Let X and Y be independent uniform random variables on (0,1). Let $Z=\lfloor{\frac{1}{X+Y}}\rfloor$
I want to find $P(Z=0)$ and $E[Z]$
Firstly, I got that:
$$f_{X+Y}(a)=\begin{cases} a&0\leq a\leq 1\\2-a&1<a<2\\ 0&\text{otherwise} \end{cases} $$
Thus for $Z=0$, $X+Y>1$ so $P(Z=0) = \frac{1}{2}$.
But I have no idea how to find $E[Z]$.