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=\(\frac{5}{34}\)