Question

In a​ five-card poker​ hand, what is the probability of being dealt exactly one ace and...

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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