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Question
From a lot of 15 bulbs which include 5 defectives, a sample of 4 bulbs is drawn one by one with replacement. Find the probability distribution of number of defective bulbs. Hence find the mean of the distribution.
Solution
Total number of bulbs = 15
Number of defective bulbs, i.e., n(D) = 5
`∴ P(D) = 5/15=1/3`
Number of non-defective bulbs, n(ND) = 10
`∴ P(ND) = 10/15=2/3`
Let X be a random variable that shows the number of defective bulbs in a draw of 4 bulbs.
Clearly, X can take the values 0, 1, 2, 3 and 4.
P(X = 0) = P (no defective bulbs)
` = (2/3)^4=16/81`
P(X = 1) = P (1 defective and 3 non-defective bulbs)
`= 4 xx1/3xx(2/3)^3=32/81`
P(X = 2) = P (2 defective and 2 non-defective bulbs)
`= 6xx(1/3)^2xx(2/3)^2=24/81`
P(X = 3) = P (3 defective and 1 non-defective bulb)
` = 4xx (1/3)^3xx 2/3=8/81`
P(X = 4) = P (All bulbs are defective)
` = (1/3)^4=1/81`
Now, probability distribution is given by
| X | 0 | 1 | 2 | 3 | 4 |
| pi |
`16/81` |
`32/81` |
`24/81` |
`8/81` |
`1/81` |
`∴ Mean = ∑_i x_i p_i = 0xx16/81 + 1xx32/81 + 2xx24/81 + 3xx8/81 + 4xx1/81`
`= 0 + 32/81 + 48/81 + 24/81 + 4/81 `
`=108/81=4/3`
Hence, the mean of the distribution is `4/3.`
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