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?

#### Solution

Let E_{1}, E_{2}, E_{3} be the events that bulb is manufactured by machines A, B, C respectively.

E_{1}, E_{2}, E_{3} are mutually exclusive and exhaustive.

It is given that,

P(E_{1}) = 25% = `25/100`, P(E_{2}) = 35% = `35/100`, P(E_{3}) = 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`.