Multiple choice

Out of (2n+1) tickets consecutively numbered,three are drawn at random. The chance that the numbers on them are in A.P. is

  1. $\displaystyle \frac{n}{n^{2}-1}$
  2. $\displaystyle \frac{3n}{n^{2}-1}$
  3. $\displaystyle \frac{3n}{4n^{2}-1}$
  4. $\displaystyle \frac{3n}{4n^{2}+2n-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The total number of ways to choose 3 tickets from 2n+1 is C(2n+1, 3). For the numbers to be in A.P., the sum of the first and third must be even, meaning they must have the same parity; there are n+1 odd and n even numbers. Calculating the favorable outcomes leads to the probability 3n / (4n^2 - 1).

AI explanation

Three numbers chosen from 2n+1 consecutive numbers will form an arithmetic progression if their sum is even, which requires them to be either all odd, all even, or two odd with one even. The total number of ways to choose 3 numbers is (2n+1)C3 = n(2n+1)(2n-1)/3. The number of favorable APs is n(2n+1), making the probability n(2n+1) / [n(2n+1)(2n-1)/3]. Simplifying this expression by canceling n(2n+1) and multiplying by 3 gives the result 3n / (4n^2 - 1).

Keep practicing — related questions