In a five-card poker hand, what is the probability of being dealt exactly one ace and no picture cards?
The probability is?
(Round to four decimal places as needed.)
we know that total cards = 52
12 are picture cards, 4 are aces and 36 are normal 1 to 9 digit cards.
So, exactly one ace = combination(4,1) or C(4,1) (picking one ace out of 4)
no picture cards = C(12,0) (picking 0 cards out of 12)
and we have to pick 5 cards from 36 normal 1 to 9 digit cards = C(36,5)
number of ways of selecting 5 cards out of 52 = C(52,5)
therefore, required probability = [C(4,1)*C(12,0)*C(36,4)]/C(52,5)0.0907
= (4*1*58905)/(2598960)
= 0.0907
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