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 $2/3$.
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);
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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
\[\int_{0}^{2/3} \int_{2/3}^{1} 1 \, da \, db = \int_{0}^{2/3} (b - 2/3) \, db = \boxto{0.0555555556}.\]
We start by considering the probability that interval $I$ contains the number $2/3.$
Since the four numbers $x_1,$ $x_2,$ $x_3,$ $x_4$ are chosen at random in the interval $[0,1],$ the probability that any one of them is equal to $2/3$ is $\frac{1}{3}.$ Therefore, the probability that both $x_1$ and $x_2$ are equal to $2/3$ is $\left( \frac{1}{3} \right)^2 = \frac{1}{9}.$
Similarly, the probability that both $x_3$ and $x_4$ are equal to $2/3$ is $\left( \frac{1}{3} \right)^2 = \frac{1}{9}.$
Now, consider the probability that intervals $I$ and $J$ both contain the number $2/3.$ If interval $I$ contains $2/3,$ then $x_1$ must be less than $2/3,$ and $x_2$ must be greater than $2/3.$ Similarly, if interval $J$ contains $2/3,$ then $x_3$ must be less than $2/3,$ and $x_4$ must be greater than $2/3.$
So the probability that both intervals contain $2/3$ is the product of the probability that interval $I$ contains $2/3$ and the probability that interval $J$ contains $2/3,$ which is $\frac{1}{9} \cdot \frac{1}{9} = \boxed{\frac{1}{81}}.$
To find the probability that both intervals $I$ and $J$ contain the number $2/3,$ we need to determine the probability of choosing four numbers $x_1,$ $x_2,$ $x_3,$ $x_4$ in the interval $[0,1]$ such that $2/3$ falls within the intervals $I$ and $J.$
Let's break down the problem step by step:
Step 1: Define the interval $I$
The interval $I$ is defined by the two numbers $x_1$ and $x_2.$ To contain the number $2/3,$ we need $x_1 \le 2/3$ and $x_2 \ge 2/3.$
Step 2: Define the interval $J$
The interval $J$ is defined by the two numbers $x_3$ and $x_4.$ To contain the number $2/3,$ we need $x_3 \le 2/3$ and $x_4 \ge 2/3.$
Step 3: Determine the probability for $I$ and $J$ to contain $2/3$
To calculate the probability, we need to determine the length of the intervals $I$ and $J,$ and then find the probability that the length of both intervals is less than or equal to $2/3.$
The length of $I$ is $x_2 - x_1,$ and the length of $J$ is $x_4 - x_3.$
Since all four numbers $x_1,$ $x_2,$ $x_3,$ and $x_4$ are chosen randomly from the interval $[0,1],$ each of them has a uniform distribution between $0$ and $1.$
Therefore, the lengths of intervals $I$ and $J$ are independent of each other and follow the uniform distribution on $[0,1].$
Step 4: Calculate the probability
To calculate the probability that both intervals $I$ and $J$ contain $2/3,$ we need to find the probability that both lengths are less than or equal to $2/3.$
The probability that the length of interval $I$ is less than or equal to $2/3$ is given by the ratio of the length of the subinterval $[0,2/3]$ to the length of the entire interval $[0,1],$ which is $\frac{2/3}{1} = \frac{2}{3}.$
Similarly, the probability that the length of interval $J$ is less than or equal to $2/3$ is also $\frac{2}{3}.$
Since the lengths of intervals $I$ and $J$ are independent, the probability that both lengths are less than or equal to $2/3$ is the product of their individual probabilities: $\frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9}.$
Therefore, the probability that intervals $I$ and $J$ both contain the number $2/3$ is $\boxed{\frac{4}{9}}.$