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?

Answer 1

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