SOLUTION: 7% of the items produced on a machine in the past have been defective. If 15 of the items produced during a day are selected at random, calculate the probability that:
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eatleast
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Question 728408: 7% of the items produced on a machine in the past have been defective. If 15 of the items produced during a day are selected at random, calculate the probability that:
eatleast two of the items is deffective
This is my attempt: 2/15 (7%) + 3/15 (7%) = 0.093 + 0.014 = 0.107 Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 7% of the items produced on a machine in the past have been defective. If 15 of the items produced during a day are selected at random, calculate the probability that: at least two of the items are defective
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P(2<= x <=15) = 1 - P(0<= x <=1) = 1 - [(0.93)^15 + (15(0.07)(0.93)^14]
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= 1 - [0.7168]
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= 0.2832
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Cheers,
Stan H.
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