Question

Suppose an item is manufactured by three machines X, Y and Z. All the machines have equal capacity and are operated at the same rate. It is known that the percentage of defective items produced by X, Y and Z are 2, 7 and 12 percent respectively. All the items produced by X,Y and Z are put into one bin.from this bin one item is drawn at random and is found to be defective. What is the probability that this item was produced by Y?.

Answer 1

we are already given that the the item picked is defective .

probability = no of favorable outcome /total number of outcomes

And our favorable outcome is that the item is from y, i.e. p(y).

Thus, p(y)= (no of defectives from y)/(total number of defective items)

p(y)= 7/( 2+7+12)

p(y) = 7/21 = 1/3

p(y)= 1/3 = 0.3333