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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 6% of the tubes produced on each of machines E1 and E2 is defective and that 5% of those produced on E3 is defective. If one tube is picked up at random from a day’s production, calculate the probability that it is defective.

A
23400
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B
110
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C
380
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D
980
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Solution

The correct option is A 23400
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 machinesE1, E2 and E3 respectively.
Then, by total probability theorem
P(D) = P(A1)P(D|A1)+P(A2)P(D|A2) + P(A3)P(D|A3) ...(1)
Now, P(A1) = 50100 = 12,P(A2) = 14 and P(A3) = 14Given P(D|A1)=P(D|A2)= 6100 = 350and P(D|A3) = 5100 = 120
Putting these values in (1), we get
P(D)= 12×350+14×350+14×120P(D)= 23400

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