We perform the experiment "deal a five card poker hand". What is the probability that you are dealt a hand that contains the Ace, Two, Three, and Four of Diamonds?
Question 22 answers
4 / 2,598,960
48 / 2,598,960
1 / 2,598,960
1 / 52
1/52
There are C(52,5) or 2598960 ways to get a 5 card hand.
The one you describe is one of those
(I assume all the cards you mentioned are diamonds)
so prob is 1/2 598 960
To calculate the probability of being dealt a hand that contains the Ace, Two, Three, and Four of Diamonds, we need to determine the total number of possible five-card hands and the number of hands that meet the given condition.
First, let's find the total number of possible five-card hands. In a standard deck of 52 cards, there are 52 choices for the first card, 51 choices for the second card (since one card has already been chosen), 50 choices for the third card, 49 choices for the fourth card, and 48 choices for the fifth card. Therefore, the total number of possible hands is calculated as:
52 * 51 * 50 * 49 * 48 = 311,875,200
Now, let's determine the number of hands that contain the Ace, Two, Three, and Four of Diamonds. Since there is only one Ace of Diamonds, one Two of Diamonds, one Three of Diamonds, and one Four of Diamonds in the deck, there is only one possible combination of these specific cards. Therefore, the number of hands that meet the condition is:
1
Finally, we can calculate the probability by dividing the number of hands that meet the condition by the total number of possible hands:
1 / 311,875,200
Simplifying this fraction, the probability of being dealt a hand that contains the Ace, Two, Three, and Four of Diamonds is approximately:
1 / 311,875,200