Given: A lot of 30 bulbs which include 6 defectives.

Then number of non-defective bulbs = 30 – 6 = 24

As 4 bulbs are drawn at random with replacement.

Let X denotes the number of defective bulbs from the selected bulbs.

Clearly, X can take the value of 0, 1, 2, 3 or 4.

P(X = 0) = P(4 are non defective and 0 defective)

P(X = 1) = P(3 are non defective and 1 defective)

P(X = 2) = P(2 are non defective and 2 defective)

P(X = 3) = P(1 are non defective and 3 defective)

P(X = 4) = P(0 are non defective and 4 defective)

Hence, the required probability distribution is,