$X,Y$ and $Z$ are uniformly distributed random variables on $(0,1)$ What is the probability that $X+Y>Z$?
I tried to do it geometrically and find the volume $x+y$ in the given limits i.e. $0$ to $1$ for both $x$ and $y$. But that gives $P(x+y>z)=1$ which means no such examples exist for which $x+y< z$. Where am I going wrong?
You tried to find $P(z<x+y)$ by $$ \int_0^1dx \int_0^1 dy \int_0^{x+y} 1\, dz $$ which does work out to 1. However, the range of $z <x+y$ is not $z\in(0,x+y)$ because $z$ cannot be more than 1. When $y>1-x$,$z$ can never exceed $x+y$, but you should integrate only up to $1$, not $x+y$. This explains the extra magnitude that brings that integral up to 1; the volume you shouldn't have added when $z>1$ amounts to \frac{1}{6}.
The correct integral to compute $P(z>x+y)$ is $$ \int_0^1dx \int_0^1 dy \int_{z=x+y}^1 1\, dz = \int_0^1dx \int_0^1\left[ 1-x-y] \right]dy = \int_0^1 \left[(1-x)-x(1-x)-\frac{(1-x)^2}{2}\right]dx =\frac{1}{6} $$