Two points are selected randomly on a line of length L 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, L/2) and Y is uniformly distributed over (L/2, L).] Find the probability that the distance between the two points is greater than L/3.

P(Y-X>L/3)=7/9

Step-by-step explanation:

We know  that  X is uniformly distributed over (0, L/2) and Y is uniformly distributed over (L/2, L).

We know that:

f(x,y)=f(x)·f(y)=\frac{1}{L/2}·\frac{1}{L/2}=4/L²

Therefore, the probability is  

P(Y-X>L/3)=∫∫ f(x,y) dx dy

P(Y-X>L/3)=∫∫ 4/L²   dx dy

We conclude that is  

P(Y-X>L/3)=1-P(Y-X≤L/3)

We know  that for the region  (0, L/2)x(L/2, L) where is satisfied Y-X≤L/3

is a right triangle with bases x∈(L/6, L/2) and y∈(L/2, 5L/6).

We calculate the area of that triangle:

p=1/2 · (L/3)²= L²/18

Now, we have

P(Y-X≤L/3)=4/L² ·  L²/18=4/18=2/9

Therefore, the probability is    

P(Y-X>L/3)=1-P(Y-X≤L/3)=1 - 2/9

P(Y-X>L/3)=7/9

idk sry

step-by-step explanation:

d is the correct answer

step-by-step explanation:

x= 8 because 3x = 3 * 8, and 3*8 = 24