Three different machines M1, M2, and M3 were used for producing a large batch of

Question

Three different machines M1, M2, and M3 were used for producing a large batch of similar manufactured items. Suppose that 20% of the items were produced by machine M1, 30% by machine M2, and the rest by machine M3. Suppose further that 1% of the items produced by machine M1 are defective, 2% are defective from M2, and 3% are defective from M3. Suppose that one item is selected at random from the entire batch, and it is found that the item is defective. What are the probabilities that the item was produced by each of the three machines?

Explanation / Answer

defective parts from machine x are

.5 * .03 = .015
defective parts from machine y are

.3 * .04 = .012
defective parts from machine z are

.2 * .05 = .01

total defective parts from all 3 machines = .015 + .012 + .01 = .037

if a part is found to be defective, then:

probability that it comes from machine x = .015 / .037 = .4054
probability that it comes from machine y = .012 / .037 = .3243
probability that it comes from machine z = .01 / .037 = .2702

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