A) \[\frac{241}{1456}\]
B) \[\frac{164}{4165}\]
C) \[\frac{451}{884}\]
D) None of these
Correct Answer: B
Solution :
There are four aces and 48 other cards. Therefore the required probability \[=\frac{48\cdot 47\cdot ....\cdot 39}{52\cdot 51\cdot ....\cdot 43}.\frac{4}{42}=\frac{164}{4165}.\]
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