Question

A company produces $10000$ items per day. On a particular day $2500$ items were produced on machine A, $3500$ on machine $B$ and $4000$ on machine $C$. The probability that an item produced by the machines $A, B, C$ to be defective is respectively $2 \%, 3 \%$ and $5 \%$. If one item is selected at random from the output and is found to be defective, then the probability that it was produced machine $C ,$ is

• A. $\frac{10}{71}$
• B. $\frac{16}{71}$
• C. $\frac{40}{71}$
• D. $\frac{21}{71}$

Answer 1

Answer: (C)

Total number of items produced in a day $=10000$.
Let $E_{1}, E_{2}, E_{3}$ be the events of selecting a item produced by Machines $A , B , C$ respectively. Then,
$P\left(E_{1}\right)=\frac{2500}{10000}=\frac{25}{100}=\frac{5}{20} $
$P\left(E_{2}\right)=\frac{3500}{10000}=\frac{35}{100}=\frac{7}{20} $
$P\left(E_{3}\right)=\frac{4000}{10000}=\frac{40}{100}=\frac{8}{20}$
Let $E$ be the event of selecting a defective item. Then,
$P\left(\frac{E}{E_{1}}\right)=\frac{2}{100}, P\left(\frac{E}{E_{2}}\right)=\frac{3}{100}, P\left(\frac{E}{E_{3}}\right)=\frac{5}{100} $
So, required probability $=P\left(\frac{E_{3}}{E}\right)$
$=\frac{P\left(E_{3}\right) \cdot P\left(\frac{E}{E_{3}}\right)}{P\left(E_{1}\right) \cdot P\left(\frac{E}{E_{1}}\right)+P\left(E_{2}\right) \cdot P\left(\frac{E}{E_{2}}\right)+P\left(E_{3}\right) \cdot\left(\frac{E}{E_{3}}\right)} $
$=\frac{\left(\frac{8}{20} \times \frac{5}{100}\right)}{\left(\frac{5}{20} \times \frac{2}{100}\right)+\left(\frac{7}{20} \times \frac{3}{100}\right)+\left(\frac{8}{20} \times \frac{5}{100}\right)}$
$=\frac{40}{10+21+40}=\frac{40}{71}$