Question

Suppose $X$ follows a binomial distribution with parameters $n$ and $p$, where $0<p<1$. If $\frac{P(X=r)}{P(X=n-r)}$ is independent of $n$ and $r$, then $p$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{2}$

$\frac{P(X=r)}{P(X=n-r)}=\frac{{}^n{C}_r{p}^r{(1-p)}^{n-r}}{{}^n{C}_{n-r}{p}^{n-r}{(1-p)}^r}=\frac{{(1-p)}^{n-2r}}{p^{n-2r}}={(\frac{1}{p}-1)}^{n-2r}$
For ${(\frac{1}{p}-1)}^{n-2r}$ to be independent of $n$ and $r,\frac{1}{p}-1=1$
$\Rightarrow \frac{1}{p}=1\Rightarrow p=\frac{1}{2}.$