If $x$, $y$, $z$ are selected independently uniformly from $[0, 1]$, compute the probability that $x \ge yz$

Solution 1:

Since $yz$ is in $[0,1]$ almost surely and $x$ is independent of $(y,z)$, $$P[x\geqslant yz\mid y,z]=\left.P[x\geqslant t]\right|_{t=yz}=\left.1-t\right|_{t=yz}=1-yz\ \text{almost surely}$$ hence, by the law of total expectation, $$P[x\geqslant yz]=E[P[x\geqslant yz\mid y,z]]=E[1-yz]=1-E[y]\cdot E[z]=1-\tfrac12\cdot\tfrac12$$ that is,

$$P[x\geqslant yz]=\tfrac34$$

All this assumes that the random variables $x$, $y$ and $z$ are (independent, as made explicit in the post, but also) uniformly distributed on $[0,1]$ (as was not made explicit in the post).

More generally, if $x$, $y$ and $z$ are i.i.d. with density $at^{a-1}$ on $(0,1)$, for some positive $a$, then the CDF of $x$ is $P[x\leqslant t]=t^a$ hence $P[x\geqslant yz\mid y,z]=1-(yz)^a$ almost surely, and $E[y^a]=E[z^a]=\frac12$, hence one gets the same result that $$P[x\geqslant yz]=\tfrac34$$ This coincidence with the first result (when $a=1$) is due to the fact that $[x\geqslant yz]=[x^a\geqslant y^az^a]$ and that the random variables $x^a$, $y^a$ and $z^a$ are i.i.d. uniformly distributed on $[0,1]$.

Solution 2:

Let's call $$R=\bigl\{\langle x,y,z\rangle\in[0,1]^3:x\ge yz\bigr\}.$$ We want to know what fraction of the volume of the cube $[0,1]^3$ is occupied by $R$, but the cube has volume $1$, so the problem amounts to determining the volume of $R.$ Given any $y,z\in[0,1]$, the only $x\in[0,1]$ such that $x\ge yz$ lie in the interval $[yz,1].$ Our problem then reduces to evaluating the triple integral $$\begin{align}\int_0^1\int_0^1\int_{yz}^11\,dx\,dy\,dz &= \int_0^1\int_0^1(1-yz)\,dy\,dz\\ &= \int_0^1\int_0^11\,dy\,dz-\int_0^1z\int_0^1y\,dy\,dz\\ &= \left(\int_0^11\,dy\right)\left(\int_0^11\,dz\right)-\left(\int_0^1y\,dy\right)\left(\int_0^1z\,dz\right)\\ &= 1\cdot 1-\frac12\cdot\frac12\\ &=\frac34.\end{align}$$ As Did pointed out in the comments below, this is the same approach as Did's answer, put into (marginally) different terms. (A tip of my proverbial hat to Did for pointing out the reason that my original approach was so tedious.)