Posted by chinthaka on .
Two machines A and B produce respectively 60% and 40% of the total number of items of a factory. The percentages of defective output for A and B are 3% and 2% respectively. Suppose an item is selected at random, and is found to be defective. The probability that the item was produced by B, is equal to
.4*.02=.008 Pr reject and B
.6*.03=.018 Pr reject and A
What is .008/(.008+.018)=8/26 ?