Probability that a number is greater than another number Suppose we pick a real number $x \in [0,5]$ and an arbitrary real number $y \in [0,2]$. What is the probability that $x$ is greater than $y$?
How do I tackle this problem? We know that $x$ is greater than $y$ if $x>2$. Hence I would have to compute $P(X>2)$. But that does not help very much.
EDIT: I should have mentioned the following possibilities were given. However, I was curious to the answers without giving these possibilities. 
The following possibilities were given:
(A) 40%
(B) 60%
(C) 70%
(D) 75%
(E) 80%
A hint would be appreciated.
Thanks in advance!
 A: Like everyone else assuming uniformity and independence.
Making a picture can help to see what is going on while keeping it light and informal.

The entire rectangle represents all possible points $(x,y)$.
The darker area (which is above the line $y=x$ ) are the points where $y>x$ and the lighter area (which is below the line $y=x$) are the points where $x>y$.
I hope this helps to get a simple picture of the problem.
A: In the absence of more information, one naturally assumes uniform distributions:
$$X\sim \operatorname{Unif}[0,5]$$
$$Y\sim\operatorname{Unif}[0,2]$$
which means
$$f_X(x) = \begin{cases}\tfrac15,&0\leq x\leq 5\\0,&\textrm{otherwise}\end{cases}$$
$$f_Y(y) = \begin{cases}\tfrac12,&0\leq y\leq 2\\0,&\textrm{otherwise}\end{cases}$$
To find $\Pr(X>Y)$, condition on $Y$ and integrate over all possible values of $Y$:
$$\Pr(X>Y)=\int_{y=0}^{2}\Pr(X>Y|Y=y)f_Y(y)\; dy$$
$$=\int_{y=0}^{2}\Pr(X>y)(\tfrac12)\; dy$$
$$=\int_{y=0}^{2}(\tfrac{5-y}{5})(\tfrac12)\; dy$$
$$=\tfrac1{10}\left.\left(5y-\tfrac{y^2}{2}\right)\right|_{y=0}^{y=2}$$
$$=0.8$$
So the correct answer is 80%.
