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 contain the number $2/3$.

Answer 1

We place the numbers $x_1,$ 2/3, and $x_2$ on a number line. Without loss of generality, suppose $x_1 < x_2,$ so $I$ is the interval from $x_1$ to $x_2.$

[asy]
unitsize(6 cm);

draw((0,0)--(1,0));
dot((0,0.2));
dot((0.8,0.2));
dot((0.4,0.2));
label("$x_1$", (0,0.2), NW);
label("$\frac{2}{3}$", (0.4,0.2), N);
label("$x_2$", (0.8,0.2), NE);

label("$I$", (0.2,0.05), N);
[/asy]

Let $a = x_1,$ so $I = (a,2/3),$ except that $I$ may reduce to a point interval or be empty.

We place the numbers $x_3,$ 2/3, and $x_4$ on a number line. Without loss of generality, suppose $x_3 < x_4,$ so $J$ is the interval from $x_3$ to $x_4.$

[asy]
unitsize(6 cm);

draw((0,0)--(1,0));
dot((0.2,0.2));
dot((0.8,0.2));
dot((0.4,0.2));
label("$x_3$", (0.2,0.2), NW);
label("$\frac{2}{3}$", (0.4,0.2), N);
label("$x_4$", (0.8,0.2), NE);

label("$J$", (0.5,0.05), N);
[/asy]

Let $b = x_4,$ so $J = (2/3,b),$ except that $J$ may reduce to a point interval or be empty.

Note that $I$ and $J$ can only both contain $2/3$ if $a < 2/3 < b.$ Thus, the probability that this occurs is
\[\int_{0}^{2/3} \int_{2/3}^{1} 1 \, da \, db = \int_{0}^{2/3} (b - 2/3) \, db = \boxto{0.0555555556}.\]