Question

Suppose $X$ follows a binomial distribution with parameters $n$ and $p$, where $0<p<1.1,\frac{P(X=r)}{P(X=n-r)}$ is independent of $n$ for every $r$, then $p=$

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{8}$

Answer 1

Answer: (A)

$\frac{1}{2}$

$\frac{p(x=r)}{P(x=n-r)}=\frac{{}^n{C}_r{P}^r{q}^{n-r}}{{}^n{C}_{n-r}{P}^{n-r}{Q}^r}$

$=(\frac{q}{p}{)}^{n-2r}$

As it is independent of $n$ for every $r$ so

$\left(\frac{q}{p}\right)=1$

$\Rightarrow q=p$

and as $q+p=1$

$\Rightarrow p=q=\frac{1}{2}$