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.
$\therefore n(E) = \,{}^{n+1}C_2 + \,{}^nC_2$
$=\frac{(n+1)n}{2} + \frac{n(n-1)}{2} = n^2$
$\therefore $ Required probability $ P(E) = \frac{n(E)}{n(S)} $
$ = \frac{n^2}{\frac{n(4n^2 -1)}{3}} = \frac{3n}{4n^2 - 1}$