Question

Two numbers are selected at random from the number $1,2,...,n$. Let $p$ denote the probability that the difference between the first and second is not less than $m$ (where $0<m<n$). If $n=25$ and $m=10$, find $5p$

Answer 1

Let the first number be $x$ and the second be $y$.
Let $A$ denote the event that the difference between the first and second numbers is at least $m$.
Let $E_x$ denote the event that the first number chosen is $x$.
We must have $x-y\ge m$ or $y\le x-m$.

Therefore $x>m$ and $y<n-m$.
Thus, $P\left(E_x\right)=0$ for $0<x\le m$ and $P\left(E_x\right)=1/n$ for $m<x\le n$.

Also, $P\left(A|E_x\right)=\frac{(x-m)}{(n-1)}$.
Therefore,
$P\left(A\right)=\sum _{x=1}^{n}P\left(E_x\right)P\left(A|E_x\right)$
$=\sum _{x=m+1}^{n}P\left(E_x\right)P\left(A|E_x\right)=\sum _{x=m+1}^n\frac{1}{n}\cdot \frac{x-m}{n-1}$
$=\frac{1}{n(n-1)}[1+2+...+(n-m\left)\right]=\frac{(n-m)(n-m+1)}{2n(n-1)}$
Putting $n=25$ and $m=10$,
$P\left(A\right)=\frac{(25-10)(25-10+1)}{2.25(25-1)}=\frac{15.16}{\mathrm{2.25.24}}=\frac{1}{5}$
$\Rightarrow 5p=1$