A box contains 6 defective and 14 nondefective items. If 3 items are selected without replacement, what is the probability that exactly 1 is defective?
Correct answer: A. 273/760
- A. 273/760
- B. 189/760
- C. 91/380
- D. 7/20
Explanation
This is a hypergeometric probability because selection occurs without replacement. The required probability is [C(6,1)C(14,2)]/C(20,3) = 273/760.
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