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A factory has three machines, X,Y and Z, producing 1000,2000 and 3000 bolts per day respectively. The machine X produces 1% defective bolts, Y produces 1.5% defective bolts and Z produces 2% defective bolts. At the end of the day, a bolt is drawn at randomand it is found to be defective. What is the probability that this defective bolt has been produced by the machine X? |
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Answer» <P> Solution :Total number of bolts produced in DAY`=(1000+2000+3000)=6000`. Let `E_1,E_2and E_3` b the events of drawing a bolt produced by MACHINES X, Y and Z respectively. Then, `P(E)=1000/6000=1/6,P(E_2)=2000/6000=1/3andP(E_3)=3000/6000=1/2`. 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 X `=1/100`. `P(E//E_2)`= probability of drawing a defective bolt, given that it is produced by the machine Y `1.5/100=15/1000=3/200`. `P(E//E_3)`=probability of drawing a defective bolt, given that it is produced by the machine Z `=2/100=1/50`. Required probability `=P(E_1//E)` = probability that the bolt drawn is produced by X, given that it is defective `(P(E_1).P(E//E_1))/(P(E_1).P(E//E_1)+P(E_2).P(E//E_2)+P(E_3).P(E//E_3))` `=((1/6xx1/100))/((1/6xx1/100)+(1/3xx3/200)+(1/2xx1/50))` `=(1/600xx600/10)=1/10=0.1`. Hence, the required probability is 0.1. |
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