Problem 4 - Entrance Test

There are 52 cards in a standard deck, and 4 of them are aces. Method 1: Permutations The total number of ways to choose an ordered pair of cards for the top and bottom positions is P(52, 2) = 52 * 51 = 2652. The number of ways to choose an ordered pair of aces for the top and bottom positions is P(4, 2) = 4 * 3 = 12. The probability is the ratio of favorable outcomes to the total possible outcomes: Probability = 12 / 2652 = 1 / 221. Method 2: Conditional Probability Let A be the event that the top card is an ace. Let B be the event that the bottom card is an ace. We want to find P(A and B) = P(A) * P(B|A). The probability that the top card is an ace is P(A) = 4/52 = 1/13. If the top card is an ace, then there are 51 cards remaining, and 3 of them are aces. The probability that the bottom card is an ace, given that the top card is an ace, is P(B|A) = 3/51 = 1/17. So, the probability that both the top and bottom cards are aces is P(A and B) = (1/13) * (1/17) = 1/221.