Question 1137982: a stockist has 20 items in a slot out of which 12 are non defective and 8 defective. A customer selects 3 items from the slot. (i) what is the probability that all two three items are non defective. (ii) what is the probability that out of these three, two are non defective and one is defective?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! all 3 non-defective is 12C3/20C3, number of ways to choose 3 non-defective from 12 and 20.
That is 0.1930
or, probability of first is 12/20, then 11/19 then 10/18, and that product is the same
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2 are non-defective 3 ways to have that happen so 3*12C2/20C3=0.1737
or 3 ways to have a numerator of 12*11*11/20*19*18
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