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

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

Answer 1

The correct option is A $p=\frac{1}{2}$

We have $\frac{P(X=r)}{P(X=n-r)}=\frac{{}^n{C}_r.p^r.{(1-p)}^{n-r}}{{}^n{C}_{n-r}.p^{n-r}.{(1-p)}^r}$
$=\frac{{(1-p)}^{n-2r}}{p^{n-2r}}={\left(\frac{1-p}{p}\right)}^{n-2r}={(\frac{1}{p}-1)}^{n-2r}$
and $\frac{1}{p}-1>0\therefore$ the ratio will be independent of $n$ and $r$ is $\frac{1}{p}-1=1$ or $p=\frac{1}{2}$