I have the following problem:
Let $X,Y$ be two random variables which are indipendent and uniformly distributed on $[0,1]$. I need to compute $\Bbb{P}\left(Y\leq \frac{X}{2}\right)$
Since they are uniformly distributed I know that their density maps are $\Bbb{1}_{[0,1]}$. Then I thought that $$\Bbb{P}\left(Y\leq \frac{X}{2}\right)=\Bbb{P}(2Y-X\leq 0)=F_{2Y-X}(0)$$. But I'm not sure if this works.
Could maybe someone help me?
I think your approach is correct, just compute the area of $\{(x,y)\in\mathbb{R}^2:2y-x\leq0\}\cap[0,1]^2$.
If you want to use integration, you would have to compute $\int_0^1\int_0^{\frac{x}{2}}dydx$.