First let us look in ~ the probability that the very first card handle is an ace and also the other four non-aces...

There space #4# aces in #52# cards, therefore the probability the the an initial card dealt being an ace is:

#4/52 = 1/13#

The remaining pack is composed of #51# cards of which #3# space aces, and also #48# not. For this reason the probability the the second card being a non-ace is:

#48/51 = 16/17#

The probability that the third card gift a non-ace is:

#47/50#

The probability of the fourth card gift a non-ace is:

#46/49#

The probability that the 5th card being a non-ace is:

#45/48 = 15/16#

So the probability of one ace adhered to by #4# non-aces is:

#1/13*color(red)(cancel(color(black)(16)))/17*47/50*46/49*15/color(red)(cancel(color(black)(16))) = 32430/541450 = 3243/54145#

The other four possible sequences of ace vs non-ace which an outcome in exactly one ace gift dealt space mutually exclusive, and will have the very same probability as this an initial case.

So the probability of precisely one ace being dealt is:

#5 * 3243/54145 = 3243/10829#

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

Answer link

MathFact-orials.blogspot.com

Feb 12, 2017

#P=(778,320)/(2,598,960)~=29.9%#

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Explanation:

An alternative way to do this is to use equations because that combinations, the basic formula for which is:

#C_(n,k)=(n!)/((k)!(n-k)!)# with #n="population", k="picks"#

We want our hand come have exactly one Ace. Due to the fact that there are four aces in a deck, we can set #n=4, k=1#, and so:

#C_(4,1)#

But we also need to account because that the other 4 cards in our hand. There room 48 cards to pick from, so us can set #n=48, k=4#, and so:

#C_(48,4)#

We main point them together to discover the total variety of ways we can get precisely one Ace in our hand:

#C_(4,1)xxC_(48,4)=(4!)/((1)!(4-1)!)(48!)/((4)!(48-4)!)=>#

#(cancel(4!)(48!))/((3!)cancel(4!)(44!))=(cancel(48)^8xx47xx46xx45xxcancel44!)/(cancel(3xx2)xxcancel44!)=>#

#8xx47xx46xx45=778,320#

~~~~~

To number out the probability, we also need to understand the total variety of 5-card hand possible:

#C_(52,5)=(52!)/((5)!(52-5)!)=(52!)/((5!)(47!))#

Let"s advice it!

#(52xx51xxcancelcolor(orange)(50)^10xx49xxcancelcolor(red)48^2xxcancelcolor(brown)(47!))/(cancelcolor(orange)5xxcancelcolor(red)(4xx3xx2)xxcancelcolor(brown)(47!))=52xx51xx10xx49xx2=2,598,960#