Question

A sample of 4 bulbs is drawn at random with replacement from a lot of 30 bulbs which includes 6 defective bulbs. Find the probability distribution of the number of defective bulbs.

Answer 1

Let X denote the number of defective bulbs.

∴ Possible values of X are 0, 1, 2, 3, 4.

Let P(getting a defective bulb) = p = `(6)/(30) = (1)/(5)`

∴ q = 1 – p = `1 - (1)/(5) = (4)/(5)`

∴ P(X = 0) = P(no defective bulb)

= qqqq = q⁴ = `(4/5)^4`

P(X = 1) = P(one defective bulb)

= qqqp + qqpq + qpqq + pqqq

= 4pq³

= `4 xx (1)/(5) xx (4/5)^3 = (4/5)^4`

P(X = 2) = P(two defective bulbs)

= ppqq + pqqp + qqpp + pqpq + qpqp + qppq

= 6p²q²

= `6(4/5)^2(1/5)^2`

P(X = 3) = P(three defective bulbs)

= pppq + ppqp + pqpp + qppp

= 4qp³

= `4(4/5)(1/5)^3`

P(X = 4) = P(four defective bulbs)

= pppp = p⁴ = `(1/5)^4`

∴ Probability distribution of X is as follows:

X	0	1	2	3	4
P(X = x)	`(4/5)^4`	`(4/5)^4`	`6(4/5)^2(1/5)^2`	`4(4/5)(1/5)^3`	`(1/5)^4`