Maharashtra State BoardHSC Commerce 12th Board Exam
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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. - Mathematics and Statistics

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Sum

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.

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Solution

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`
Concept: Random Variables and Its Probability Distributions
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