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Probability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsProbability1 Mark
MCQ
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 :
  • ✓
    $\frac{1}{2}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{1}{5}$
  • D
    $\frac{1}{7}$
✓

Answer

Correct option: A.
$\frac{1}{2}$
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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