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?
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=\(\frac{5}{34}\)
Bayes’ Theorem is a part of the conditional probability that helps in finding the probability of an event, based on previous knowledge of conditions that might be related to that event.
Mathematically, Bayes’ Theorem is stated as:-
\(P(A|B)=\frac{P(B|A)P(A)}{P(B)}\)
where,
Events A and B are mutually exhaustive events.
P(A) and P(B) are the probabilities of events A and B, respectively.
P(A|B) is the conditional probability of the happening of event A, given that event B has happened.
P(B|A) is the conditional probability of the happening of event B, given that event A has already happened.
This formula confines well as long as there are only two events. However, Bayes’ Theorem is not confined to two events. Hence, for more events.
