Multiple choice

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

  1. $ \displaystyle \frac { \left( n-m \right) \left( n-m+1 \right) }{ n\left( n-1 \right) } $
  2. $ \displaystyle \frac { \left( n-m \right) \left( m-n+1 \right) }{ n\left( n-1 \right) } $
  3. $ \displaystyle \frac { \left( n-m \right) \left( n-m-1 \right) }{ n\left( n-1 \right) } $

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total ways to choose 2 tickets is n(n-1)/2. Favorable ways where difference > m: (n-m)(n-m-1)/2. Probability = [(n-m)(n-m-1)/2] / [n(n-1)/2] = (n-m)(n-m-1) / n(n-1).

Keep practicing — related questions