Question

A bag contains $n$ tickets marked $1,2,3,...,n$. If two tickets are drawn, then the chance that the difference of the numbers on the tickets exceed $m<(n-1)$ is

• A. $\frac{(n-m)(n-m+1)}{n(n-1)}$
• B. $\frac{(n-m)(m-n+1)}{n(n-1)}$
• C. $\frac{(n-m)(n-m-1)}{n(n-1)}$
• D. None of these

Answer 1

The correct option is C $\frac{(n-m)(n-m-1)}{n(n-1)}$

Two tickets out of $n$ can be picked up in ${}^n{C}_2=\frac{n(n-1)}{2}$ ways.

Now, we find all such pairs of numbers from $1$ to $n$ whose difference is more than $m$.

Such numbers are $(1,m+2),(1,m+3),(1,m+4),...,(1,n);(2,m+3),(2,m+4),...,(2,n);$

$(3,m+4),(3,m+5),...,(3,n);...(n-m-2,n-1),(n-m-2,n);(n-m-1,n)$

The number of such pairs

$=(n-m-1)+(n-m-2)+(n-m-3)+...+2+1$

(An A.P. of $(n-m-1)$ terms)

$=\frac{(n-m-1)(n-m-1+1)}{2}=\frac{(n-m)(n-m-1)}{2}$

Hence the required probability $=\frac{(n-m)(n-m-1)}{n(n-1)}$