$P(X+Y > Z)$ where $X,Y,Z \sim U(0,1)$ and are IID. I am trying to solve this problem Compute $P(X+Y>Z)$ where $X, Y$ and $Z$ are independent, uniform random variables in the interval $[0,1]$., but I only want to think about this problem geometrically, but I'm struggling with visualizing this problem.
I can imagine the sample space of $X,Y,Z$ to be uniformly distributed within a $1\times1\times1$ cube. Visually, I can consider 2 disjoint sample spaces $X+Y < 1$ and $X+Y > 1$. Each represents half of the cube sliced along the $x+y=1$ diagonal plane that extends in the z-direction. For $X+Y>1$, which occurs with probability 0.5, we know that $P(X+Y>Z, X+Y > 1) = 1$. This one is simple to visualize.
I am struggling with visualizing the $X+Y < 1, X+Y > Z$ case. It seems this forms a region that starts out as a triangle cross section at $z=0$ and then evolves to a trapezoidal cross-sectional area for $z > 0$ that decreases in area as $Z \rightarrow 1$?
 A: In the unit cube, consider the vertices for which $X+Y = Z$.  These are $(0,0,0)$, $(1,0,1)$, and $(0,1,1)$.  Since we have enumerated three distinct, non-collinear points on such a plane, and these are not common to a single face of the cube, no other vertices satisfy this equation.  Therefore, the set of all points inside the cube that satisfy $X + Y > Z$ comprise those points "below" this plane.  Since this plane cuts off a pyramidal volume of $1/6$--note the fourth vertex is $(0,0,1)$--the desired probability is simply $5/6$.
A: Let $W=1-Z.$ Then

*

*$X,Y,W \sim \text{i.i.d.}\operatorname{Uniform}(0,1)$ and

*$X+Y<Z$ if and only if $X+Y+W<1.$
And so
\begin{align}
& \Pr(X+Y+W<1) \\[8pt]
= {} & \iiint\limits_{(x,y,w)\,:\, x+y+w \,<\, 1} 1 \, d(x,y,w) \\[8pt]
= {} & \int_0^1 \left( \int_0^{1-x} \left( \int_0^{1-x-y} 1 \, dz \right)\, dy \right) \, dx \\[8pt]
= {} & \frac 1 6.
\end{align}
Therefore $\Pr(Z>X+Y) = \Pr(X+Y+W)>1 = \dfrac 5 6.$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\LARGE\left. a\right)}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}
\bracks{x + y > z}\dd x\,\dd y\,\dd z} =
\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}
\bracks{z < x + y}\dd z\,\dd x\,\dd y
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}\braces{%
\bracks{x + y < 1}\int_{0}^{x + y}\dd z +
\bracks{x + y > 1}\int_{0}^{1}\dd z}\,\dd x\,\dd y
\\[5mm] = &\
\int_{0}^{1}\int_{0}^{1}\braces{\vphantom{\Large A}%
\bracks{x < 1 - y}\pars{x + y} + \bracks{x > 1 - y}}\,\dd x\,\dd y
\\[5mm] = &\
\int_{0}^{1}\bracks{%
\int_{0}^{1 - y}\pars{x + y}\dd x + \int_{1 - y}^{1}\dd x}
\,\dd x\,\dd y
\\[5mm] = &\
\int_{0}^{1}\bracks{%
\pars{{1 \over 2} - {y^{2} \over 2}} + y}\dd y =
{1 \over 2}\int_{0}^{1}\pars{1 + 2y - y^{2}}\dd y =
\bbx{5 \over 6} \\ &
\end{align}
A: When I posted an answer earlier I thought I was missing a simpler way. Now I know what it is.
First notice that each of the three random variables $X,Y,Z$ is equally likely to be the largest of the three. So
$$
\Pr(Z>X\ \&\ Z>Y) = \frac 1 3.
$$
Next it will be enough to prove that
$$
\Pr(Z>X+Y\mid Z>X\ \&\ Z>Y) = \frac 1 2, \tag 1
$$
because it would then follow that $\Pr(Z>X+Y) = \dfrac 1 3 \times \dfrac 1 2 = \dfrac 1 6.$
So the question is how to prove what it says on line $(1).$
For that it will be enough to prove that for every particular value of (lower-case) $z,$ we have
$$
\Pr(Z>X+Y\mid Z>X\ \&\ Z>Y\ \&\ Z=z) = \frac 1 2,
$$
which is the same as the simpler statement
$$
\Pr(z>X+Y\mid z>X\ \&\ z>Y) = \frac 1 2. \tag 2
$$
The difference between the meaning of line $(1)$ and that of line $(2)$ is that line $(1)$ says something about a random variable (capital) $Z$ and line $(2)$ about every particular fixed, non-random, number $z$ between $0$ and $1.$
Now just observe that the conditional distribution of $(X,Y)$ given $z>X\ \&\ z >Y$ is that they are i.i.d. uniform on the interval $(0,z).$ Then the rest is routine.
