SOLUTION: 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? b. What i

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Question 1142661: 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?
b. What is the probability that the contaminated can is selected for testing?
Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
(a)  C%5B24%5D%5E3 = %28%2824%2A23%2A22%29%2F%281%2A2%2A3%29%29 = 2024 different combinations of 3 cans.



(b)   C%5B24-1%5D%5E2 = C%5B23%5D%5E2 = %28%2823%2A22%29%2F%281%2A2%29%29 = 253 different combinations of 3 cans, containing the contaminated one.


      (You can form all possible combinations by adding any two other cans to the contaminated one).


      The probability is  253%2F2024 = 0.125.