In a manufacturing company, three machines A, B and C respectively produce $20 \%, 30 \%$ and $50 \%$ of the…
In a manufacturing company, three machines A, B and C respectively produce $20 \%, 30 \%$ and $50 \%$ of the total product. The defective products from A, B and C are respectively 5\%, $3 \%$ and $2 \%$. If an article produced by the company is selected at random and is found to be defective, then the probability that it is produced by machine $B$ is
$\frac{10}{29}$
$\frac{8}{29}$
$\frac{9}{29}$
$\frac{11}{29}$
Solution
Let events,
Production by machine $\mathrm{A}=E_1$
Production by machine $\mathrm{B}=E_2$
Production by Machine $\mathrm{C}=E_3$
and let defective production $=E$
Then, probabilities of production by machine
$\mathrm{A}, \mathrm{B}, \mathrm{C}$ respectively is
$
P\left(E_1\right)=\frac{20}{100}, P\left(E_2\right)=\frac{30}{100}, P\left(E_3\right)=\frac{50}{100}
$
Probability of defective production by,
Machine $\mathrm{A}$ is $=P\left(E / E_1\right)=\frac{5}{100}$
by machine $\mathrm{B}$ is $=P\left(\frac{E}{E_2}\right)=\frac{3}{100}$
by machine $\mathrm{C}$ is $=P\left(\frac{E}{E_3}\right)=\frac{2}{100}$
An article is selected at random and is found to be defective. Then, the probability that it is produce by machine $\mathrm{B}$ is
$\begin{aligned} & P\left(E_2 / E\right) \\ & =\frac{P\left(E_2\right) \cdot P\left(E / E_2\right)}{P\left(E_1\right) \cdot P\left(E / E_1\right)+P\left(E_2\right) \cdot P\left(E / E_2\right)+P\left(E_3\right) \cdot P\left(E / E_3\right)} \\ & =\frac{\frac{30}{100} \times \frac{3}{100}}{\frac{20}{100} \times \frac{5}{100}+\frac{30}{100} \times \frac{3}{100}+\frac{50}{100} \times \frac{2}{100}} \\ & =\frac{90 / 10000}{290 / 10000}=\frac{90}{290}=\frac{9}{29} .\end{aligned}$