Question

Suppose $X$ follows 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$=$

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

Answer 1

Answer: (A)

$\frac{1}{2}$

$P(x=r)={}^n{C}_r{p}^r(1-p{)}^{n-r}$
$P(x=n-r)={}^n{C}_{n-r}{p}^{n-r}(1-p{)}^r$
Now, ${}^n{C}_r={}^n{C}_{n-r}$

Hence, $\frac{P(x=r)}{P(x=n-r)}$

$=\frac{p^r(1-p{)}^{n-r}}{p^{n-r}(1-p{)}^r}$

Is independent of $n$.
The above expression is true for $x=\frac{1}{2}$ only.