Q:

$P(|X-Y|>\frac{1}{3}/X\geq1/2)=?$

Let X and Y are two independent uniformly distributed random variable on [0,1].The value of $P(|X-Y|>\frac{1}{3}/X\geq1/2)=?$

My try
$P(|X-Y|>\frac{1}{3}/X\geq1/2)=\frac{P(|X-Y|>\frac{1}{3},X\geq1/2)}{P(X\geq1/2)}$
Here I solved the denominator part that is $P(X\geq1/2)$,But the numerator part $P(|X-Y|>\frac{1}{3},X\geq1/2)$ I am not getting any idea how to start?

A:

Hint: See the following image. The area of the shaded parts is your answer:

$$P(|X-Y|>\frac{1}{3},X\geq1/2) = \left(\frac{1}{2}\times (\frac{2}{3} - \frac{1}{2})\times (1-\frac{1}{2}-\frac{1}{3})\right) + \left(\frac{1}{2}\times((\frac{1}{2}-\frac{1}{3})+(1-\frac{1}{3}))\times(1-\frac{1}{2})\right)=\frac{2}{9}$$

