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2 bulbs out of a sample of 10 bulbs manufactured by a company are defective. The probability that 3 out of 4 bulbs bought by a customer will not be defective is
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For Binomial distribution to be used , these conditions should be followed :

a) In such trial we have either success or failure ..

b) Sampling should be from infinite population(here it is referred to no of bulbs)

c) Sampling from finite population but with replacement

So here we cannot use binomial distribution as none of b) and c) is satisfied..As no of bulbs is finite and nothing is given so we consider without replacement in such scenario by default..

So we use hypergeometric distribution here..Let me give an idea about this

We have 8 non defective bulbs ..But in outcome we require 3 such bulbs..So this can be done in 8C3  ways..And similarly for defective bulbs we have 2C1  ways..

So favorable outcomes of choosing desired bulbs =  8C3  2C1  ways

And we can choose 4 bulbs out of 10 bulbs in 10C4 ways which is the total outcome number..

Hence P(3 non defective out of 4 bulbs chosen)  =  (8C3  2C1) / 10C4

                                                                                               =  8 / 15

Hence 8 / 15 is the correct answer..

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