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a bin contains a total of 20 batteries, of which 8 are defective. If you select 2 at random, without replacement, find the probability that at least one is good.
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For at least one good, we either have 1 good/1 bad or both good. For the former, we know the probability of the first but not the last being good is the same as \(\frac{12}{20}\times\frac8{19}\); the second being good but the first not is given by \(\frac8{20}\times\frac{12}{19}\). The probability of both being good is \(\frac{12}{20}\times\frac{11}{19}\). Sum all of these products.
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