A) \[\frac{4}{7}\]
B) \[\frac{3}{4}\]
C) \[\frac{2}{7}\]
D) \[\frac{37}{56}\]
Correct Answer: D
Solution :
Let\[{{E}_{1}}\]be the event that copper coin is drawn from the bag A and\[{{E}_{2}}\]be the event that copper coin is drawn from the bag B and E be the event of drawing a copper coin. Since both the events are exhaustive. Therefore, \[P({{E}_{1}})=P({{E}_{2}})=\frac{1}{2}\] \[P\left( \frac{E}{{{E}_{1}}} \right)=\frac{4}{7},P\left( \frac{E}{{{E}_{2}}} \right)=\frac{6}{8}\] \[P(E)=P({{E}_{1}})P\left( \frac{E}{{{E}_{1}}} \right)+P({{E}_{2}})P\left( \frac{E}{{{E}_{2}}} \right)\] \[=\frac{1}{2}.\frac{4}{7}+\frac{1}{2}.\frac{6}{8}=\frac{2}{7}+\frac{3}{8}\] \[=\frac{16+21}{56}=\frac{37}{56}\]
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