Mathematics - Probability Question with Solution | TestHub

$x$ and $y$ can be any one of (2n+1) numbers given. $|x-y|\le n\Rightarrow x-n\le y\le x+n$ Hence number of possibilities of $y$, for $x=0,1,2,3,\cdots -\mathrm{n}-1,\mathrm{n},\mathrm{n}+1-\cdots -2\mathrm{n}$ are $\mathrm{n}+1,\mathrm{n}+2,\mathrm{n}+3\mathrm{n}-\cdots ---2\mathrm{n},2,+1,2\mathrm{n},2,-1,-\cdots -\mathrm{n}+1$ respectively. $\therefore \text{ Probability }=\frac{2(\overline{n+1}+\overline{n+2}+-----+\overline{2n})+2n+1}{(2n+1{)}^2}=\frac{3n^2+3n+1}{(2n+1{)}^2}$