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-2 r} $
For independent of $n, \frac{q}{p}=1$
$\Rightarrow q=p $
$\because p+q=1$
$\therefore 2 p=1 $
$\Rightarrow p=\frac{1}{2}$