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Question
In a bolt factory, three machines A, B, and C manufacture 25%, 35% and 40% of the total production respectively. Of their respective outputs, 5%, 4% and 2% are defective. A bolt is drawn at random from the total production and it is found to be defective. Find the probability that it was manufactured by machine C.
Solution
| Machine A | Machine B | Machine C | |
| % of total production | 25% | 35% | 40% |
| % defective of their output | 5% | 4% | 2% |
∴ P(A) = `(25)/(100) = (1)/(4)`,
P(B) = `(35)/(100) = (7)/(20)`
and P(C) = `(40)/(100) = (2)/(5)`
P(D/A) = `(5)/(100) xx (1)/(4) = (1)/(80)`
P(D/B) = `(4)/(100) xx (7)/(20) = (7)/(500)`
P(D/C) = `(2)/(100) xx (2)/(5) = (1)/(125)`
P(C/D) = `("P"("C") ·"P"("D"/"C"))/( ("P"("A") ·"P("D"/"A") + "P"("B") ·"P"("D"/"B) +"P"("C") ·"P"("D"/"C"))`
= `((2)/(5) xx (1)/(125))/ ((1)/(4) xx (1)/(80) + (7)/(20) xx (7)/(500) + (2)/(5) xx (1)/(125))`
= `((2)/(625))/((250 + 392 + 256)/(80000))`
= `(2xx 80000)/(625 xx 898) = (128)/(449) = 0.2851`
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