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.
[NCERT EXEMPLAR, CBSE 2015]

Answer 1

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

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

So, the probability that the picked tube is defective is $\frac{17}{400}$.