# $X,Y,Z$ are iid ~ $U(0,1)$, find $P(X>YZ)$ and $P(X<Y<Z)$

$X,Y,Z$ are iid ~ $U(0,1)$, find $P(X>YZ)$ and $P(X<Y<Z)$

I have no idea how to solve this problem, anyone could help me? Thanks

For the first one, simply evaluate the integral

\begin{align*} P(X>YZ) &= \int_0^1 \int_0^1 \int_{yz}^1 dx dy dz = \int_0^1 \int_0^1 (1-yz) dy dz\\ & = \int_0^1 \left(1-\frac{z}{2}\right) dz = 1 - \frac{1}{4} = \frac{3}{4}. \end{align*}

As for the second probability, since the problem is symmetric with respect to $X$,$Y$ and $Z$, each ordering of these variables is equally likely. Since there are $3!$ orderings of $X$,$Y$ and $Z$,

$$P(X<Y<Z) = \frac{1}{3!} = \frac{1}{6}.$$

• it's really helpful to me, thanks. – Jakoer Dec 10 '14 at 4:57