The numbers $x_1,$ $x_2,$ $x_3,$ $x_4$ are chosen at random in the interval $[0,1].$ Let $I$ be the interval between $x_1$ and $x_2,$ and let $J$ be the interval between $x_3$ and $x_4.$ Find the probability that intervals $I$ and $J$ both have length greater than $3/4$.

Answer 1

Without loss of generality, suppose $x_1 \le x_2$ and $x_3 \le x_4.$ Then the probability that $I$ has length greater than $3/4$ is $\frac{1}{4}.$

If $I$ has length greater than $3/4,$ then the probability that $J$ has length greater than $3/4$ is the probability that both $x_3$ and $x_4$ lie in the interval
\[\left[ \frac{3}{4},1 \right] \setminus \left[ x_1 + \frac{3}{4}, x_2 \right] = \left[ \frac{3}{4},x_1 + \frac{1}{4} \right] \cup \left[ x_2 + \frac{1}{4},1 \right].\]Since $x_1,$ $x_2,$ $x_3,$ $x_4$ are chosen at random in the interval $[0,1],$ each of $x_1,$ $x_2,$ $x_3,$ $x_4,$ is equally likely to be in each subinterval of length $\frac{1}{4}.$ Thus, the probability that both $x_3$ and $x_4$ lie in $\left[ \frac{3}{4},x_1 + \frac{1}{4} \right]$ is
\[\left( \frac{1}{4} \right)^2 = \frac{1}{16},\]and the probability that both $x_3$ and $x_4$ lie in $\left[ x_2 + \frac{1}{4},1 \right]$ is also $\frac{1}{16}.$ Hence, the probability of interest is
\[\frac{1}{4} \cdot \left( \frac{1}{16} + \frac{1}{16} \right) = \boxed{\frac{1}{32}}.\]