conditional probabilityprobabilityprobability theory
Two points are selected randomly on a line of length $L$ so as to be on the opposite sides of the midpoint of the line. In other words, two points X and Y are independent random variables such that X is uniformly distributed over $(0,L/2)$ and Y is so over $(L/2,0)$. Find the probability that the distance between these two points is greater than $2L/3$.
Here's what I have done.
Can I get some help to finish this.
Since you are integrating a constant function, it's going to be so much easier if you draw yourself a picture of the area and use simple geometry, rather than integrating an analytic expression:
And dividing the hatched area by the total square area, the resulting probability is $\frac{7}{36}/\frac{1}{4} = \frac{7}{9}$
Probability is $3/4$ if points $x$ and $y$ are chosen with uniform probability. That corresponds to the area in color in the picture below.
Best Answer
Does this make sense to you? I think this aligns with what Ethan has commented.