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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.
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` |
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