Question

A manufacturer has three machine operators A,B and C.The first operator A produces 1% defective items,whereas the other two B and C produces 5%and 7% defective items respectively.A is on the job for 50%of the time,B on the job for 30%of the time and C on the job for 20% of the time.A defective item is produced,what is the probability that it was produced by A?

Answer 1

Let E1=the item is manufactured by the operator A, E2=the item is manufactured by the operator B, E3=the term is manufactured by the operator C and A=the item is defective
Now P(E1)=$\frac{50}{100}$,P(E2)=$\frac{30}{100}$,P(E3)=$\frac{20}{100}$
Now P(A|E1)=P(item drawn is manufactured by operator A)=$\frac{1}{100}$
Similarly,P(A|E2)=$\frac{5}{100}$ and P(A|E3)=$\frac{7}{100}$
Now required probability =probability that the item is manufactured by operator A given that the item drawn is defective
$P(E_1|A)=\frac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)+P(E_3)P(A|E_3)}$
=\frac{50}{100}$$×\frac{1}{100}.\frac{50}{100}×$\frac{1}{100}$+$\frac{30}{100}$×$\frac{5}{100}$+$\frac{20}{100}$×$\frac{7}{100}$=$\frac{50}{50}$+150+140=$\frac{5}{34}$