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In a bolt factory, there machines A, B, C, manufature 25%,35% and 40% of the total production respectively. Of their respective outputs, 5%, 4% and 2% are defective. A bolt is drawn at random from the total product and it is found to be defective. Find the probability that it was manufactured by the machine C. |
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Answer» Solution :Let `E_1,E_2 and E_3` be the EVENTS of drawing a bolt produced by machine A, B and C RESPECTIVELY. Then, `=P(E_1)=25/100=1/4,P(E_2)=35/100=7/20,and P(E_3)=40/100=2/5`. Let E be the event of drawing a DEFECTIVE bolt. Then, `P(E//E_1)` = probability of drawing a defective bolt, given that it is produced by the machine A `=5/100=1/20`. `P(E//E_2)` = probabilityof drawing a defective bolt, given that it is produced by the machine B `=4/100=1/25`. `P(E//E-3)` =probability of drawing a defective bolt, given that it is produced by the machine C `=2/100=1/50`. Probability that the bolt drawn is manufactured by C, given that it is defective `=P(E_3//E)` `(P(E//E_3)P(E_3))/(P(E_1).P(E//E_1)+P(E_2).P(E//E_2)+P(E_3).P(E//E_3))` [by Bayes's theorem] `((1/50xx2/5))/((1/20xx1/4)+(1/25xx7/20)+(1/50xx2/5))=(1/125xx2000/69)=16/69`. Hence, the REQUIRED probability is `16/69`. |
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