0
$\begingroup$

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.

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

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.)

$\endgroup$
6
  • $\begingroup$ So should it be, 48!/ (48-10)! ? $\endgroup$
    – Belf
    Commented Mar 12, 2017 at 20:26
  • $\begingroup$ @Belf. If you meant $\frac{48!}{10!(48-10)!}$, then yes. $\endgroup$
    – Graham Kemp
    Commented Mar 12, 2017 at 20:34
  • $\begingroup$ Yes, that's what I meant and I got ~41% probability! $\endgroup$
    – Belf
    Commented Mar 12, 2017 at 20:36
  • $\begingroup$ @Belf. It is approximately that, yes. $246/595 =0.4\dot{\overline{134453781512605042016806722689075630252100840336}}$. $\endgroup$
    – Graham Kemp
    Commented Mar 13, 2017 at 4:15
  • $\begingroup$ 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. $$\dfrac{42}{52}\dfrac{41}{51}\dfrac{40}{50}\dfrac{39}{49}\quad\approx\quad \left(\dfrac 45\right)^4$$ $\endgroup$
    – Graham Kemp
    Commented Mar 13, 2017 at 4:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .