A) \[\frac{13}{49}\]
B) \[\frac{10}{49}\]
C) \[\frac{3}{49}\]
D) \[\frac{1}{49}\]
Correct Answer: B
Solution :
Total 13 cards are present in suit of club if 3 cards are removed, then 10 cards of clubs are remaining. \[\therefore \] Probability of getting a club \[=\frac{10}{49}\]
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