JEE MAIN - Mathematics (2023 - 8th April Morning Shift - No. 5)
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 :
$$\frac{2}{7}$$
$$\frac{9}{28}$$
$$\frac{5}{14}$$
$$\frac{3}{7}$$
Explanation
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, $\mathrm{P}(\mathrm{C} / \mathrm{E})$
$$ \begin{aligned} & =\frac{P\left(\frac{E}{C}\right) \times P(C)}{P\left(\frac{E}{A}\right) \times P(A)+P\left(\frac{E}{B}\right) \times P(B)+P\left(\frac{E}{C}\right) \times P(C)} \\\\ & =\frac{\frac{5}{10} \times \frac{2}{100}}{\frac{3}{100} \times \frac{2}{10}+\frac{4}{100} \times \frac{3}{10}+\frac{2}{100} \times \frac{5}{10}} \\\\ & =\frac{10}{6+12+10}=\frac{10}{28}=\frac{5}{14} \end{aligned} $$
$$ P(B)=\frac{30}{100}=\frac{3}{10} $$
$$ P(C)=\frac{50}{100}=\frac{5}{10} $$
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, $\mathrm{P}(\mathrm{C} / \mathrm{E})$
$$ \begin{aligned} & =\frac{P\left(\frac{E}{C}\right) \times P(C)}{P\left(\frac{E}{A}\right) \times P(A)+P\left(\frac{E}{B}\right) \times P(B)+P\left(\frac{E}{C}\right) \times P(C)} \\\\ & =\frac{\frac{5}{10} \times \frac{2}{100}}{\frac{3}{100} \times \frac{2}{10}+\frac{4}{100} \times \frac{3}{10}+\frac{2}{100} \times \frac{5}{10}} \\\\ & =\frac{10}{6+12+10}=\frac{10}{28}=\frac{5}{14} \end{aligned} $$
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