Binomial Distribution — Maths STD 12 Science — Question
Gujarat BoardEnglish MediumSTD 12 ScienceMathsBinomial Distribution4 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).
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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 $\text{P(X = r})=\text{ }^{\text{n}}\text{c}_{\text{r}}\text{p}^{\text{r}}\text{q}^{\text{n}-\text{r}}$ $\text{P(X= r})\text{ }^8\text{c}_{\text{r}}\big(\frac{1}{2}\big)^{\text{r}}\big(\frac{1}{2}\big)^{8-\text{r}}$ $\text{P(X=}1)$ $=\text{ }^8\text{c}_{1}\big(\frac{1}{2}\big)^{1}\big(\frac{1}{2}\big)^{8-1}$ $=8\big(\frac{1}{2}\big)^8$ $=\big(\frac{1}{2}\big)^5$ $=\frac{1}{32}$ $\text{P(X}=1)=\frac{1}{32}$
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