Question

Suppose $X$ follows a binomial distribution with parameter $\mathrm{n}$ and $\mathrm{p}$. If $\frac{\mathrm{P}\left(\mathrm{X}=\mathrm{r}\right)}{\mathrm{P}\left(\mathrm{X}=\mathrm{n}-\mathrm{r}\right)}$ is independent of $\mathrm{n}$ and $\mathrm{r}$,then

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

Answer 1

The correct option is A

$\mathrm{p}=\frac{1}{2}$

Explanation for the correct option:

Given that,

$$\begin{gathered} \frac{\mathrm{P}\left(\mathrm{X}=\mathrm{r}\right)}{\mathrm{P}\left(\mathrm{X}=\mathrm{n}-\mathrm{r}\right)}=\frac{{}^n{C}_r{p}^r{q}^{n-r}}{{}^n{C}_{n-r}{p}^{n-r}{q}^r} \\ ={\left(\frac{q}{p}\right)}^{n-2r} \end{gathered}$$

We know that, $\frac{\mathrm{P}\left(\mathrm{X}=\mathrm{r}\right)}{\mathrm{P}\left(\mathrm{X}=\mathrm{n}-\mathrm{r}\right)}$ is the independent of $\mathrm{n}$ and $\mathrm{r}$ as it is given,

So, $\mathrm{n}-2\mathrm{r}=0$

$$\begin{gathered} \therefore {\left(\frac{q}{p}\right)}^0=1 \\ \Rightarrow q=p \end{gathered}$$

We know that,

$$\begin{gathered} \Rightarrow \mathrm{p}+\mathrm{q}=1 \\ \Rightarrow \mathrm{p}+\mathrm{p}=1[\because p=q] \\ \Rightarrow 2\mathrm{p}=1 \\ \Rightarrow \mathrm{p}=\frac{1}{2} \end{gathered}$$

Hence, the correct answer is option (A)