Let X denote the random variable, which represents the number of defective bulbs.
∴ X can assumes the values 0, 1, 2, 3, 4.
Total number of bulbs = 30
Number of defective bulbs = 6.
p = ‘probability of not getting defective bulb’ =

24

30

q = ‘probability of getting defective bulb’ =

6

30

∴ P(X = 0) = P(No defective bulb) = p × p × p × p
=

24

30

×

24

30

×

24

30

×

24

30

=

256

625

P(X = 1) = P(1 defective bulb) = 4(p × p × p × q)
=4(

6

30

×

24

30

×

24

30

×

24

30

) =

256

625

P(X = 2) = P(2 defective bulbs) = 6(p × p × q × q)
=6(

6

30

×

6

30

×

24

30

×

24

30

) =

96

625

P(X = 3) = P(3 defective bulbs) = 4(q × q × q × p)
=4(

6

30

×

6

30

×

6

30

×

24

30

) =

16

625

P(X = 4) = P(4 defective bulbs) = q × q × q × q
=

6

30

×

6

30

×

6

30

×

6

30

=

1

625

Hence, the probability distribution is :