$x,y,z$ independent and uniform random on $[0,1]$, $P(x \geq  yz)$? $x,y,z$ are independent and uniform random on $[0,1]$.
I know that $P(x \geq yz) = 3/4$ by triple integrating the triple density function.
However, is there a easier way to get this solution? Without triple integration.
 A: The conditional probability given $z \in [0,1]$,  $P(x \ge yz | z)$, is the area of the region of the square $[0,1]^2]$ below the line $y = x/z$.  Since the region above the line
is a right triangle with sides $1$ and $z$, we have $P(x \ge y z | z) = 1 - z/2$.
Since $E[z] = 1/2$,  the unconditional probability is $E[P(x \ge y z | z)] = E[1 - z/2] = 1 - E[z]/2 = 3/4$.   
A: $$
\mathrm P(X\leqslant YZ\mid Y,Z)=YZ\qquad\quad \mathrm E(YZ)=\mathrm E(Y)\cdot\mathrm E(Z)=1/4$$
A: Here is an alternative way to see it.  
If $U$ is uniformly distributed on $(0,1)$ random variable, $X= -\log(U)$ follows the exponential distribution with unit rate. Thus:
$$
   \mathbb{P}\left( U_1 \geqslant U_2 U_3 \right) = \mathbb{P}\left(-\log(U_1) \leqslant -\log(U_2) -\log(U_3) \right) = \mathbb{P}\left(X_1 \leqslant X_2 + X_3 \right) = 
   \mathbb{P}\left(\frac{X_1}{X_1+X_2+X_3} \leqslant \frac{1}{2} \right)
$$
The ratio $\frac{X_1}{X_1+X_2+X_3}$ follows Beta distribution with parameters $(1,2)$ (e.g., see here, or in "Question about order statistic") meaning that
$$
  \mathbb{P}\left( U_1 \geqslant U_2 U_3 \right) = \int_0^{\frac{1}{2}} \frac{1-x}{\operatorname{B}(1,2)} \mathrm{d} x  \stackrel{\operatorname{B}(1,2)=\frac{1}{2}}{=} \int_\frac{1}{2}^1 2 x \mathrm{d} x  = 1^2 - \left(\frac{1}{2}\right)^2 = \frac{3}{4}
$$
