In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is (A) 10-1 (B) `(1/2)^5` (C) `(9/(10))^5` (D) `9/(10)`

Explanation

Let `X` denote the number of defective bulbs out of a sample of 5 bulbs.


P(getting a defective bulb), `p=10/100 = 1/10`


`therefore q=1 - p = 1 - 1/10 = 9/10`


Clearly, X has binomial distribution with `n=5` and `p=1/10`


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