We choose $10$ random cards from a normal deck of cards($52$ cards). What is the probability that we get:
a. $0$ aces
b. maximum $3$ aces
c. at least $1$ ace and at least one face card
That's the problem. I thought that since we're choosing $10$ cards from a deck, the sample space should be
$$
\binom{10}{52} = \frac{52}{10!(52-10)!} \qquad \text{ (1)}
$$
Then about a. question I thought that basically if $4$ aces are picked we should write $\dfrac{4}{52}\cdot \dfrac{3}{51}\cdot \dfrac{2}{50}\cdot\dfrac{1}{49}$ and then dividing this with (1) and then getting the derived set of this, we get the result..
I really don't know if I'm getting anything right here, so I'm in need of your insight.
Best Answer
You have the size of the sample space correct. You have counted the possible selections of 10 cards from 52.
For (a) you wish to count the selections of 10 cards from 48 non-aces, then divide by the size of the sample space.
(What you were doing was counting selections of 4 aces from fifty two-cards. You don't want to do that.)