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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:
  • A
    $\frac{1}{2}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{1}{5}$
  • D
    $\frac{1}{7}$
✓

Answer

  1. $\frac{1}{2}$

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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