Answer:
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
