From a lot of $\displaystyle 30$ bulbs which include $\displaystyle 6$ defective bulbs, a sample of $\displaystyle 2$ bulbs is drawn at random one by one with replacement. Find the probability distribution of the number of defective bulbs and hence find the mean number of defective bulbs.

Let X be the random variable which denotes the number of defective bulbs drawn from a sample of $\displaystyle 2$ bulbs. Here X may take values $\displaystyle 0,1$ or $\displaystyle 2$ . Let A and B be the event of drawing a defective bulb and non defective bulb respectively\begin{gathered} \mathrm{P}(\mathrm{~A})=\frac{6}{30}=\frac{1}{5} ; \mathrm{P}(\mathrm{~B})=1-\frac{1}{5}=\frac{4}{5} \text { Now } \mathrm{P}(\mathrm{X}=0)=\frac{4}{5} \times \frac{4}{5}=\frac{16}{25} \mathrm{P}(\mathrm{X}=1)=2\left[\frac{4}{5} \times \frac{1}{5}\right]=\frac{8}{25} \mathrm{P}(\mathrm{X}=2)=\frac{1}{5} \times \frac{1}{5}=\frac{1}{25} \end{gathered} $$ Thus P.D. of X is