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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?
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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Mr. X goes to office by Auto, Car, and train. The probabilities him travelling by these modes are `2/7, 3/7, 2/7` respectively. The chances of him being late to the office are `1/2, 1/4, 1/4` respectively by Auto, Car, and train. On one particular day, he was late to the office. Find the probability that he travelled by car.
Solution: Let A, C and T be the events that Mr. X goes to office by Auto, Car and Train respectively. Let L be event that he is late.
Given that P(A) = `square`, P(C) = `square`
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P(L) = P(A ∩ L) + P(C ∩ L) + P(T ∩ L)
`="P"("A")*"P"("L"//"A") + "P"("C")*"P"("L"//"C") + "P"("T")*"P"("L"//"T")`
`= square * square + square * square + square * square`
`= square + square + square`
`= square`
`"P"("C"//"L") = ("P"("L" ∩ "C"))/("P"("L"))`
= `("P"("C") * "P"("L"//"C"))/("P"("L"))`
`= (square * square)/square`
`= square`
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