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)
$\qquad=(\large\frac{1}{5})^4$
$\qquad=\large\frac{1}{625}$