# 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 $P\{X_1 + X_2 \leq X_3\}$?

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*}

• I know this question was answered months ago, but I just ran across it searching for something similar. Could you please explain what convolution is? – pocketlizard Dec 3 '14 at 20:30
• Hi, read about convolution in this book, has several good problems. – gar Dec 8 '14 at 11:10

Hint: What is the volume of the region in $\mathbb R^3$ satisfying $0 \le x \le 1$, $0 \le y \le 1$, $x+y \le z \le 1$?
