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Question
Three machines E1, E2, E3 in a certain factory produced 50%, 25% and 25%, respectively, of the total daily output of electric tubes. It is known that 4% of the tubes produced one each of machines E1 and E2 are defective, and that 5% of those produced on E3 are defective. If one tube is picked up at random from a day’s production, calculate the probability that it is defective.
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Solution
Let D be the event that the picked up tube is defective
Let A1, A2 and A3 be the events that the tube is produced on machines E1, E2 and E3, respectively.
P(D) = P(A1) P(D|A1) + P(A2) P(D|A2) + P(A3) P(D|A3) ......(1)
P(A1) = `50/100 = 1/2`
P(A2) = `1/4`
P(A3) = `1/4`
Also P(D|A1) = P(D|A2)
= `4/100`
= `1/25`
P(D|A3) = `5/100`
= `1/20`
Putting these values in (1), we get
P(D) = `1/2 xx 1/25 + 1/4 xx 1/25 + 1/4 xx 1/20`
= `1/50 + 1/100 + 1/80`
= `17/400`
= 0.0425
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