Probability of the distance between the two points on a line.

Consider the one end of the rod at origin and other end is at $(L,0)$

Let $p$ be the probability where distance between two points will be $\leq \dfrac{L}{3}$

Two random variables are $x,y$

So, to evaluate $p$ the $x$ should be between $(\dfrac{L}{6},\dfrac{L}{2})$

The variable $y$ is dependent upon the selection of $x$

So, $y$ varies from $\dfrac{L}{2} \to (\dfrac{L}{2} +x - \dfrac{L}{6})$

The probability $p=\displaystyle \dfrac{\int_{\dfrac{L}{6}}^{\dfrac{L}{2}} dx \int_{\dfrac{L}{2}}^{\dfrac{L}{2} +x - \dfrac{L}{6}} dy}{\int_{0}^{L/2} dx \int_{L/2}^{L} dy} \\ =\dfrac{2}{9}$

We need to evaluate the distance between two points greater than $\dfrac{L}{3}$ which is $1-p=1-\dfrac{2}{9}=\dfrac{7}{9}$