Question

In a 2-card hand, what is the probability of holding 2 kings?

Answer 1

To find the probability of holding 2 kings in a 2-card hand, you need to calculate the number of favorable outcomes and divide it by the total number of possible outcomes. In this case, there are 6 favorable outcomes out of 1,326 possible outcomes, resulting in a probability of approximately 0.0045 or 0.45%.

Step-by-step explanation:

To find the probability of holding 2 kings in a 2-card hand, we first need to determine the total number of possible 2-card hands that can be dealt from a standard 52-card deck. The number of combinations of choosing 2 cards from 52 is given by the formula C(52, 2) = 52! / (2! * (52-2)!) = 1,326.

Next, we need to determine the number of favorable outcomes, which is the number of ways to choose 2 kings from the 4 kings in the deck. This can be calculated as C(4, 2) = 4! / (2! * (4-2)!) = 6.

Finally, we can calculate the probability by dividing the number of favorable outcomes by the total number of possible outcomes: P(2 kings) = 6/1,326 ≈ 0.0045 or 0.45%.

Answer 2

A pack of cards has 52 cards, 4 of which are kings, so the probability of the first card being a king is (4/52). Since you have already chosen a card, there are only 51 cards left to choose from in the entire deck (the first card is not counted anymore because it has already been pulled from the deck), and since the card you chose was a king, you are left with only 3 more kings, meaning that the probability of the second card being a king is (3/51). Since we are interested in both happening simultaneously, we must multiply the probability of the first by the probability of the second: (4/52) • (3/51), which leaves us with: * 12/2,652 or 0.0045249 *