First let united state look in ~ the probability the the very first card handle is one ace and also the other 4 non-aces...

There are #4# aces in #52# cards, therefore the probability of 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 #48# not. So the probability that the second card being a non-ace is:

#48/51 = 16/17#

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

#47/50#

The probability the the 4th card gift a non-ace is:

#46/49#

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

#45/48 = 15/16#

So the probability of an 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 room mutually exclusive, and will have actually the exact same probability as this an initial case.

So the probability of precisely one ace gift dealt is:

#5 * 3243/54145 = 3243/10829#

You are watching: What is the probability that a five-card poker hand contains exactly one ace? MathFact-orials.blogspot.com
Feb 12, 2017

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

See more: Una Hoja Membretada En Ingles, Formato De La Nota De Entrega En Inglés

Explanation:

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

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

We desire our hand to have exactly one Ace. Because 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 four cards in our hand. There space 48 cards to choose from, so we can set #n=48, k=4#, and so:

#C_(48,4)#

We main point them together to discover the total number of ways we have the right to get precisely one Ace in ours 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 additionally need to understand the total number of 5-card hands possible:

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

Let"s evaluate it!

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