Mark the correct alternative in the following question: Suppose a…

Question

Mark the correct alternative in the following question:
Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If $\frac{\text{P(X = r})}{\text{P(X = n} -\text{r})}$ is independent of n and r, then p equals:

$\frac{1}{2}$
$\frac{1}{3}$
$\frac{1}{5}$
$\frac{1}{7}$

✓

Answer

$\frac{1}{2}$

Solution:
Consider,
$\text{P(X = r) = kP(X = n}-\text{r})$
Using $\text{ }^{\text{n}}\text{C}_{\text{r}}=\text{ }^{\text{n}}\text{C}_{\text{n}-\text{r}},\text{q}=1-\text{p}$
$\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}=\text{kp}^{\text{n}-\text{r}}\text{q}^{\text{r}}$
$\text{p}^{\text{r}}(1-\text{p})^{\text{n}-\text{r}}=\text{kp}^{\text{n}-\text{r}}(1-\text{p})^{\text{r}}$
$\text{p}^{2\text{r}-\text{n}}=\text{k}(1-\text{p})^{\text{2r}-\text{n}}$
$\big(\frac{\text{p}}{\text{q}}\big)^{\text{2r}-\text{n}}=\text{k}$
when p = q then k = 1
$\Rightarrow\text{p = q}=\frac{1}{2}$

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