutes of waiting.

p=\frac{2}{15}

Model

Let A and B denote the long-wait events for the two independent vehicles.

\Pr(A)=\Pr(B)=\frac{2}{15}

\Pr(A\cap B)=\left(\frac{2}{15}\right)^2

Compute

Use inclusion–exclusion, equivalently the complement that neither vehicle arrives in the critical interval.

\Pr(A\cup B)=\frac{2}{15}+\frac{2}{15}-\frac4{225}

\Pr(A\cup B)=\frac{56}{225}=0.2488889

Answer

The probability that at least one vehicle waits more than three minutes is about 0.25.

\boxed{0.25\quad\text{(A)}}

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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.

BAdding the two marginal probabilities gives 4/15=0.2667, nearest 0.27. This double-counts the case in which both vehicles arrive in the critical interval.
CUsing a three-minute critical interval gives individual probability 3/15=0.20 and union 1-0.80 squared=0.36. An arrival with exactly three minutes remaining does not exceed three minutes, and the interval length is two.
DThe value 0.40 adds two erroneous 0.20 marginal probabilities and also omits their overlap.
EUsing the full five-minute closure gives individual probability 5/15 and union 1-(2/3) squared=5/9=0.5556, nearest 0.56. Late arrivals during the closure wait three minutes or less.

Original practice · fully worked

Original variant: exactly two reports in a secure window

Four inspection reports are submitted independently at times chosen uniformly over an eight-hour shift. Direct upload is available during one fixed 1.5-hour secure window. Calculate the probability that exactly two reports arrive during that window.

A
0.0352
B
0.1393
C
0.1875
D
0.4358
E
0.5642

Answer and calculation

Each report lands in the secure window with probability 1.5/8=3/16. The number arriving there is binomial with size 4, so the probability of exactly two is C(4,2)(3/16)²(13/16)²=0.1393, choice B.

Setup

Convert the secure window's length into one report's arrival probability.

p=\frac{1.5}{8}=\frac3{16}

Model

Independence makes the number of reports in the window binomial.

K\sim\operatorname{Binomial}\left(4,\frac3{16}\right)

Compute

Choose the two in-window reports and require the other two to fall outside.

\Pr(K=2)=\binom42\left(\frac3{16}\right)^2\left(\frac{13}{16}\right)^2

\Pr(K=2)=0.1392517

Answer

Exactly two reports arrive in the secure window with probability about 0.1393.

\boxed{0.1393\quad\text{(B)}}

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How to solve this Continuous Uniform Distribution question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: exactly two reports in a secure window

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability