From alot of 30 bulbs which include 6 defective, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

.: Let $X$ denote the random variable which represents the number of defective bulbs.

$\therefore$ $X$ can assume the values 0, 1, 2, 3, 4

Total number of bulbs $= 30$

Number of defective bulbs $= 6$

$p =$ probability of not getting defective bulb $= \cfrac{{24}}{{30}}$

$q =$ probability of getting defective bulb $= \cfrac{6}{{30}}$

$P(X = 0) = P$ (No defective bulb) $= p \times p \times p \times p$

$= \cfrac{{24}}{{30}} \times \cfrac{{24}}{{30}} \times \cfrac{{24}}{{30}} \times \cfrac{{24}}{{30}} = \cfrac{4}{5} \times \cfrac{4}{5} \times \cfrac{4}{5} \times \cfrac{4}{5} = \cfrac{{256}}{{625}}$

$P(X = 1) = P(1{\rm{ defective bulb }}) = 4(p \times p \times p \times q)$

$= 4\left( {\cfrac{6}{{30}} \times \cfrac{{24}}{{30}} \times \cfrac{{24}}{{30}} \times \cfrac{{24}}{{30}}} \right) = 4\left( {\cfrac{1}{5} \times \cfrac{4}{5} \times \cfrac{4}{5} \times \cfrac{4}{5}} \right) = \cfrac{{256}}{{625}}$

$P(X = 2) = P(2{\rm{ defective bulbs }}) = 6(p \times p \times q \times q)$

$= 6\left( {\cfrac{6}{{30}} \times \cfrac{6}{{30}} \times \cfrac{{24}}{{30}} \times \cfrac{{24}}{{30}}} \right) = 6\left( {\cfrac{1}{5} \times \cfrac{1}{5} \times \cfrac{4}{5} \times \cfrac{4}{5}} \right) = \cfrac{{96}}{{625}}$

$P(X = 3) = P(3{\rm{ defective bulbs }}) = 4(q \times q \times q \times p)$

$= 4\left( {\cfrac{6}{{30}} \times \cfrac{6}{{30}} \times \cfrac{6}{{30}} \times \cfrac{{24}}{{30}}} \right) = 4\left( {\cfrac{1}{5} \times \cfrac{1}{5} \times \cfrac{1}{5} \times \cfrac{4}{5}} \right) = \cfrac{{16}}{{625}}$

$P(X = 4) = P(4{\rm{ defective bulbs }}) = q \times q \times q \times q$

$= \cfrac{6}{{30}} \times \cfrac{6}{{30}} \times \cfrac{6}{{30}} \times \cfrac{6}{{30}} = \cfrac{1}{5} \times \cfrac{1}{5} \times \cfrac{1}{5} \times \cfrac{1}{5} = \cfrac{1}{{625}}$

Hence, the probability distribution is: