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A manufacturer has three machine operators $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$. The first operator A produces 1\% defective items, where as the other two operators $\mathrm{B}$ and $\mathrm{C}$ produce $5 \%$ and $7 \%$ defective items respectively. $A$ is on the job for $50 \%$ of the time, $B$ is on the job for $30 \%$ of the time and $\mathrm{C}$ is on the job for $20 \%$ of the time. A defective item is produced, what is the probability that it was produced by $\mathbf{A}$ ?
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Let $\mathrm{E}_1, \mathrm{E}_2, \mathrm{E}_3$ and $\mathrm{A}$ be the events defined as follows:
$\mathrm{E}_1=$ the item is manufactured by the operator $\mathrm{A}$
$\mathrm{E}_2=$ the item is manufactured by the operator B
$\mathrm{E}_3=$ the item is manufactured by the operator $\mathrm{C}$ and $\mathrm{A}=$ the item is defective $P\left(E_1\right)=\frac{50}{100}, P\left(E_2\right)=\frac{30}{100}, P\left(E_3\right)=\frac{20}{100}$ $\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_1\right)=$ Probability that the item drawn is manufactured by operator $\mathrm{A}=\frac{1}{100}$
Similarly $P\left(A \mid E_2\right)=\frac{5}{100}$ and $P\left(A \mid E_3\right)=\frac{7}{100}$
Now required probability = Probability that the item is manufactured by operator A given that the item drawnis defective $=\mathrm{P}\left(\mathrm{E}_1 \mid \mathrm{A}\right)$
$$
\begin{aligned}
&P\left(E_1 / A\right)=\frac{P\left(E_1\right) P\left(A \mid E_1\right)}{P\left(E_1\right) P\left(A \mid E_1\right)+P\left(E_2\right) P\left(A \mid E_2\right)+P\left(E_3\right) P\left(A \mid E_3\right)} \\
&=\frac{\frac{50}{100} \times \frac{1}{100}}{\frac{50}{100} \times \frac{1}{100}+\frac{30}{100} \times \frac{5}{100}+\frac{20}{100} \times \frac{7}{100}}=\frac{5}{34}
\end{aligned}
$$
$\mathrm{E}_1=$ the item is manufactured by the operator $\mathrm{A}$
$\mathrm{E}_2=$ the item is manufactured by the operator B
$\mathrm{E}_3=$ the item is manufactured by the operator $\mathrm{C}$ and $\mathrm{A}=$ the item is defective $P\left(E_1\right)=\frac{50}{100}, P\left(E_2\right)=\frac{30}{100}, P\left(E_3\right)=\frac{20}{100}$ $\mathrm{P}\left(\mathrm{A} \mid \mathrm{E}_1\right)=$ Probability that the item drawn is manufactured by operator $\mathrm{A}=\frac{1}{100}$
Similarly $P\left(A \mid E_2\right)=\frac{5}{100}$ and $P\left(A \mid E_3\right)=\frac{7}{100}$
Now required probability = Probability that the item is manufactured by operator A given that the item drawnis defective $=\mathrm{P}\left(\mathrm{E}_1 \mid \mathrm{A}\right)$
$$
\begin{aligned}
&P\left(E_1 / A\right)=\frac{P\left(E_1\right) P\left(A \mid E_1\right)}{P\left(E_1\right) P\left(A \mid E_1\right)+P\left(E_2\right) P\left(A \mid E_2\right)+P\left(E_3\right) P\left(A \mid E_3\right)} \\
&=\frac{\frac{50}{100} \times \frac{1}{100}}{\frac{50}{100} \times \frac{1}{100}+\frac{30}{100} \times \frac{5}{100}+\frac{20}{100} \times \frac{7}{100}}=\frac{5}{34}
\end{aligned}
$$
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