How to find the probability $P\{X_1 + X_2 \leq X_3\}$? Suppose $X_1,X_2,X_3$ be three independent and mutually identically distributed random variable with uniform distribution on closed interval [0,1]. What is the probability $P\{X_1 + X_2 \leq X_3\}$?
 A: We may also do it by convolution of pdfs.
Suppose the random variable $Z=X_1+X_2$, then by convolution, the pdf of $Z$ is:
\begin{align*}
  f(z) &= \left\{\begin{matrix}
 z & 0\le z < 1\\ 
2-z & 1\le z < 2 \\
0 & \text{otherwise}
\end{matrix}\right.
\end{align*}
Hence
\begin{align*}
  \mathbb{P}\left(Z<X_3\right) &= \int_0^1\, \int_0^{x_3} z\, dz\, dx_3 \\
  &= \frac{1}{6}
\end{align*}
A: Hint:  What is the volume of the region in $\mathbb R^3$ satisfying $0 \le x \le 1$, $0 \le y \le 1$, $x+y \le z \le 1$?
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}&\color{#66f}{\large%
\left.\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\dd z\,\dd y\,\dd x\,
\right\vert_{z\ >\ x\ +\ y}}
=\left.\int_{0}^{1}\int_{0}^{1}\int_{x + y}^{1}\dd z\,\dd y\,\dd x\,
\right\vert_{x\ +\ y\ <\ 1}
\\[3mm]&=\left.\int_{0}^{1}\int_{0}^{1}\pars{1 - x - y}\,\dd y\,\dd x\,
\right\vert_{y\ <\ 1\ -\ x}
=\int_{0}^{1}\int_{0}^{1 - x}\pars{1 - x - y}\,\dd y\,\dd x
\\[3mm]&=\int_{0}^{1}\bracks{\pars{1 - x}^{2} - {\pars{1 - x}^{2} \over 2}}\,\dd x
=\half\int_{0}^{1}x^{2}\,\dd x = \color{#66f}{\large{1 \over 6}}
\end{align}
