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A shopkeeper buys a particular type of electric bulbs from three manufactures $M_1, M_2$ and $M_3$. He buys $25 \%$ of his requirement from $M_1, 45 \%$ from $M_2$, and $30 \%$ from $M_3$. Based on past experience he found that $2 \%$ of type $\mathrm{M}_3$ bulbs are defective, where as only $1 \%$ of type $M_1$ and type $\mathrm{M}_2$ are detective. If a bulb chosen by him at random is defective, then the probability that it was of type $M_3$ is
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Verified Answer
The correct answer is:
$\frac{6}{13}$
Let $D=$ Chosen bulbs are defective.
$$
\begin{aligned}
& \therefore \mathrm{P}\left(M_3 / D\right) \\
& =\frac{P\left(D / M_3\right) \cdot P\left(M_3\right)}{P\left(D / M_1\right) \cdot P\left(M_1\right) P\left(D / M_2\right) \cdot P\left(M_2\right)+P\left(D / M_3\right) \cdot P\left(M_3\right)} \\
& =\frac{\frac{2}{100} \times \frac{30}{100}}{\frac{1}{100} \times \frac{25}{100}+\frac{1}{100} \times \frac{45}{100}+\frac{2}{100} \times \frac{30}{100}}=\frac{6}{13}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \mathrm{P}\left(M_3 / D\right) \\
& =\frac{P\left(D / M_3\right) \cdot P\left(M_3\right)}{P\left(D / M_1\right) \cdot P\left(M_1\right) P\left(D / M_2\right) \cdot P\left(M_2\right)+P\left(D / M_3\right) \cdot P\left(M_3\right)} \\
& =\frac{\frac{2}{100} \times \frac{30}{100}}{\frac{1}{100} \times \frac{25}{100}+\frac{1}{100} \times \frac{45}{100}+\frac{2}{100} \times \frac{30}{100}}=\frac{6}{13}
\end{aligned}
$$
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