Find $ P(Z>X+Y)$ where $X,Y,Z \sim U(0,1)$ independently I'm trying to follow a line in a derivation for $P(Z&gt;X+Y)$ where $X,Y,Z$ are independent continuous random variables distributed uniformly on $(0,1)$.

I've already derived the pdf of $X+Y$ using the convolution theorem, but there's a line in the answer that says:

$P(Z&gt;X+Y) = \mathbb{E}[\ P(Z&gt;X+Y\ |\  X+Y )\  ]$ where $\mathbb{E}$ is the expectation.

I'm not familiar with this result. Could anyone give a pointer to a similar result if one exists?

Thanks.
This is not an answer to your question about the justification for the 
equation that is puzzling you, but I think the geometrical method
described below for solving the problem that may give you a different 
insight into the calculation of the desired probability $P\{Z &gt; X+Y\}$.

The random point $(X,Y,Z)$ is uniformly distributed in the interior of the unit cube with diagonally opposite vertices $(0,0,0)$ and $(1,1,1)$.  The cube has unit volume and so the probability that $(X,Y,Z)$ is in some region is just the volume of that region.  Thus, $P\{Z &gt; X+Y\}$ is the volume of the tetrahedron with vertices $(0,0,0)$, $(1,0,1)$, $(0,1,1)$ and $(0,0,1)$.  If we think of this
as an inverted pyramid whose base is the right triangle with vertices
$(1,0,1)$, $(0,1,1)$ and $(0,0,1)$ and apex $(0,0,0)$ is at 
altitude $1$ "above" the base,
then since the base has area $\frac{1}{2}$, we get
the volume as
$$P\{Z &gt; X+Y\} = \frac{1}{3}\times (\text{area of base})\times(\text{altitude}) 
= \frac{1}{3}\times \frac{{3}}{2}\times1 = \frac{1}{6}.$$
Of course, if you have already computed the density of $X+Y$, then it is
straightforward to use the result given by Artiom Fiodorov to get
$$P\{Z &gt; X+Y\}= \int_0^2{P}(Z&gt;v)f_{X+Y}(v)\;dv
= \int_0^1(1-v)\cdot v\;dv = 
\left.\frac{v^2}{2}-\frac{v^3}{3}\right|_0^1 = \frac{1}{6}.$$
I don't know if this helps since Dilip has given the answer, but the distribution of X+Y is triangular on [0,2] (isosceles with peak at X+Y = 1). So P(Z>X+Y) is the probability that a uniform on [0,1] is larger than  the triangular random variable on [0,2]. If X+Y>1 then Z cannot be >X+Y and the probability that X+Y is greater than 1 is 1/2. Now this is where taking the expectation fo the conditional probability helps in my proof.
P{Z>X+Y) =E[P(Z>X+Y|X+Y)]= ∫u P(Z>u|X+Y=u)du =∫u P(Z>u)du where u is integrated from 0 to 1.  The condition X+Y=u gets dropped because Z is independent of X+Y. P(Z>u)=1-u for 0&lt;=u&lt;=1.
hence P(Z>X+Y) =∫u(1-u)du  = 1/6. Just as Dilip showed.