<PRE><B>Question 269809</B>
A case of 24 cans contains 1 can that is contaminated. Three cans are to be chosen randomly for testing. 
(a) How many different combinations of 3 cans could be selected? 
Number of different combinations.
24C3 = 2024
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(b) What is the probability that the contaminated can is selected for testing? (Round your answer to 3 decimal places.)
P(bad can included) = 1 - P(bad can not included)
= 1 - [23C3/24C3]
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= 1 - [1771/2024]
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= 0.125
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Cheers,
Stan H.
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