SOLUTION: in a set of 15 tools, 2 are defective and the remaining 13 are good. A QC person randomly selects 3 of the 15 tools without replacemnt and classifies each as defective or good. H
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Question 507805: in a set of 15 tools, 2 are defective and the remaining 13 are good. A QC person randomly selects 3 of the 15 tools without replacemnt and classifies each as defective or good. How many elementary events does this sample space have?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
in a set of 15 tools, 2 are defective and the remaining 13 are good. A QC person randomly selects 3 of the 15 tools without replacemnt and classifies each as defective or good. How many elementary events does this sample space have?
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The sample space is the list of all possible outcomes.
Ans: 15C3 = (15*14*13)/(1*2*3) = 455
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Cheers,
Stan H.
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