# If X is binomial variate with parameters $n$ and $p$ where $0 < P < 1$ such that $\large\frac{P(x=r)}{P(x=n-r)}$ is independent of n and r then p equals.

$\begin {array} {1 1} (A)\;\large\frac{1}{2} & \quad (B)\;\large\frac{1}{3} \\ (C)\;\large\frac{1}{4} & \quad (D)\;None\: of \: these \end {array}$

We have
$\large\frac{P(x=r)}{P(x=n-r)} = \large\frac{n\: c_r(1-P)^{n-r}}{n\: C_{n-r} (1-P)^r}$
$= \bigg( \large\frac{1-P}{P} \bigg)^{n-2r}$
$= \bigg( \large\frac{1}{P}-1 \bigg)^{n-2r}$
For this to be independent of $n$ and $r$, we must have
$\large\frac{1}{P}-1=1$
$P = \large\frac{1}{2}$
Hence (A) is correct option
answered Jan 9, 2014