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Question

Three machines in a factory produce respectively 20%, 50% and 30% of items daily. The percentage of defective items of these machines are respectively 3, 2 and 5. An item is taken at random from the production and is found to be defective, the probability that it is produced by machine A is ____.

A
6/31
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B
3/31
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C
1/31
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D
2/31
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Solution

The correct option is A 6/31
Probability of item produced by machine AP(A)
=20
Probability of item produced by machine BP(B)
=50
Probability of item produced by machine CP(C)
=30
Given the item produced by A, probability of it being defective
P(DA)=3
Given the item produced by B, probability of it being defective
P(DB)=2
Given the item produced by C, probability of it being defective
P(DC)=5
P(AD)=P(A)×P(DA)[P(A)×P(DA)]+[P(B)×P(DB)]+[P(C)×P(DC)]
=0.2×0.03[0.2×0.03]+[0.5×0.02]+[0.3×0.05]
=0.0060.006+0.01+0.015
=0.0060.031=631

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