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Question
Solve the following:
In a factory which manufactures bulbs, machines A, B and C manufacture respectively 25%, 35% and 40% of the bulbs. Of their outputs, 5, 4 and 2 percent are respectively defective bulbs. A bulbs is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?
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Solution
Let E1, E2, E3 be the events that bulb is manufactured by machines A, B, C respectively.
E1, E2, E3 are mutually exclusive and exhaustive.
It is given that,
P(E1) = 25% = `25/100`, P(E2) = 35% = `35/100`, P(E3) = 40% = `40/100`
Let D ≡ the event that bulb is defective.
It is given that machines A, B, C have outputs of which 5% 4%, 2% are defective,
∴ `"P"("D"/"E"_1) = 5/100, "P"("D"/"E"_2) = 4/100, "P"("D"/"E"_3) = 2/100`
By Baye's Theorem, the required probability = `"P"("E"_2/"D")`
= `("P"("E"_2)*"P"("D"/"E"_2))/("P"("E"_1)*"P"("D"/"E"_1) + "P"("E"_2)*"P"("D"/"E"_2) + "P"("E"_3)*"P"("D"/"E"_3))`
= `((35/100)*(4/100))/((25/100)*(5/100) + (35/100)*(4/100) + (40/100)*(2/100))`
= `140/(125 + 140 + 80)`
= `140/345`
= `28/69`.
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