Question 1205466
.
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.
~~~~~~~~~~~~~~~~~~



        The problem is posed/worded not perfectly (and far from to be perfect).
        I will not even try to re-formulate it in a perfectionist's way.
        I will simply assume that it is formulated perfectly by somebody else,
        and will solve it for this "ideal" case.



<pre>
In order for x_1 and x_2 satisfy  the condition 

    "[x_1,x_2} is an interval and the length of the interval [x_1,x_2] is greater than 3/4",

these conditions must be satisfied

    x_1 <= 1/4;   (1)

    x_1 <= x_2,  x_2-x_1 >= 3/4.    (2)


x_2 |
    | 
8/8 + . . . . .|---|---|---|---|---|
    | . . . 
7/8 + . .
    | .
6/8 +  
    |
5/8 +
    |
4/8 +
    |
3/8 +
    |
2/8 +
    |
1/8 +
    |
    +---|---|---|---|---|---|---|---|
  0    1/8 2/8 3/8 4/8 5/8 6/8 7/8 8/8 <<<---x_1


In order for you could see the domain, which is good for points (x_1,x_2)) under conditions (1) and (2),
I prepared this plot above. It has horizontal axis x_1 and vertical axis x_2.


The points that satisfy conditions (1) and (2), are the point close to the upper left corner of the square 

    [0 <= x_1 <= 1, 0 <= x_2 <= 1].


It is the right-angled triangle with the legs 1/4 in each direction, so the area of this triangle is {{{(1/2)*(1/4)*(1/4)}}} = {{{1/32}}}.


Thus the probability to get a "good" point (x_1,x_2) randomly is 1/32.


The probability to get the other "good" point (x_3,x_4) randomly is AGAIN 1/32.


Therefore, since the events to get a "good" point (x_1,x_2) and a "good" point (x_3,x_4) are independent,
the probability under the problem's question is the product 

    P = {{{(1/32)*(1/32)}}} = {{{1/1024}}} = {{{1/2^10}}}.    <U>ANSWER</U>
</pre>

Solved.