Answer to Question #182080 in Statistics and Probability for amna rao

Question

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.

Answer 1

Three different machines $A_1,A_2,A_3$ were used for producing a large batch of similar manufactured items. Suppose that 40% of items are produced by machine $A_1,$ 20% by $A_2,$ and 40 % by $A_3.$ Suppose further that 1% of the items produced by $A_1,$ 2% by $A_2,$ and 3% by $A_3$ 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 $A_2.$

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

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

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

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

We know that $P(A_1)=0.4,P(A_2)=0.2,$ and $P(A_3)=0.4.$

Also

$P(D|A_1)=0.01,P(D|A_2)=0.02, P(D|A_3)=0.02$

By the Bayes’ theorem

$$P(A_2|D)$$

$$=\dfrac{P(A_2)P(D|A_2)}{P(A_1)P(D|A_1)+P(A_2)P(D|A_2)+P(A_3)P(D|A_3)}$$

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

The probability that this item was produced by machine $A_2$ is $0.2.$