Q.6.18 - A First Course in Probability

Question

Two points are selected randomly on a line of length L so as to be on opposite sides of the midpoint of the line. [In other words, the two points X and Y are independent random variables such that X is uniformly distributed over (0, L/2) and Y is uniformly distributed over (L/2, L).] Find the probability that the distance between the two points is greater than L/3 

Step-by-Step Solution

Answer

The required probability is $\frac{7}{9}$

Step 1: Content Introduction

In an equation, a variable is a symbol that represents an unknown numerical value.

Step 2: Explanation

We are given that $X~Unif (0, L/2)$. We are required to find $P(Y-X>L/3)$. Firstly, we will construct joint pdf . Since we have chosen points arbitrary, we have that

$f(x,y)=f_X\left(x\right)f_Y\left(y\right)=\frac{1}{{(L/2)}^2}=\frac{4}{L^2}$

For $(x,y) \in (0,L/2) \times (L/2,L)$ hence the required probability is

$P(Y-X>\frac{L}{3})=1-P(Y-X\le \frac{L}{3})$

The region within $(0, L/2)\times (L/2, L)$ where is satisfied $y-x\le \frac{L}{3}$ is a right angle triangle with base $x\in (L/6, L/2) and y\in (L/2, 5L/6)$. Hence the area of that triangle is $\frac{1}{2}$.

So,

$P(Y-X\le \frac{L}{3})=\frac{4}{L^2}.\frac{1}{2}.{\left(\frac{L}{3}\right)}^2=\frac{2}{9}$

we get $P(Y-X\le \frac{L}{3})=\frac{2}{9}.$