Question -
Answer -
Given a lot of 30bulbs which include 6 defectives.
Then number ofnon-defective bulbs = 30 тАУ 6 = 24
As 4 bulbs are drawnat random with replacement.
Let X denotes thenumber of defective bulbs from the selected bulbs.
Clearly, X can takethe value of 0, 1, 2, 3 or 4.
P (X = 0) = P (4 arenon-defective and 0 defective)
P (X = 1) = P (3 arenon-defective and 1 defective)
P (X = 2) = P (2 arenon-defective and 2 defective)
P (X = 3) = P (1 arenon-defective and 3 defective)
P (X = 4) = P (0 arenon-defective and 4 defective)
Hence, the requiredprobability distribution is,
| X | 0 | 1 | 2 | 3 | 4 |
| P (X) | 256/625 | 256/625 | 96/625 | 16/625 | 1/625 |