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Probability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSProbability1 Mark
Question
If X is a binomial variate with parameters n and p, where 0 < p < 1 such that $\frac{\text{P(X = r)}}{\text{P(X = n - r})}$ is independent of n and r, then p equals:
✓

Answer

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

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