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 the 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, then calculate the probability that it is defective.

Answer 1

Let A be the event that the tube picked is defective.

\[\text{ We have } , \]
\[P\left( E_1 \right) = 50 % = \frac{50}{100} = \frac{1}{2}, P\left( E_2 \right) = 25 % = \frac{25}{100} = \frac{1}{4}, P\left( E_3 \right) = 25 % = \frac{25}{100} = \frac{1}{4}, \]
\[P\left( A| E_1 \right) = 4 % = \frac{4}{100} = \frac{1}{25}, P\left( A| E_2 \right) = 4 % = \frac{4}{100} = \frac{1}{25} and P\left( A| E_3 \right) = 5 % = \frac{5}{100} = \frac{1}{20}\]
\[\text{ Now } , \]
\[P\left( A \right) = P\left( E_1 \right) \times P\left( A| E_1 \right) + P\left( E_2 \right) \times P\left( A| E_2 \right) + P\left( E_3 \right) \times P\left( A| E_3 \right)\]
\[ = \frac{1}{2} \times \frac{1}{25} + \frac{1}{4} \times \frac{1}{25} + \frac{1}{4} \times \frac{1}{20}\]
\[ = \frac{1}{50} + \frac{1}{100} + \frac{1}{80}\]
\[ = \frac{8 + 4 + 5}{400}\]
\[ = \frac{17}{400}\]

So, the probability that the picked tube is defective is 

\[\frac{17}{400}\] .