Correct Answer - Option 4 : 41/70
Concept:
1st purse contains 4 copper and 3 silver coins, so the probability of selecting 1 copper coin from it,
Thus, P (copper coin from 1st purse) = \(\frac{^4C_1}{^7C_1}\) = \(\frac{4}{7}\)
2nd purse contains 6 copper and 4 silver coins, so the probability of selecting 1 copper coin from it,
Thus, P (copper coin from 2nd purse) = \(\frac{^6C_1}{^10C_1}\)= \(\frac{6}{{10}}\)
And, Probability of selecting any purse = \(\frac{1}{2}\)
Hence, the Probability of selecting a copper coin = \(\frac{1}{2} \times \left( {\frac{4}{7} + \frac{6}{{10}}} \right)\)
Probability of selecting a copper coin = \(\frac{{41}}{{70}}\)