Two points are selected randomly on a line of length 40 so as to be on opposite sides of the midpoint of the line. in other words, the two points x and y are independent random variables such that x is uniformly distributed over [0,20) and y is uniformly distributed over(20,40]. find the probability that the distance between the two points is greater than 12 .

Probability that we have an ordered pair (x,y) representing two points that satisfy the conditions is 328/400.

Step-by-step explanation:

This is a geometric probability.

set of all possible x in [0,20) and y in (20,40] we obtain a square in the x-y plane (in the first quadrant).

Area of this square is 20x20 = 400.

Set of all (x,y), that satisfy the distance being greater than 12

y > x, this means that (y - x) > 12 which is the same as the inequality y > x + 12.

For inequality, we obtain the region in the plane above the line y = x + 12.  The area of this region is

(1/2)bh = (1/2)(12)(12) = 72.

Thus the area of the set of all points in our square that satisfy the condition (y - x) > 12 is 400 - 72 = 328.

 Probability that we have an ordered pair (x,y) representing two points that satisfy the conditions is 328/400.

it could be a ..idk tho

i think asking google will you get the answer to that question