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प्रश्न
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?
उत्तर
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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