Let $x$ and $y$ be uniformly distributed, independent random variables on $[0,1]$. What is the probability that the distance between $x$ and $y$ is less than $1/2$?
My continuous probability knowledge is very slim. I know that $P(x\leq \frac{1}{2})=\frac{1}{2}$, but I don't know how to work in the $y$. I would like to see an answer involving integrals, even if its not necessary, so I can try I pick up some of this continuous probability stuff.
It helps if you plot the support, X on the x-axis, Y on the y-axis. The support is $[0,1]\times[0,1]$.
$P(|X-Y|<1/2)$ is the area between the line $y=x-1/2$ and $y=x+1/2$ in the unit square. For independent uniform, you could use geometry. For other distributions, you'll have to integrate.