Question

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.

Answer 1

It is given that out of 30 bulbs, 6 are defective. Then the remaining bulbs are non-defective. So, the number of non-defective bulbs is 30−6=24.

Since, four bulbs are drawn from the replaced lot, then let the random variable X denote the number of defective bulbs in the selected bulbs.

P( X=0 )=P( 4 non-defective and 0 defective bulbs ) = C 4 0 × 4 5 × 4 5 × 4 5 × 4 5 =1× 256 625 = 256 625

P( X=1 )=P( 3 non-defective and 1 defective bulbs ) = C 4 1 × 1 5 × 4 5 × 4 5 × 4 5 =4× 64 625 = 256 625

P( X=2 )=P( 2 non-defective and 2 defective bulbs ) = C 4 2 × 1 5 × 1 5 × 4 5 × 4 5 =6× 16 625 = 96 625

P( X=3 )=P( 1 non-defective and 3 defective bulbs ) = C 4 3 × 1 5 × 1 5 × 1 5 × 4 5 =4× 4 625 = 16 625

P( X=4 )=P( 0 non-defective and 4 defective bulbs ) = C 4 4 × 1 5 × 1 5 × 1 5 × 1 5 =1× 1 625 = 1 625

Thus, the probability distribution is as follows:

X	0	1	2	3	4
P( X )	256 625	256 625	96 625	16 625	1 625