Question

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 1

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

P(X₄|X₁) = P(tube produced by machine E₁ is defective)

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

⇒ P(X₄|X₁) = \(\cfrac4{100}\)

\(\therefore\) P(X₄|X₁) = \(\cfrac1{25}\)

Similarly,  P(X₄|X2) = \(\cfrac1{25}\)

We also have

P(X₄|X₁) = P(tube produced by machine E₃ is defective)

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

 ⇒ P(X₄|X3) = \(\cfrac5{100}\)

\(\therefore\) (X₄|X3) = \(\cfrac1{20}\)

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 \(\cfrac{17}{400}.\)