Question

If X is a binomial variate with parameters n and p, where 0 < p < 1 such that $\frac{P\left(X=r\right)}{P\left(X=n-r\right)}\mathrm{is}$ independent of n and r, then p equals
(a) 1/2
(b) 1/3
(c) 1/4
(d) None of these

Answer 1

(a) 1/2

Given that P(X=r) = k P(X=n$-$r), where k is independent of n and r .

$$\begin{gathered} {}^{n}C_r\mathit{}{p}^r{q}^{n-r}=k{}^{n}C_{n\mathit{-}r}{p}^{n\mathit{-}r}{q}^r \\ \mathrm{We}\mathrm{have}{}^{n}C_{\mathrm{r}}={}^{n}C_{n\mathit{-}r}\mathrm{and}\mathrm{also}\mathit{}q=1-p \\ \mathrm{Hence},\mathrm{the}\mathrm{equation}\text{changes to the following:} \\ {p}^{\mathrm{r}}(1-p{)}^{n-r}=k{p}^{n-r}(1-p{)}^r \\ \Rightarrow (1-p{)}^{n-2r}=\mathit{}k\mathit{}{p}^{n-2r} \\ \Rightarrow {\left(\frac{q}{p}\right)}^{n-2r}=k\mathit{} \\ \mathrm{This}\mathrm{is}\mathrm{possible}\mathrm{when}p=q\mathrm{and}k\mathrm{becomes}1. \\ \mathrm{Hence},p=q=\frac{1}{2} \end{gathered}$$