What is the probability that a random selected poker hand contains exactly 3 aces given that it contains at least 2 aces?

Answer 1

An integer is chosen at random from the first 20,000 positive integers . What is the probability that the integer chosen is divisible by atleast one of the numbers 5,7,8? *

Answer 2

To calculate the probability of a random selected Poker hand containing exactly 3 aces given that it contains at least 2 aces, we first need to determine the total number of ways to choose a poker hand with at least 2 aces. This can be done by considering the different combinations of choosing 2, 3, or 4 aces from the 4 available in a standard deck of 52 cards. Once we have the total number of ways to choose at least 2 aces, we then calculate the number of ways to choose exactly 3 aces from the selected hand. Finally, we divide the number of ways to choose exactly 3 aces by the total number of ways to choose at least 2 aces to obtain the probability.

Answer 3

Well, isn't that a happy little question! If we already have at least 2 aces in our poker hand, there are only 2 ways to get exactly 3 aces out of the remaining cards. So the probability would be the number of ways to choose those 3 aces divided by the number of ways to choose any 3 cards from the remaining cards. Just a matter of counting and you'll have your answer!

Answer 4

I will assume that you mean a five card poker hand. We can label the cards C1, C2, C3, C4, and C5. We are basically told already that C1 and C2 are both aces. So we have to find the probability that exactly one of C3, C4, and C5 is an ace. Knowing that the first two cards in our hand are both aces means that there are only 50 cards left in the deck. The probability that C3 is an ace and that C4 and C5 are both not aces is (2/50)(48/49)(47/48)=0.03836734694. The same probability also applies to each of C4 and C5, considered independently of each other. Therefore, our final probability is 3* 0.03836734694=0.1151020408