# Probability of picking 10 cards from a deck of cards

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.

• @Belf. If you meant $\frac{48!}{10!(48-10)!}$, then yes. – Graham Kemp Mar 12 '17 at 20:34
• @Belf. It is approximately that, yes. $246/595 =0.4\dot{\overline{134453781512605042016806722689075630252100840336}}$. – Graham Kemp Mar 13 '17 at 4:15
• It is not unsurprising. Consider that it is the probability that for the fifty two places in a shuffled deck, all four aces do not occupy any from the top ten. That is that all of them lie approximately within the lower four fifths of the deck. $$\dfrac{42}{52}\dfrac{41}{51}\dfrac{40}{50}\dfrac{39}{49}\quad\approx\quad \left(\dfrac 45\right)^4$$ – Graham Kemp Mar 13 '17 at 4:21