In a bolt factory, machines A, B and C manufacture respectively [20 %, 30 %] and [50 %] of…

Question

In a bolt factory, machines $A, B$ and $C$ manufacture respectively $$20 \%, 30 \%$$ and $$50 \%$$ of the total bolts. Of their output 3, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. If the bolt drawn is found the defective, then the probability that it is manufactured by the machine $C$ is :

Accepted Answer

Given : $P(A)=\frac{20}{100}=\frac{2}{10}$

$$
P(B)=\frac{30}{100}=\frac{3}{10} $$

$$ P(C)=\frac{50}{100}=\frac{5}{10}
$$

Let $\mathrm{E} \rightarrow$ Event that the bolt is defective.

$$
\text { So, } P(E / A)=\frac{3}{100}, $$

$$P\left(\frac{E}{B}\right)=\frac{4}{100}, P\left(\frac{E}{C}\right)=\frac{2}{100}
$$

So, $\mathrm{P}(\mathrm{C} / \mathrm{E})$

$$
\begin{aligned}
& =\frac{P\left(\frac{E}{C}\right) \times P(C)}{P\left(\frac{E}{A}\right) \times P(A)+P\left(\frac{E}{B}\right) \times P(B)+P\left(\frac{E}{C}\right) \times P(C)} \\
& =\frac{\frac{5}{10} \times \frac{2}{100}}{\frac{3}{100} \times \frac{2}{10}+\frac{4}{100} \times \frac{3}{10}+\frac{2}{100} \times \frac{5}{10}} \\
& =\frac{10}{6+12+10}=\frac{10}{28}=\frac{5}{14}
\end{aligned}
$$

Solutions · 1