Question

Three machines $E_1,E_2,E_3$ in a certain factory produced 50%, 25% and 25%, respectively, of the total daily output of electric tubes. It is known that 6% of the tubes produced on each of machines $E_1$ and $E_2$ is defective and that 5% of those produced on $E_3$ is defective. If one tube is picked up at random from a day’s production, calculate the probability that it is defective.

• A. $\frac{23}{400}$
• B. $\frac{1}{10}$
• C. $\frac{3}{80}$
• D. $\frac{9}{80}$

Answer 1

The correct option is A $\frac{23}{400}$

Let D be the event that the picked up tube is defective.
Let $A_1,A_2\mathrm{and}{A}_3$ be the events that the tube is produced on machines$E_1,E_2\mathrm{and}{E}_3$ respectively.
Then, by total probability theorem
$P\left(D\right)=P\left(A_1\right)P\left(D|A_1\right)+P\left(A_2\right)P\left(D|A_2\right)+P\left(A_3\right)P\left(D|A_3\right)...\left(1\right)$
$\mathrm{Now},P\left(A_1\right)=\frac{50}{100}=\frac{1}{2},P\left(A_2\right)=\frac{1}{4}\mathrm{and}P\left(A_3\right)=\frac{1}{4}\mathrm{Given}P\left(D|A_1\right)=P\left(D|A_2\right)=\frac{6}{100}=\frac{3}{50}\mathrm{and}P\left(D|A_3\right)=\frac{5}{100}=\frac{1}{20}$
Putting these values in (1), we get
$P\left(D\right)=\frac{1}{2}\times \frac{3}{50}+\frac{1}{4}\times \frac{3}{50}+\frac{1}{4}\times \frac{1}{20}P\left(D\right)=\frac{23}{400}$