A) 3/7
B) 4/7
C) 3/4
D) 37/56
Correct Answer: D
Solution :
[d]: Let \[{{E}_{1}}=\] Event of choosing purse I \[{{E}_{2}}=\] Event of choosing purse II A = Event that the coin drawn is of copper. \[\therefore \]\[P({{E}_{1}})=\frac{1}{2}=P({{E}_{2}})\] \[\therefore \]\[P(A/{{E}_{1}})=\frac{4}{7},P(A/{{E}_{2}})=\frac{3}{4}\] By theorem of total probability, \[P(A)=P({{E}_{1}})P(A/{{E}_{1}})+P({{E}_{2}})P(A/{{E}_{2}})\] \[=\frac{1}{2}\times \frac{4}{7}+\frac{1}{2}\times \frac{3}{4}=\frac{37}{56}\]You need to login to perform this action.
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