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

It is given that out of $30$ bulbs,$6$ are defective.
$\Rightarrow$ Probability of drawing a defective bulb,$p = \frac{6}{30} = \frac{1}{5}$.
$\Rightarrow$ Probability of drawing a non-defective bulb,$q = 1 - p = 1 - \frac{1}{5} = \frac{4}{5}$.
Since $4$ bulbs are drawn with replacement,this follows a binomial distribution $B(n, p)$ where $n = 4$ and $p = \frac{1}{5}$.
The probability of $X$ defective bulbs is given by $P(X = k) = ^{4}C_{k} \cdot (p)^{k} \cdot (q)^{4-k}$.
$P(X=0) = ^{4}C_{0} \cdot (\frac{1}{5})^{0} \cdot (\frac{4}{5})^{4} = 1 \cdot 1 \cdot \frac{256}{625} = \frac{256}{625}$.
$P(X=1) = ^{4}C_{1} \cdot (\frac{1}{5})^{1} \cdot (\frac{4}{5})^{3} = 4 \cdot \frac{1}{5} \cdot \frac{64}{125} = \frac{256}{625}$.
$P(X=2) = ^{4}C_{2} \cdot (\frac{1}{5})^{2} \cdot (\frac{4}{5})^{2} = 6 \cdot \frac{1}{25} \cdot \frac{16}{25} = \frac{96}{625}$.
$P(X=3) = ^{4}C_{3} \cdot (\frac{1}{5})^{3} \cdot (\frac{4}{5})^{1} = 4 \cdot \frac{1}{125} \cdot \frac{4}{5} = \frac{16}{625}$.
$P(X=4) = ^{4}C_{4} \cdot (\frac{1}{5})^{4} \cdot (\frac{4}{5})^{0} = 1 \cdot \frac{1}{625} \cdot 1 = \frac{1}{625}$.

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$\frac{256}{625}$	$\frac{256}{625}$	$\frac{96}{625}$	$\frac{16}{625}$	$\frac{1}{625}$