91Ó°ÊÓ

Since X and Y are independent random variables and uniformly distributed over their respective intervals, their pdfs are given by: \(f_X(x) = \frac{1}{L/2}\) for \(0 < x < L/2\), and \(f_Y(y) = \frac{1}{L/2}\) for \(L/2 < y < L\). We can find the joint pdf of X and Y by multiplying their individual pdfs, as they are independent: \(f_{X,Y}(x, y) = f_X(x) \cdot f_Y(y) = \frac{1}{(L/2)^2} = \frac{4}{L^2}\) for \(0 < x < L/2\) and \(L/2 < y < L\).

We want to find the probability that the distance between the two points is greater than L/3, that is, |X - Y| > L/3. To find this probability, we need to integrate the joint pdf of X and Y over the appropriate region, denoted as R: \(P(|X - Y| > L/3) = \int\int_R f_{X,Y}(x, y) dxdy\) The region R corresponds to the area where |X - Y| > L/3. In this case, since 0 < x < L/2 and L/2 < y < L, we have: X - Y < -L/3 (1) and Y - X > L / 3 (2) Let's express these inequalities in terms of y: From (1), we have \(y > x + L/3\). From (2), we have \(y > x + L/3\).

To calculate the integral, we need to find the limits of integration for x and y. We can visualize the region R in the xy-plane, where the horizontal axis represents X and the vertical axis represents Y. We have two inequalities for y in terms of x, which define the lower bounds of R. The upper bound of R is given by the line \(y = L\). Now, we can set up the integral with the appropriate limits: \(P(|X - Y| > L/3) = \int_0^{L/2} \int_{x + L/3}^{L} f_{X,Y}(x, y) dy dx\) \(= \frac{4}{L^2} \int_0^{L/2} \int_{x + L/3}^{L} dy dx\) \(= \frac{4}{L^2} \int_0^{L/2} (L - x - L/3) dx\) \(= \frac{4}{L^2} \int_0^{L/2} (2L/3 - x) dx\) Now, integrate with respect to x: \(= \frac{4}{L^2} \left[2Lx/3 - x^2/2\right]_0^{L/2}\) \(= \frac{4}{L^2} \left(2L(L/2)/3 - (L/2)^2/2\right)\) \(= \frac{4}{L^2} \left(\frac{L^2}{2} - \frac{L^2}{8}\right)\) \(= \frac{4}{L^2} \cdot \frac{3L^2}{8}\) Finally, we have the required probability: \(P(|X - Y| > L/3) = \frac{3}{2}\)