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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