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.
let `x` be the event of defective bulb found
`p(x=0) = 4/5*4/5*4/5*4/5 =256/625`
`p(x=1) = .^4C_1(1/5)(4/5)^3 = 256/625`
`p(x=2) = .^4C_2(1/5)^2(4/5)^2 = 96/625`
`p(x=3) = .^4C_3(1/5)^3(4/5) = 16/625`
`p(x=4) = .^4C_4(1/5)^4 = 1/625`
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.