Is it $\cfrac{\binom{4}{1} \cdot \binom{48}{4}}{\binom{52}{5}}$?
The way to choose 1 of 4 aces * the way to choose 4 cards from the remaining non-aces, divided by 52 choose 5 (total)?
I know it's the total of the probability of the event over the probability of the total, but I'm not sure about the top part.
On
Seems to me that both answers are correct. The first method is just written on a variant of the hypergeometric distribution pmf formular.enter image description here
If I'm not mistaken this form is equel to the one suggested above:enter image description here
On
I should perhaps note that the accepted solution appears incorrect.
The distribution of the number of ace cards in a 5-card hand is a problem in sampling without replacement ... which leads to a Hypergeometric distribution.
Imagine an urn contains a total of $T$ balls, $r$ of which are red $(r<T)$. The experiment proceeds by drawing one-by-one a sample of $n$ balls from the urn without replacement $(n<T)$. Interest lies in determining the pmf of $X$, where $X$ is the number of red balls drawn. Then, the pmf of $X$, $P(X=x)$ is:
$$f(x)=\frac{\left( \begin{array}{c} n \\ x \\ \end{array} \right) \left( \begin{array}{c} T-n \\ r-x \\ \end{array} \right)}{\left( \begin{array}{c} T \\ r \\ \end{array} \right)}$$
In our example, the 'urn' is the deck of $T=52$ playing cards, and the 'red balls' are the ace cards, so $r=4$. There are $n=5$ cards in a hand. Therefore, the pmf of the number of ace cards in a hand of 5 cards is given by:
$$f(x)=\frac{\left( \begin{array}{c} 5 \\ x \\ \end{array} \right) \left( \begin{array}{c} 47\\ 4-x \\ \end{array} \right)}{\left( \begin{array}{c} 52 \\ 4 \\ \end{array} \right)}$$
where $X = \{0,1,2,3,4\}$.In the case of just one ace, $P(X=1) = \frac{3243}{10829} \approx 0.299474$.
The following diagram plots the pmf of the number of aces in a hand of 5 cards:
Do you want the probability of $exactly$ one ace? Or just the probability of an ace appearing in a 5-card hand?
If you want exactly one ace, then your answer is correct. $\binom{52}{5}$ is the number of 5-card hands in the deck, and you have 4 choices for which ace to include (hence, $\binom{4}{1}$), and 48 choose 4 choices for the other 4 cards (hence, $\binom{48}{4}$).
If, instead, you want the probability of at least one ace appearing in a 5-card hand, we do things differently. The easiest answer is to find the probability of getting $no$ aces in a 5-card hand.
This probability is $$\frac{\binom{48}{5}}{\binom{52}{5}},$$ for we have 48 choose 5 possible hands with no aces.
Then the solution to the problem - that is, the probability of at least one ace appearing in a 5-card hand - is one minus the complement: $$ 1 - \frac{\binom{48}{5}}{\binom{52}{5}}.$$