Question

An urn contains $3$ white and $6$ red balls. Four balls are drawn one by one with replacement from the urn. Find the probability distribution of the number of red balls drawn. Also find mean and variance of the distribution

Answer 1

$P\left(w\right)=\frac{3}{9}=\frac{1}{3}$

$P\left(R\right)=\frac{6}{9}=\frac{2}{3}$

Probability distribution of number of red balls.

$\Rightarrow$ Mean $=\sum xP\left(x\right)=nP$ ( in case of binomial distribution )

$\Rightarrow \mu =4\times \frac{2}{3}=\frac{8}{3}$

Variance $\left({\sigma }^2\right)=\sum x^{2}P\left(x\right)-{[\sum xP\left(x\right)]}^2=npq$ ( in case of binomial distribution )

$=4\times \frac{1}{3}\times \frac{2}{3}$

$=\frac{8}{9}$

Hence, the answer is $\frac{8}{9}.$