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 $3/4$.
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
\[\frac{1}{4} \cdot \left( \frac{1}{16} + \frac{1}{16} \right) = \boxed{\frac{1}{32}}.\]
To find the probability that intervals $I$ and $J$ both have length greater than $3/4,$ we need to determine the valid region in the 4-dimensional space defined by $x_1,$ $x_2,$ $x_3,$ and $x_4.$
First, let's consider the probability that interval $I$ has length greater than $3/4.$ The length of interval $I$ is given by $x_2 - x_1.$ Since $x_1$ and $x_2$ are chosen randomly from the interval $[0,1],$ this means $x_2 - x_1$ can take any value in the interval $[0,1].$ To find the probability that $x_2 - x_1 > 3/4,$ we need to find the fraction of the interval $[0,1]$ for which this condition holds.
Since the total length of the interval $[0,1]$ is 1, the length of the sub-interval for which $x_2 - x_1 > 3/4$ is $1 - 3/4 = 1/4.$ Therefore, the probability that $x_2 - x_1 > 3/4$ is $1/4.$
Similarly, for interval $J,$ the length of interval $J$ is given by $x_4 - x_3.$ Again, since $x_3$ and $x_4$ are chosen randomly from the interval $[0,1],$ we have $x_4 - x_3 \in [0,1].$ The probability that $x_4 - x_3 > 3/4$ is also $1/4.$
To find the probability that both $x_2 - x_1 > 3/4$ and $x_4 - x_3 > 3/4,$ we multiply the probabilities of the two events because they are independent. Therefore, the probability that intervals $I$ and $J$ both have length greater than $3/4$ is $(1/4) \times (1/4) = 1/16.$
Hence, the probability that intervals $I$ and $J$ both have length greater than $3/4$ is $1/16.$
Let's start by visualizing the problem. Consider a number line representing the interval $[0,1].$ We want to find the probability that both intervals $I$ and $J$ have length greater than $3/4.$
To help visualize this, let's consider lengths of intervals rather than the intervals themselves. We can think of $x_1,$ $x_2,$ $x_3,$ and $x_4$ as dividing the interval $[0,1]$ into 5 smaller sub-intervals.
There are a few situations to consider:
- Suppose $x_1,$ $x_2,$ $x_3,$ and $x_4$ are all within the first or the last sub-interval. In this case, both $I$ and $J$ will have lengths less than or equal to $3/4.$
- Suppose $x_1$ and $x_2$ fall within the first or the last sub-interval, while $x_3$ and $x_4$ fall within different sub-intervals. In this case, the lengths of both $I$ and $J$ will be greater than $3/4.$
- Suppose $x_1$ falls within the first or the last sub-interval, while $x_2,$ $x_3,$ and $x_4$ fall within different sub-intervals. Again, both $I$ and $J$ will have lengths greater than $3/4.$
- Suppose $x_1$ and $x_2$ fall within different sub-intervals, while $x_3$ and $x_4$ fall within the first or the last sub-interval. In this case, both $I$ and $J$ will have lengths greater than $3/4.$
- Finally, suppose $x_1,$ $x_2,$ $x_3,$ and $x_4$ all fall within different sub-intervals. In this case, both $I$ and $J$ will have lengths less than or equal to $3/4.$
From these situations, we can see that the only way for both $I$ and $J$ to have lengths greater than $3/4$ is if $x_1$ and $x_2$ fall within the first or last sub-interval, and $x_3$ and $x_4$ fall within different sub-intervals.
- The probability that $x_1$ falls within the first or last sub-interval is $2/5.$
- Given that $x_1$ is in the first or last sub-interval, the probability that $x_2$ is also in the first or last sub-interval is $2/4 = 1/2.$
- Given that $x_1$ and $x_2$ are in the first or last sub-interval, the probability that $x_3$ is in a different sub-interval is $3/3 = 1.$
- Given that $x_1$ and $x_2$ are in the first or last sub-interval, and $x_3$ is in a different sub-interval, the probability that $x_4$ is also in a different sub-interval is $2/2=1.$
Therefore, the probability that both $I$ and $J$ have lengths greater than $3/4$ is $\left(\frac{2}{5}\right)\left(\frac{1}{2}\right)\left(\frac{1}{1}\right)\left(\frac{1}{1}\right) = \frac{1}{5}.$
Hence, the probability is $\boxed{\frac{1}{5}}.$