cards: probability of holding all 4 aces in a 5-card hand

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Hello, i am totaly lost on this problem and have already been reduced to punching randome numbers in the calculator. Could someone please tell me how to start this problem?

What is the probability of holding all 4 aces in a 5 card hand dealt from a standard 52 card deck?

Thanks
 
\(\displaystyle \L \frac{48} {52 \choose 5}\)
WHY? Can you explain?
 
If a five-card deal has all four aces then how many ways can there be a fifth card?
 
52-4? So that would give you the 48 right?

Sorry if i sound dumb but its hard to think straight when panicking, the test on this is friday. :cry:
 
Do not panic! Just think!
Yes you are correct: 52-4=48.
 
Re: cards: probability of holding all 4 aces in a 5-card han

Hello, adon!

Okay, some baby-talk may be in order . . .

What is the probability of holding all 4 aces in a 5-card hand dealt from a standard 52-card deck?
Click to expand...
First, there are \(\displaystyle \,\begin{pmatrix}52\\5\end{pmatrix}\,= \,2,598,960\) possible hands.

Now, how many 5-card hands will contain the four Aces?

There is only 1 way to have the four Aces.
The fifth card can be any of the remaining 48 cards.
\(\displaystyle \;\;\)Hence, there are: \(\displaystyle \,1\,\times\,48\:=\:48\) hands that contain the four Aces.

Therefore: \(\displaystyle \,P(\text{4 Aces})\;=\;\L\frac{48}{2,598,960}\;=\;\frac{1}{54,145}\)
 
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