Uniform distribution problem of finding overlapping region

Question

I found $\displaystyle \int_{5}^{15}\dfrac{dx}{22}$ Can somebody just verify is it correct?.

Answer 1

Let $x,y$ be the midpoints of the first and second segments, respectively.

For the sample space we can take the set $S$ of points $(x,y)\in\mathbb{R}^2$ such that $$ \left\{ \begin{align*} &0\le x\le 14\\[4pt] &6\le y\le 20\\[4pt] \end{align*} \right. $$ which is a $14{\times}14$ square region.

The subregion of $S$ for which the segments overlap is the set $$ A=\{(x,y)\in S{\,:\,}|x-y|\le 2\} $$ We can break up $A$ into two subregions:

• If$\;4\le x\le 8\;$then $(x,y)\in A\iff 6\le y\le x+2$.$\\[4pt]$
• If$\;8\le x\le 14\;$then$\;(x,y)\in A\iff x-2\le y\le x+2$.

hence the area of $A$ is equal to $$ \int_4^8 \Bigl((x+2)-6\Bigr)\,dx + \int_8^{14} \Bigl((x+2)-(x-2)\Bigr)\,dx $$ so the probability that the segments overlap is $$ \frac{8+24}{14^2} = \frac{32}{196} = \frac{8}{49} $$