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A purse contains 4 copper coins and 3 silver coins. A second purse contains 6 copper coins and 4 silver coins. A purse is chosen randomly and a coin is taken out of it. What is the probahility that it is a copper coin
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The correct answer is:
$41 / 70$
$\begin{array}{l}
\mathrm{P}_{1}: 4 \text { copper coins } \quad 3 \text { silver coins } \\
\mathrm{P}_{2}: 6 \text { copper coins } & 4 \text { silver coins } \\
\mathrm{E}=\text { Event of copper coin } \\
\mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{P}_{1}\right) . \mathrm{P}\left(\mathrm{E} / \mathrm{P}_{1}\right)+\mathrm{P}\left(\mathrm{P}_{2}\right) \cdot \mathrm{P}\left(\mathrm{E} / \mathrm{P}_{2}\right) \\
=\frac{1}{2} \times \frac{4}{7}+\frac{1}{2} \times \frac{6}{10} \\
=\frac{41}{70}
\end{array}$
\mathrm{P}_{1}: 4 \text { copper coins } \quad 3 \text { silver coins } \\
\mathrm{P}_{2}: 6 \text { copper coins } & 4 \text { silver coins } \\
\mathrm{E}=\text { Event of copper coin } \\
\mathrm{P}(\mathrm{E})=\mathrm{P}\left(\mathrm{P}_{1}\right) . \mathrm{P}\left(\mathrm{E} / \mathrm{P}_{1}\right)+\mathrm{P}\left(\mathrm{P}_{2}\right) \cdot \mathrm{P}\left(\mathrm{E} / \mathrm{P}_{2}\right) \\
=\frac{1}{2} \times \frac{4}{7}+\frac{1}{2} \times \frac{6}{10} \\
=\frac{41}{70}
\end{array}$
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