<PRE><B>Question 469599</B>
Suppose you just received a shipment of 13 televisions, 3 of the televisions are defective. If 2 televisions are randomly selected,
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P(defective) = 3/13 ; P(not defective) = 10/13
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Compute the probability that both televisions work
Ans: 10C2/13C2 = 0.5769
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What is the probability at lease one does not work?
Ans: P(both not work) = 1-P(both work) = 1-0.5769 = 0.4231
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Cheers,
Stan H.</PRE>