In a bolt factory, machines A, B and C manufacture $$60$$%, $$25$$%and $$15$$% respectively. Of the total of their output $$1$$%, $$2$$%, $$1$$% is defective bolts. A bolt is drawn at random from the total production and found to be defective. From which machine, the defective bolt is most expected to have been manufactured

Question

In a bolt factory, machines A, B and C manufacture $$60$$%, $$25$$%and $$15$$%  respectively. Of the total of their output $$1$$%, $$2$$%, $$1$$% is defective bolts. A  bolt is drawn at random from the total production and found to be defective.  From which machine, the defective bolt is most expected to have been  manufactured

Infinity Learn · Accepted Answer

Let $$E1$$,$$E2$$,$$E3$$     be the events of drawing a bolt produced by machine A, B  and C respectively. Let D the event of drawing a defective bolt. We have $$P(E1)=60100$$,$$P(E2)=25100$$,$$P(E3)=15100$$       $$P(D/E1)=1100$$,$$P(D/E2)=2100$$,$$P(D/E3)=1100$$      The events $$E1$$,$$E2$$     and $$E3$$     are mutually exclusive and exhaustive.$$P(defective$$ bolt is produced by $$A)=P(E1/D)=P(E1)P(D/E1)P(E1)P(D/E1)+P(E2)P(D/E2)+P(E3)P(D/E3)$$                               $$=60100$$×$$110060100$$×$$1100+25100$$×$$2100+15100$$×$$1100$$     $$=60125=1225$$     [using baye’s theorem]Similarly, $$P(defective$$ bolt is produced by $$B)=P(E2/D)=P(E2)P(D/E2)P(E1)P(D/E1)+P(E2)P(D/E2)+P(E3)P(D/E3)$$     $$=25100$$×$$210060100$$×$$1100+25100$$×$$2100+15100$$×$$1100=50125=1025$$     $$P(defective$$ bolt is produced by $$C)=P(E3/D)=P(E3)P(D/E3)P(E1)P(D/E1)+P(E2)P(D/E2)+P(E3)P(D/E3)$$                             $$=15100$$×$$110060100$$×$$1100+25100$$×$$2100+15100$$×$$1100=15125=325$$     From the above $$3$$ probabilities it is clear that the machine A has the  highest probability of producing a defective bolt.

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