Is the probability of this event: $$\frac{{4\choose 3}\cdot4\cdot4}{52\choose 5}$$
What is the probability of getting 3 aces, a king and a queen
0
$\begingroup$
probability
combinatorics
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0Yes, correct, and it is obvious where the numerator comes from. You might be expected to write a few words. – 2011-09-26