1. ## probability/cards

Suppose that 13 cards are selected at random from a regular deck of 52 playing cards. If it is known that at least one ace gas been selected, what is the probability that at least two aces have been selected?
And if it is known that an ace of hearts has been selected, what is the probability that at least two aces have been selected?

2. Originally Posted by tanichkap
Suppose that 13 cards are selected at random from a regular deck of 52 playing cards. If it is known that at least one ace gas been selected, what is the probability that at least two aces have been selected?
And if it is known that an ace of hearts has been selected, what is the probability that at least two aces have been selected?
Let X be the random variable number of aces selected.

Then X ~ Hypergeometric(N = 52, n = 13, m = 4). See Hypergeometric distribution - Wikipedia, the free encyclopedia

1. $\displaystyle \Pr(X \geq 2 | X \geq 1) = \frac{\Pr(X \geq 1) \, \text{and} \, \Pr(X \geq 2)}{\Pr(X \geq 1)} = \frac{\Pr(X \geq 2)}{\Pr(X \geq 1)}$ .....

2. $\displaystyle \Pr(X \geq 2 | \text{Ace of hearts is selected}) = \frac{\Pr(\text{Ace of hearts is selected}) \, \text{and} \, \Pr(X \geq 2)}{\Pr(\text{Ace of hearts is selected})}$ .....

3. Is the Pr(Ace of hearts) = (13 choose 4) *( 39 choose 9) / (52 choose 13)?

Thanks for the help!

4. Originally Posted by tanichkap
Is the Pr(Ace of hearts) = (13 choose 4) *( 39 choose 9) / (52 choose 13)?

Thanks for the help!
It's $\displaystyle \frac{{1 \choose 1} {51 \choose 12}}{{52 \choose 13}} = \frac{1}{4}$.

5. For the second question of the exercise I need the Pr(Ace of hearts) to calculate the Pr(X>2|Ace of hearts), don't I?

6. Originally Posted by tanichkap
For the second question of the exercise I need the Pr(Ace of hearts) to calculate the Pr(X>2|Ace of hearts), don't I?
Yes. My mistake. I made a typo in my first reply and that caused a mistake in my second. Both have been edited.

7. ## probabilty/cards

originally bytanichkap.
you are dealt 13 cars from a 52 card deck;
question is pr of getting 2 or more aces knowing that the hand contains 1 or more aces.
The pr of 2 or more is fixed. The info does not help.If the cards are turned over one by one and an ace shows up between the first and the 12th new prs can be calculated based on where an ace appears.

BJ