Question: In a factory, machines M1, M2 and M3 manufacture respectively, 30, 30 and 40 percent of the total output. Of their output 1, 3, and 2 percent are defective items. An item is drawn from day’s output and is found defective. What is the probability that it was manufactured by M1 by M2, by M3?

Solution:

(P) that an item is manufactured by M1 = 30/100 = .3, and the item is defective:
.3 x .01 = .003.
(P) that an item is manufactured by M2 = 30/100 = .3, and the item is defective:
.3 x .03 = .009.
(P) that an item is manufactured by M3 = 40/100 = .4, and the item is defective:
.4 x .02 = .008.
Probability of defective item = .003+.009+.008 = .02
Probability that the defective item is manufactured by M1 = .003/.02 = or 3/20.
Probability that the defective item is manufactured by M2 = .009/.02 = or 9/20.
Probability that the defective item is manufactured by M3 = .008/.02 = or 8/20.