Solution:

Let $X$ denote the number of success. Clearly $X$ can take the values $0,1,2$ or $3\$ .

$ \text { Probability of success }=\frac{2}{6}=\frac{1}{3} ; \quad \text { Probability of failure }=1-\frac{1}{3}=\frac{2}{3} \\ $

$ P(X=0)=P(\text { no success })=P \text { (all } 3 \text { failures })=\frac{2}{3} \times \frac{2}{3} \times \frac{2}{3}=\frac{8}{27}\\ $

$ P(X=1)=P \text { (one success and } 2 \text { failures) }={ }^{3} C_{1} \times \frac{1}{3} \times \frac{2}{3} \times \frac{2}{3}=\frac{12}{27}\\ $

$ P(X=2)=P(\text { two successes and one failure })={ }^{3} C_{2} \times \frac{1}{3} \times \frac{1}{3} \times \frac{2}{3}=\frac{6}{27}\\ $

$ P(X=3)=P(\text { all } 3 \text { successes })=\frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}=\frac{1}{27}\\ $

$\therefore$

The probability distribution of the random variable $X$ is, $$ \begin{array} {ccccc}\mathrm{X}: & 0 & 1 & 2 & 3\ \mathrm{P}(\mathrm{X}): & \frac{8}{27} & \frac{12}{27} & \frac{6}{27} & \frac{1}{27}\ \end{array} $$

Mean,

$ \mu=\Sigma p_{i} x_{i}=1 $

Variance,

$\quad \sigma^{2}=\Sigma p_{i} x_{i}{ }^{2}-\mu^{2}=\frac{5}{3}-1=\frac{2}{3}$.