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A store has 24 computers in stock of which 3 are defective. If 4 computers are sold, what is the probability that all three of the defective ones are among the ones that are sold.
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There are 3 Choose 3 ways or C(3,3) ways to choose the defective ones. There are 21 Choose 1 ways to select a good computer. So there are C(3,3)*C(21,1) = 21 ways to choose 4 computers of which exactly 3 are defective. The number of ways to select any 4 computers is 24 Choose 4 or C(24,4) = 10 626. The probability of selecting a group of 4 computers, 3 of which are defective is: 21/ 10626 = .00197 approx
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