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

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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

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