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In a battery manufacturing factory, machines P,Q and R manufacture $20 \%, 30 \%$ and $50 \%$ respectively of the total output. The chances that a defective battery is produced by these machines are $1 \%, 1.5 \%$ and $2 \%$ respectively. If a battery is selected as random from production, then the probability that it is defective is
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Verified Answer
The correct answer is:
$\frac{33}{2000}$
Let $E_1$ be an event that the battery is produced by machine $\mathrm{P}$
$E_2$ be an event that the battery is produced by machine Q.
$E_3$ be an event that the battery is produced by machine R.
and $\mathrm{A}$ be an event that the battery is defective.
Now,
$$
\begin{aligned}
& P\left(E_1\right)=20 \%=\frac{2}{10}, \quad P\left(E_2\right)=30 \%=\frac{3}{10} \\
& P\left(E_3\right)=50 \%=\frac{5}{10} \text { and } P\left(\frac{A}{E_1}\right)=1 \%=\frac{1}{100} \\
& P\left(\frac{A}{E_2}\right)=1.5 \%=\frac{1.5}{100}, \quad P\left(\frac{A}{E_2}\right)=2 \%=\frac{2}{100}
\end{aligned}
$$
The probability that the selected battery is defective
$$
\begin{aligned}
& P(A)=P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)+P\left(E_3\right) \cdot P\left(\frac{A}{E_3}\right) \\
& =\frac{2}{10} \times \frac{1}{100}+\frac{3}{10} \times \frac{1.5}{100}+\frac{5}{10} \times \frac{2}{100}=\frac{2+4.5+10}{1000} \\
& =\frac{16.5}{1000}=\frac{165}{10 \times 1000}=\frac{165}{2 \times 1000}=\frac{33}{2000}
\end{aligned}
$$
$E_2$ be an event that the battery is produced by machine Q.
$E_3$ be an event that the battery is produced by machine R.
and $\mathrm{A}$ be an event that the battery is defective.
Now,
$$
\begin{aligned}
& P\left(E_1\right)=20 \%=\frac{2}{10}, \quad P\left(E_2\right)=30 \%=\frac{3}{10} \\
& P\left(E_3\right)=50 \%=\frac{5}{10} \text { and } P\left(\frac{A}{E_1}\right)=1 \%=\frac{1}{100} \\
& P\left(\frac{A}{E_2}\right)=1.5 \%=\frac{1.5}{100}, \quad P\left(\frac{A}{E_2}\right)=2 \%=\frac{2}{100}
\end{aligned}
$$
The probability that the selected battery is defective
$$
\begin{aligned}
& P(A)=P\left(E_1\right) P\left(\frac{A}{E_1}\right)+P\left(E_2\right) \cdot P\left(\frac{A}{E_2}\right)+P\left(E_3\right) \cdot P\left(\frac{A}{E_3}\right) \\
& =\frac{2}{10} \times \frac{1}{100}+\frac{3}{10} \times \frac{1.5}{100}+\frac{5}{10} \times \frac{2}{100}=\frac{2+4.5+10}{1000} \\
& =\frac{16.5}{1000}=\frac{165}{10 \times 1000}=\frac{165}{2 \times 1000}=\frac{33}{2000}
\end{aligned}
$$
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