Question #182080

Three different machines A1, A2, A3 were used for producing a large batch of similar manufactured items. Suppose that 10% of items are produced by machine A1, 20% by A2, and 40 % by A3. Suppose further that 1% of the items produced by A1, 2% by A2, and 3% by A3 machine are defective. Suppose one item is selected from the entire batch and it is found to be defective. Calculate the probability that this item was produced by machine A2.


Expert's answer

Three different machines A1,A2,A3A_1,A_2,A_3 were used for producing a large batch of similar manufactured items. Suppose that 40% of items are produced by machine A1,A_1, 20% by A2,A_2, and 40 % by A3.A_3. Suppose further that 1% of the items produced by A1,A_1, 2% by A2,A_2, and 3% by A3A_3 machine are defective. Suppose one item is selected from the entire batch and it is found to be defective. Calculate the probability that this item was produced by machine A2.A_2.


Let D={item is defective}D=\{item\ is \ defective\}

A1={item is produces by machine A1}A_1=\{item\ is \ produces\ by\ machine\ A_1\}

A2={item is produces by machine A2}A_2=\{item\ is \ produces\ by\ machine\ A_2\}

A3={item is produces by machine A3}A_3=\{item\ is \ produces\ by\ machine\ A_3\}

We know that P(A1)=0.4,P(A2)=0.2,P(A_1)=0.4,P(A_2)=0.2, and P(A3)=0.4.P(A_3)=0.4.

Also

P(D∣A1)=0.01,P(D∣A2)=0.02,P(D∣A3)=0.02P(D|A_1)=0.01,P(D|A_2)=0.02, P(D|A_3)=0.02

By the Bayes’ theorem


P(A2∣D)P(A_2|D)

=P(A2)P(D∣A2)P(A1)P(D∣A1)+P(A2)P(D∣A2)+P(A3)P(D∣A3)=\dfrac{P(A_2)P(D|A_2)}{P(A_1)P(D|A_1)+P(A_2)P(D|A_2)+P(A_3)P(D|A_3)}


=0.2(0.02)0.4(0.01)+0.2(0.02)+0.4(0.03)=0.2=\dfrac{0.2(0.02)}{0.4(0.01)+0.2(0.02)+0.4(0.03)}=0.2

The probability that this item was produced by machine A2A_2 is 0.2.0.2.



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