A bin contains 20 batteries, 6 are defective. If you select two at random determine the probability:

a) None of the batteries are good

b) At least one of the batteries is good.

14 batteries are good

(a) selection of defective battery can be made by = 6c2 ways

6c2 = 6!/[4!*2!] = 15

And number of ways to select 2 batteries = 20c2 = 20!/[18!*2!] = 190

Hence probability = 15/190 = 3/38

(b) if none of the battery is good then probability = 3/38

hence for at least one good battery = 1-3/38 =35/38

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