The means to select 1 of 4 aces * the method to choose 4 cards from the staying non-aces, split by 52 pick 5 (total)?

I understand it"s the full of the probcapacity of the event over the probcapacity of the total, yet I"m not sure around the peak part.

You are watching: What is the probability that a five-card poker hand contains exactly one ace

Do you desire the probability of $exactly$ one ace? Or simply the probcapacity of an ace showing up in a 5-card hand?

If you desire precisely one ace, then your answer is correct. $inom525$ is the variety of 5-card hands in the deck, and also you have 4 options for which ace to include (hence, $inom41$), and 48 select 4 choices for the various other 4 cards (for this reason, $inom484$).

If, rather, you desire the probability of at leastern one ace showing up in a 5-card hand also, we perform points differently. The simplest answer is to uncover the probcapacity of getting $no$ aces in a 5-card hand.

This probcapacity is $$fracinom485inom525,$$ for we have actually 48 select 5 feasible hands with no aces.

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Then the solution to the problem - that is, the probcapability of at leastern one ace appearing in a 5-card hand also - is one minus the complement: $$ 1 - fracinom485inom525.$$

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edited May 21 "14 at 13:33

answered May 21 "14 at 13:26

Alex K.Alex K.

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Seems to me that both answers are correct.The first strategy is simply written on a variant of the hypergeometric circulation pmf formular.enter photo description here

If I"m not mistaken this develop is equel to the one suggested above:enter photo description here

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answered Sep 24 "16 at 21:51

SimonSimon

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I should perhaps note that the embraced solution appears incorrect.

The circulation of the variety of ace cards in a 5-card hand is a problem in sampling *without replacement* ... which leads to a Hypergeometric circulation.