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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?

Answer»

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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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