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Probability — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSProbability1 Mark
Question
Choose the correct answer from the given four options.
Suppose a random variable X follows the binomial distribution with parameters n and p, where 0 < p < 1. If $\frac{\text{P}(\text{x}=\text{r})}{\text{P}(\text{x}=\text{n}–\text{r})}$ is independent of n and r, then p equals:
✓

Answer

  1. $\frac{1}{2}$
Solution:
$\text{P}(\text{x}=\text{r})={^\text{n}}\text{C}_\text{r}(\text{p})^\text{r}(\text{q})^{\text{n}-\text{r}}$
$=\frac{\text{n}!}{(\text{n}-\text{r})!\text{r}!}(\text{p})^\text{r}(1-\text{p})^{\text{n}-\text{r}}[\therefore\text{q}=1-\text{p}]$
Now, $\frac{\text{P}(\text{x}=\text{r})}{\text{P}(\text{x}=\text{n}-\text{r})}=\frac{{^\text{n}\text{C}_\text{r}\text{p}^\text{r}(1-\text{p})^{\text{n}-\text{r}}}}{{{^\text{n}}\text{C}}_{\text{n}-\text{r}}\text{p}^{\text{n}-\text{r}}(1-\text{p})^{\text{r}}}$
$=\frac{\text{P}^\text{r}(1-\text{P})^{\text{n}-\text{r}}}{\text{p}^{\text{n}-\text{r}}(1-\text{p})^\text{r}}$ $\big[\text{as}{^\text{n}}\text{C}_\text{r}={^\text{n}}\text{C}_{\text{n}-\text{r}}\big]$
$=\Big(\frac{1-\text{p}}{\text{p}}\Big)^{\text{n}-2\text{r}}$
Above expression is independent of n and r, if 
$\frac{1-\text{p}}{\text{p}}=1\Rightarrow\text{p}=\frac{1}{2}$

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