Difference of numbers in a unit interval $x,y,z\in \mathbb{R}$ are chosen at random from the unit interval $[0, 1]$. 
What is the probability that
$$\max(x,y,z) - \min(x,y,z) \leq \frac{2}{3}$$
EDIT- Solutions not using calculus would be appreciated, as this problem appeared on a test where calculus was not meant to be used.
 A: Let $E$ be the event $\{ \max(x, y, z) - \min(x, y, z) \le \frac{2}{3}\}$. Our goal is to find $\Pr[E]$. Now let's split $E$ into 6 events $E_{xy}, E_{yx}, E_{xz}, E_{zx}, E_{yz}, E_{zy}$, where
$$E_{xy} = \{ (x = \max(x, y, z)) \land (y = \min(x, y, z)) \land (x - y \le 2/3)\},$$
and the rest are defined similarly. Note that $E$ is the union of these six events and the probability of intersection of any two of these events is $0$.   Also, due to symmetry, all these 6 events have the same probability. Hence $\Pr[E] = 6\Pr[E_{xy}]$. 
It remains to  compute $\Pr[E_{xy}]$. Note that $E_{xy}$ happens iff $y\le x\le y + 2/3$ and $z\in[y, x]$. I.e.
\begin{align*}
\Pr[E_{xy}] &= \int\limits_0^1 dy \int\limits_y^{\min(y + \frac{2}{3}, 1)} dx \int\limits_y^x dz 
=  \int\limits_0^1 dy \int\limits_y^{\min(y + \frac{2}{3}, 1)} (x - y) dx \\
&= \int\limits_0^1 dy \cdot \frac{(x - y)^2}{2} \Bigg|_{y}^{\min(y + \frac{2}{3}, 1)}
= \int\limits_0^1 \frac{\left(\min(y + \frac{2}{3}, 1) - y\right)^2}{2} dy\\
&= \int\limits_0^{\frac{1}{3}} \frac{\left(\frac{2}{3}\right)^2}{2}dy + \int\limits_{\frac{1}{3}}^1 \frac{(1 - y)^2}{2}dy \\
&= \frac{2}{27} -\frac{(1 - y)^3}{6} \Bigg |_{\frac{1}{3}}^1 = \frac{10}{81}. 
\end{align*}
And thus the answer is $ 6\cdot \frac{10}{81} = \frac{20}{27}$
