Question 433452: Please help me with this difficult homework problem:
A company uses printer cartridges from three manufacturers. Manufacturers A, B, and C supply, respectively, 60%, 30%, and 10% of the company’s cartridges. 1% of cartridges from A, 2% from B, and 3% from C are defective. If a cartridge is randomly chosen and found to be defective, what is the probability that it is from manufacturer C?
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A company uses printer cartridges from three manufacturers.
Manufacturers A, B, and C supply, respectively, 60%, 30%, and 10% of the company’s cartridges.
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1% of cartridges from A,
2% from B,
3% from C are defective.
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P(d|A) = P(d and A)/P(A) = P(d and A)/0.6 = 0.01
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P(d|B) = P(d and B)/P(B) = P(d and B)/0.3 = 0.02
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P(d|C) = P(d and C)/P(C) = P(d and C)/0.1 = 0.03
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If a cartridge is randomly chosen and found to be defective, what is the probability that it is from manufacturer C?
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P(C|d) = P(C and d)/P(d)
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= P(C and d)/[P(d and A)+P(d and B)+P(d and A)]
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= [0.1*0.03)/[0.6*0.01+0.3*0.02+0.1*0.03]
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= 0.003/0.015
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= 0.2
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Cheers,
Stan H.
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Answer by MathLover1(20850)  (Show Source):
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