Question 1152818
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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