Q. 134.5( 4 Votes )

# Three machines E_{1},E_{2},E_{3} 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 E_{1} and E_{2} are defective, and that 5% of those produced on E_{3} 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 E_{1} produces 50% of the total output.

Machine E_{2} produces 25% of the total output.

Machine E_{3} produces 25% of the total output.

4% of the tubes produced by machine E_{1} are defective.

4% of the tubes produced by machine E_{2} are defective.

5% of the tubes produced by machine E_{3} 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 E_{1}, and then, the tube is defective

b. Tube was produced by machine E_{2}, and then, the tube is defective

c. Tube was produced by machine E_{3}, and then, the tube is defective

Let X_{1} be the event that the tube is produced by machine E_{1}, X_{2} be the event that the tube is produced by machine E_{2} and X_{3} be the event that the tube is produced by machine E_{3}.

As 50% of the total output is produced by machine E_{1}, we have

Similarly, as each of machines E_{2} and E_{3} produces 25% of the total tubes, we have

Let X_{4} denote the event that the tube is defective.

Hence, we have

4% of the tubes produced by machine E_{1} are defective.

Similarly,

We also have

5% of the tubes produced by machine E_{3} are defective.

Using the theorem of total probability, we get

P(X_{4}) = P(X_{1})P(X_{4}|X_{1}) + P(X_{2})P(X_{4}|X_{2}) + P(X_{3})P(X_{4}|X_{3})

Thus, the probability of the tube being defective is.

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