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If a toy factory, the machines $A, B$ and $C$ are used to manufacture $30 \%, 40 \%$ and $30 \%$ of the output, respectively. The probabilities of toys made by machines $A, B$ and $C$ to be defective are respectively $2 \%, 3 \%$ and $1 \%$. A toy is taken from the factory and is found to be defective. The probability that it was manufactured by the machine $B$ is
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$4 / 7$
$\begin{aligned} & P(A)=\frac{30}{100}, P(B)=\frac{40}{100}, P(C)=\frac{30}{100} \\ & P\left(\frac{E}{A}\right)=\frac{2}{100}, P\left(\frac{E}{B}\right)=\frac{3}{100}, P\left(\frac{E}{C}\right)=\frac{1}{100}\end{aligned}$
where $E$ denotes defective toys.
$\therefore$ Required probability $=P\left(\frac{B}{E}\right)$
$\begin{aligned} & =\frac{\frac{40}{100} \times \frac{3}{100}}{\frac{30}{100} \times \frac{2}{100}+\frac{40}{100} \times \frac{3}{100}+\frac{30}{100} \times \frac{1}{100}} \\ & =\frac{120}{60+120+30}=\frac{120}{210}=\frac{4}{7}\end{aligned}$
where $E$ denotes defective toys.
$\therefore$ Required probability $=P\left(\frac{B}{E}\right)$
$\begin{aligned} & =\frac{\frac{40}{100} \times \frac{3}{100}}{\frac{30}{100} \times \frac{2}{100}+\frac{40}{100} \times \frac{3}{100}+\frac{30}{100} \times \frac{1}{100}} \\ & =\frac{120}{60+120+30}=\frac{120}{210}=\frac{4}{7}\end{aligned}$
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