Q. 135.0( 1 Vote )

A manufacturer has three machine operators A,B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 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 :

Let us assume U1, U2, U3 and A be the events as follows:


U1 = choosing Operator A


U2 = choosing Operator B


U3 = choosing Operator C


A = Producing defective item


From the problem:





P(A|U1) = P(Producing defective item by operator A)



P(A|U2) = P(Producing defective item by operator B)



P(A|U3) = P(Producing defective item by operator C)



Now we find


P(U1|A) = P(The defective item is produced by operator A)


Using Baye’s theorem:






The required probability is .


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