$X,Y,Z$ are i.i.d $U(0,1)$. What is $P(X+Y+Z > 1)$? 
Assume that $X,Y$ and $Z \sim U(0,1)$. Assume further $X,Y$ and $Z$ are mutually independent.
  What is $P(X+Y+Z>1)$?


I've written down the density function of $X,Y$ and $Z$. I've drawn a picture. I understand the range of a new variable $U=X+Y+Z$ should be between $0$ and $3$. But I'm stuck on how to find the distribution of $U$.

The PDF of $U(0,1)$ is $f(x) = \chi_{[0,1]}(x)$. In other words, $P(X\le a) = a$ if $a\in[0,1]$ and $0$ else.
Now
$$P(X+Y+Z > 1) = 1 - P(X+Y+Z \le 1)$$
What do you know about the joint PDF of iid random variables?

We know $f_{X,Y,Z}(x,y,z) = f_X(x) f_Y(y) f_Z(z)$ so in this case we have
$$P(U \le a) = \int_0^a \int_0^{a-x} \int_0^{a-x-y} 1\cdot 1\cdot 1 \ \mathrm dz \ \mathrm dy \ \mathrm dx$$
Can you take it from here?
