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Question
A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that was produced by A?
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Solution
Let E1, E2, and E3 be the event that the machine is operated by A, B, and C, respectively.
Let A be the event of producing defective items.
∴ `P (E_1) = 50% = 1/2`
`P (E_2) = 30% = 3/10`
`P (E_3) = 20% = 1/5`
Now,
`P(A/E_1) = 1% = 1/100`
`P(A/E_2) = 5% = 5/100`
`P(A/E"_2) = 7% = 7/100`
Using Bayes' theorem, we get
Required probability = `P (E_1/A) = (P(E_1)P(A/E_1))/(P(E_1)P(A/E_1)+P(E_2)P(A/E_2)+P(E_2)P(A/E_2)`
= `(1/2 xx 1/100)/(1/2 xx 1/100 + 3/10 xx 5/100 + 1/5 xx 7/100)`
= `(1/2)/(1/2 + 15/10 + 7/5)`
= `(1/2)/((5 + 15 + 14)/10)`
= `5/34`
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