Question 1152818: When purchasing bulk orders of batteries, a toy manufacturer uses this acceptance sampling plan: Randomly select and test 37 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 3 batteries do not meet specifications. A shipment contains 5000 batteries, and 2% of them do not meet specifications. What is the probability that this whole shipment will be accepted? Will almost all such shipments be accepted, or will many be rejected?
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! You calculate the probability(P) that one is defective, two are defective, three are defective and the probability that none are defective and add them together.
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P(that none are defective) = (0.98)^37 = 0.4735 (0.98 is the probability that any individual battery is not defective and you need this to happen 37 times in a row, so ^37)
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(0.02)*(0.98)^36 represents the probability that a specific battery will be the only defective battery. You then have to multiply by 37, because there are 37 possible batteries that could be defective, so you get
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((0.02)*(0.98)^36)*37 = 0.3576
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Two and three defective batteries follow a similar procedure
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Note 37C2 = 37!/(2! * (37-2)!) = 666
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((0.02)^2*(0.98)^35)*666 = 0.1340
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Note 37C3 = 37!/(3! * (37-3)!) = 7770
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((0.02)^3*(0.98)^34)*7770 = 0.0313
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Therefore, 0.4735 + 0.3576 + 0.1340 + 0.0313 = 0.9964
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The probability shows that about 99.64% of all shipments will be accepted
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