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   

detailed solution

Correct option is A

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.