) Suppose a company makes 40 baskets, 5 of which are defective.

If 7 baskets are chosen at random, what is the probability that at least one basket is defective?
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The probability of selecting a defective basket is 5/40=⅛, so the probability of selecting a non-defective basket is ⅞. The binomial probability distribution for a 7-basket trial is given by (⅛+⅞)⁷=1. The probability of selecting 7 OK baskets is ⅞⁷, so 1-⅞⁷ is the probability of selecting at least one defective basket. This evaluates to 0.6073 or 60.73% approx.

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