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Question
A company has three machines A, B, C which produces 20%, 30% and 50% of the product respectively. Their respective defective percentages are 7, 3 and 5. From these products, one is chosen and inspected. If it is defective what is the probability that it has been made by machine C?
Solution
The probability of the product produced by machine A is P(A) = `20/100`
The probability of the product produced by machine B is P(B) = `30/100`
The probability of the product produced by the machine C is P(C) = `50/100`
Let D be the event of selecting a defective product.
Then `"P"("D"/"A")` = The probability of selecting a defective product produced by the machine A = `7/100`
`"P"("D"/"B")` = The probability of selecting a defective product produced by the machine B = `9/100` and
`"P"("D"/"C")` = The probability of selecting a detective product produced by the machine C = `5/100`
`"P"("C"/"D") = ("P"("C") xx "P"("D"/"C"))/("P"("A")"P"("D"/"A") + "P"("B")"P"("D"/"B") + "P"("C")"P"("D"/"C"))`
= `(50/100 xx 5/100)/(20/100 xx 7/100 + 30/100 xx 3/100 + 50/100 xx 5/100)`
= `(50 xx 5)/(20 xx 7 + 30 xx 3 + 50 xx 5)`
= `(5 xx 5)/(2 xx 7 + 3 xx 3 + 5 xx 5)`
= `25/(14 + 9 + 25)`
= `25/48`
= 0.5208
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