From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs

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»

It is given that out of 30 bulbs, 6 are defective.

Number of non-defective bulbs = 30 – 6 = 24

4 bulbs are drawn from the lot with replacement.

Let p = P(obtaining a defective bulb when a bulb is drawn) = 6/30 = 1/5

and q = P(obtaining a non-defective bulb when a bulb is drawn) = 24/30 = 4/5

Using Binomial distribution, we have

P(X = 0) = P (no defective bulb in the sample) = ⁴C₀p⁰q⁴ = (4/5)⁴ = 256/625

P(X = 1) = P (one defective bulb in the sample) = ⁴C₁p¹q³ = 4(1/5)¹(4/5)³ = 256/625

P(X = 2) = P (two defective and two non-defective bulbs are drawn) = ⁴C₂p²q² = 6(1/5)²(4/5)² = 96/625

P(X = 3) = P (three defective and one non-defective bulbs are drawn) = ⁴C₃p³q¹ = 4(1/5)³(4/5) = 16/625

P(X = 4) = P (four defective bulbs are drawn) = ⁴C₄p⁴q⁰= (1/5)⁴= 1/625

Therefore, the required probability distribution is as follows.

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