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`(1)/(13)``(12)/(13)``(4)/(13)`none of these

Answer :

ASolution :

Clearly, <br> Required probability <br> =Probability of getting 3 non-ace cards in first three draws and getting an ace in fourth draw <br> `=(.^(12)C_(3))/(.^(13)C_(3))xx(1)/(10)=(1)/(13)`