Question

Out of $(2n+1)$ consecutively numbered tickets, three tickets are drawn at random. The probability that the numbers on them are in arithmetic progression is

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

Answer 1

Answer: (C)

Let $S$ be the sample space and $E$ be that event of favourable cases
$\therefore n(S)={}^{2n+1}{C}_3$
$=\frac{(2n+1)2n(2n-1)}{1\cdot 2\cdot 3}=\frac{n(4n^2-1)}{3}$
Let three numbers $a,b,c$ be drawn where
$a<b<c$ and given $a,b,c$ be in $AP$.
$\therefore 2b+a+c...(i)$
It is clear from Eq. $(i)$, a and c both are even or odd.
Out of $(2n+1)$ tickets consecutively numbers either $(n+1)$ of them will be odd and n of them will be even, or $(n+1)$ of them will be even and $n$ of them will be odd.
$\therefore n(E)={}^{n+1}{C}_2+{}^n{C}_2$
$=\frac{(n+1)n}{2}+\frac{n(n-1)}{2}=n^2$
$\therefore$ Required probability $P(E)=\frac{n(E)}{n(S)}$
$=\frac{n^2}{\frac{n(4n^2-1)}{3}}=\frac{3n}{4n^2-1}$