Q. 134.5( 4 Votes )

Three machines E1,E2,E3 in a certain factory produce 50%, 25% and 25% respectively, of the total daily output of electric bulbs. 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.

Answer :

Given:


Machine E1 produces 50% of the total output.


Machine E2 produces 25% of the total output.


Machine E3 produces 25% of the total output.


4% of the tubes produced by machine E1 are defective.


4% of the tubes produced by machine E2 are defective.


5% of the tubes produced by machine E3 are defective.


There are three mutually exclusive ways to pick up a defective tube produced by one of the three machines –


a. Tube was produced by machine E1, and then, the tube is defective


b. Tube was produced by machine E2, and then, the tube is defective


c. Tube was produced by machine E3, and then, the tube is defective


Let X1 be the event that the tube is produced by machine E1, X2 be the event that the tube is produced by machine E2 and X3 be the event that the tube is produced by machine E3.


As 50% of the total output is produced by machine E1, we have




Similarly, as each of machines E2 and E3 produces 25% of the total tubes, we have




Let X4 denote the event that the tube is defective.


Hence, we have



4% of the tubes produced by machine E1 are defective.



Similarly,


We also have



5% of the tubes produced by machine E3 are defective.



Using the theorem of total probability, we get


P(X4) = P(X1)P(X4|X1) + P(X2)P(X4|X2) + P(X3)P(X4|X3)








Thus, the probability of the tube being defective is.


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