A purse contains 4 copper and 3 silver coins another purse contains 6 copper and 2 silver coins. One coin is drawn from any one of these two purses. The probability that it is a copper coin, is 

Consider, the events ${\epsilon }_1,{\epsilon }_2$  and  A 

   ${\epsilon }_1=$   selection of 1st purse 

  ${\epsilon }_2=$   selection of 2nd purse 

 A   = drawn coin is copper  

  $\begin{array}{r}P\left({\epsilon }_1\right)=\frac{1}{2}\\ P\left({\epsilon }_2\right)=\frac{1}{2}\\ P\left(A/{\epsilon }_1\right)=\frac{4}{7}\\ P\left(A/{\epsilon }_2\right)=\frac{6}{8}\end{array}$

 $\therefore$     Required probability   $P\left(A\right)=P\left({\epsilon }_1\right)\times P\left(A/{\epsilon }_1\right)+P\left({\epsilon }_2\right)\times P\left(A/{\epsilon }_2\right)$

    $=\frac{1}{2}\times \frac{4}{7}+\frac{1}{2}\times \frac{6}{8}=\frac{37}{56}$

  Hence, option (d) is correct.