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$X,Y,Z$ 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

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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 $Z$,

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

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

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