Step 1:

Let $P(S)$ be the probability of getting a defective bulb

$P(S) =\large\frac{20}{100}=\frac{1}{5}$

Let $P(F)$ be the probability of getting a good bulb

$P(F) =\large\frac{80}{100}=\frac{4}{5}$

Clearly $X$ can take values 0,1,2,3,4

Step 2:

P(X=0)=P(FFFF)

$\qquad=\large\frac{4}{5}\times \frac{4}{5}\times \frac{4}{5}\times\frac{4}{5}$

$\qquad=\large\frac{256}{625}$

P(X=0)=P(SFFF)+P(FSFF)+P(FFSF)+P(FFFS)

$\qquad=4\times \large\frac{1}{5}\times (\large\frac{4}{5})^3$

$\qquad=\large\frac{256}{625}$

P(X=2)=P(SSFF)+P(SFSF)+P(FFSS)+P(FSFS)

$\qquad=4\times (\large\frac{1}{5})^2\times (\large\frac{4}{5})^2$

$\qquad=\large\frac{64}{625}$

P(X=3)=P(SSSF)+P(SFSS)+P(FSSS)+P(SSFS)

$\qquad=4\times (\large\frac{4}{5})\times (\large\frac{1}{5})^3$

$\qquad=\large\frac{16}{625}$

P(X=4)=P(SSSS)

$\qquad=(\large\frac{1}{5})^4$

$\qquad=\large\frac{1}{625}$