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 $n$ for every $r$ then $p$ is equal to

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

Answer 1

Answer: (A)

Given $P(X=r)={}^n{C}_r{p}^r{q}^{n-r}$
$\therefore \frac{P(X=r)}{P(X=n-r)}=\frac{{}^n{C}_r{p}^r{q}^{n-r}}{{}^n{C}_{n-r}{p}^{n-r}{q}^r}$
$={\left(\frac{q}{p}\right)}^{n-2r}$
For independent of $n,\frac{q}{p}=1$
$\Rightarrow q=p$
$\because p+q=1$
$\therefore 2p=1$
$\Rightarrow p=\frac{1}{2}$