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Question
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.
Solution 1
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 X be the random variable that denotes the number of defective bulbs in the selected bulbs.
∴ X can take the value 0, 1, 2, 3, 4.
∴ P (X = 0) = P (4 non-defective and 0 defective) = `""^4C_0 xx 4/5 xx 4/5 xx 4/5 xx 4/5 = 256/625`
P (X = 1) = P (3 non-defective and 1defective) =`""^4C_1 xx (1/5) xx (4/5)^3 = 256/625`
P (X = 2) = P (2 non-defective and 2defective) =`""^4C_2 xx (1/5)^2 xx (4/5)^2= 96/625`
P (X = 3) = P (1 non-defective and 3defective) =`""^4C_3 xx (1/5)^3xx (4/5) = 16/625`
P (X = 4) = P (0 non-defective and 4defective) =`""^4C_4 xx (1/5)^4xx (4/5)^0 = 1/625`
Therefore, the required probability distribution is as follows.
| X = x | 0 | 1 | 2 | 3 | 4 |
| P (X = x) | `256/625` | `256/625` | `96/625` | `16/625` | `1/625` |
Solution 2
Here, the number of defective bulbs is the random variable.
Let the number of defective bulbs be denoted by X.
∴ X can take the value 0, 1, 2, 3, 4.
Since the draws are done with replacement, therefore the four draws are independent experiments.
Total number of bulbs is 30 which include 6 defectives.
∴ P (X = 0) = P (0) = P (all 4 non-defective bulbs)
`= 24/30 xx 24/30xx 24/30 xx 24/30 = 256/625`
P (X = 1) = P (1) = P (1 defective and 3 non-defective bulbs)
`= 6/30 xx 24/30 xx 24/30 xx 24/30 + 24/30 xx 24/30 xx 6/30 xx 24/30 + 24/30 xx 24/30 xx 6/30 xx 24/30 + 24/30 xx 24/30 xx 24/30 xx 24/30 xx 6/30 = 256/ 625`
P (X = 2) = P(2) = P (2 defective and 2 non-defective)
`= 6/30 xx 6/30 xx 24/30 xx 24/30 + 24/30 xx 6/30 xx 6/30 xx 24/30 + 24/30 xx 24/30 xx 6/30 xx6/30 + 6/30 xx 24/30 xx 6/30 xx24/30 + 6/30 xx 24/30 xx 24/30 xx 6/30 + 24/30 xx 6/30 xx 24/30 xx 6/30 = 96/625`
P (X = 3) = P (3) = P (3 defectives and 1 non-defective)
`= 6/30 xx 6/30 xx 6/30 xx 24/30 + 6/30 xx 6/30 xx ***24/30 xx6/30 + 6/30 xx 24/30 xx 6/30 xx6/30 + 24/30 xx6/30 xx6/30 xx6/30`
`= 16/625`
P (X = 4) = P (4) = P (all 4 defectives)
= `6/30 xx 6/30 xx 6/30 xx 6/30 = 1/625`
∴ The required probability distribution is
| X = x | 0 | 1 | 2 | 3 | 4 |
| P (X = x) | `(256)/625` | `256/625` | `96/625` | `16/625` | `1/625` |
Solution 3
Let X denotes 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`
= `256/625`
P(X = 1) = P(one defective bulb)
= qqqp + qqpq + qpqq + pqqq
= 4pq3
= `4 xx 1/5 xx (4/5)^3`
= `256/625`
P(X = 2) = P(two defective bulbs)
= ppqq + pqqp + qqpp + pqpq + qpqp + qppq
= 6p2q2
= `6(1/5)^2 (4/5)^2`
= `96/625`
P(X = 3) = P(three defective bulbs)
= pppq + ppqp + pqpp + qppp
= 4qp3
= `4(4/5)(1/5)^3`
= `16/625`
P(X = 4) = P(four defective bulbs)
= pppp
= p4
= `(1/5)^4`
= `1/625`
∴ Probability distribution of X is as follows:
| X | 0 | 1 | 2 | 3 | 4 |
| P(X = x) | `256/625` | `256/625` | `96/625` | `16/625` | `1/625` |
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