Question

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

• A. $\frac{n}{n^2-1}$
• B. $\frac{3n}{n^2-1}$
• C. $\frac{3n}{4n^2-1}$
• D. $\frac{3n}{4n^2+2n-1}$

Answer 1

Answer: (C)

Out of $2n+1$ numbers,$n+1$ would be odd and n would be even or vice-versa. If we select any 2 of the $n+1$ numbers, the third number will be automatically decided so as to have the three in AP and the same reasoning would be valid if we select any 2 of the $n$ numbers (for example, if we select 3 and 11, the third number has to be 7 OR if we select 6 and 14, the third number has to be 10).
Thus, the probability = $\frac{{}^{n+1}{C}_2{+}^n{C}_2}{{}^{2n+1}{C}_3}=\frac{\left[\right(n+1)\ast n+n\ast (n-1\left)\right]\ast 3!}{2!\ast (2n+1)\ast 2n\ast (2n-1)}=\frac{3n}{4n^2-1}$