Q.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.

The lot has 30 bulbs with 6 defective, so on each draw (with replacement, independent) $p = 6/30 = 1/5$ is defective and $q=4/5$ is good. With 4 draws, $X\sim Binomial(4, 1/5)$: $P(X=0)=q^4=256/625$; $P(X=1)=\binom41 pq^3=256/625$; $P(X=2)=\binom42 p^2q^2=96/625$; $P(X=3)=\binom43 p^3q=16/625$; $P(X=4) …

Because sampling is with replacement, draws are independent Bernoulli trials with p = 6/30 …

Treating the sampling as without replacement (which would need a hypergeometric distribution instead) when t …