Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSProbability5 Marks
Question
If the mean and variance of a random variable X having a binomial distribution are 4 and 2 respectively, find P (X = 1).
✓
Answer
Let n and p be the parmeters of binomial distribution,
Given,
$\text{Mean = np}=4$
$\text{Variance = npq}=2$
Dividing equation (2) by (1),
$\frac{\text{npq}}{\text{np}}=\frac{2}{4}$
$\text{q}=\frac{1}{2}$.
$\text{p}=1-\frac{1}{2}$
$\text{p}=\frac{1}{2}$
Put the value of p in equation (1),
$\text{np}=4$
$\text{n}\big(\frac{1}{2}\big)=4$
$\text{n}=8$
Hence, binomial distribution is given by
$\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$
$\text{P(X= r})\text{ }^8\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{8-\text{r}}$
$\text{P(X=}1)$
$=\text{ }^8\text{c}_{1}\big(\frac{1}{2}\big)^{1}\big(\frac{1}{2}\big)^{8-1}$
$=8\big(\frac{1}{2}\big)^8$
$=\big(\frac{1}{2}\big)^5$
$=\frac{1}{32}$
$\text{P(X}=1)=\frac{1}{32}$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.