Independent solution

How to solve this Continuous Uniform Distribution question

Setup

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=215p=\frac{2}{15}

Model

Model

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

Pr(A)=Pr(B)=215\Pr(A)=\Pr(B)=\frac{2}{15}
Pr(AB)=(215)2\Pr(A\cap B)=\left(\frac{2}{15}\right)^2

Compute

Compute

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

Pr(AB)=215+2154225\Pr(A\cup B)=\frac{2}{15}+\frac{2}{15}-\frac4{225}
Pr(AB)=56225=0.2488889\Pr(A\cup B)=\frac{56}{225}=0.2488889

Answer

Answer

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

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