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 $=P\left(\frac{E_3}{E}\right)$
$=\frac{P\left(E_3\right)\cdot P\left(\frac{E}{E_3}\right)}{P\left(E_1\right)\cdot P\left(\frac{E}{E_1}\right)+P\left(E_2\right)\cdot P\left(\frac{E}{E_2}\right)+P\left(E_3\right)\cdot \left(\frac{E}{E_3}\right)}$
$=\frac{\left(\frac{8}{20}\times \frac{5}{100}\right)}{\left(\frac{5}{20}\times \frac{2}{100}\right)+\left(\frac{7}{20}\times \frac{3}{100}\right)+\left(\frac{8}{20}\times \frac{5}{100}\right)}$
$=\frac{40}{10+21+40}=\frac{40}{71}$