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Q.

A manufacturer has three machine operators A, B and C. The first operator A produces1% of defective items, whereas the other two operators B and C produces 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job 30% of the time and C on the job for 20% of the time. All the items are put into one stock file and then one item is chosen at random from this and is found to be defective. What ist he probability that it was produced by A?

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Detailed Solution

Let,

E1= The event that machine is operated by A,

E2= The event that machine is operated by B,

E3= The event that machine is operated by C,

A= the event of producing defective items

P(E1)=50%=12      P(E2)=30%=310     P(E3)=20%=15

Now,

PA/E1=1%=1100 PA/E2=5%=5100 PA/E2=7%=7100

Using the Bayes’ theorem, we get

Required probability =PE1/A=P(E1)PA/E1PE1PA/E1+PE2PA/E2++PE2PA/E2                               =12×110012×1100+310×5100+15×7100                               =534

Therefore, the probability of the items are put into one stock pile and then one item is chosen at random from this and is found to be defective and that it was produced by A is 534.

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