Question

A deck of 52 playing cards, containing all 4 aces, is randomly divided into 4 piles of 13 cards each. Define events \(E_{1}, E_{2}, E_{3}\), and \(E_{4}\) as follows: \(E_{1}=\\{\) the first pile has exactly 1 ace \(\\}\), \(E_{2}=\\{\) the second pile has exactly 1 ace \(\\}\), \(E_{3}=\\{\) the third pile has exactly 1 ace \(\\}\), \(E_{4}=\\{\) the fourth pile has exactly 1 ace \(\\}\) Use Exercise 23 to find \(P\left(E_{1} E_{2} E_{3} E_{4}\right)\), the probability that each pile has an ace.

Short answer

The probability for each of the events \(E_1\), \(E_2\), \(E_3\), and \(E_4\) is the same and can be calculated as: \( P(E_i) = \frac{\bigl(\binom{48}{12} \cdot \frac{40!}{(13!)^3}\bigr)}{\frac{52!}{(13!)^4}}\: (i = 1, 2, 3,\: or\: 4)\)

Step-by-step solution

Analyze the events E1, E2, E3, and E4

Since we have a random deck of 52 cards, with 4 aces, and we want to find the probability of each event, we will analyze events E1, E2, E3 and E4 as follows: \(E_1\) means exactly 1 Ace in the first pile, and the probability of this event occurring can be written as \(P(E_1)\). Similarly, we can write the probabilities for events E2, E3, and E4 as \(P(E_2)\), \(P(E_3)\), and \(P(E_4)\), respectively.

Calculate the total number of possible outcomes

To find the total number of possible outcomes, let's consider the fact that there are 52 cards, which we have to distribute into 4 piles (13 cards in each pile). Using the multinomial coefficient formula, we can calculate the total number of outcomes: Total number of outcomes \(= \frac{52!}{(13!)^4}\)

Calculate the number of favorable outcomes for each event

Now, let's calculate the number of favorable outcomes for each event: For \(E_1\), there are 12 non-ace cards left to be chosen for pile 1, which can be selected in \(\binom{48}{12}\) ways. After this, the remaining 40 cards can be distributed into the other 3 piles in \(\frac{40!}{(13!)^3}\) ways. Therefore, the number of possible ways for \(E_1\) is \(\binom{48}{12} \cdot \frac{40!}{(13!)^3}\). Similarly, we can calculate the possible ways of events E2, E3 and E4: - For \(E_2\): \(\binom{48}{12} \cdot \frac{40!}{(13!)^3}\) - For \(E_3\): \(\binom{48}{12} \cdot \frac{40!}{(13!)^3}\) - For \(E_4\): \(\binom{48}{12} \cdot \frac{40!}{(13!)^3}\)

Calculate the probability for each event

Now that we have calculated the number of possible outcomes and favorable outcomes for each event, we can calculate the probability for each event: - \( P(E_1) = \frac{\bigl(\binom{48}{12} \cdot \frac{40!}{(13!)^3}\bigr)}{\frac{52!}{(13!)^4}}\) - \( P(E_2) = \frac{\bigl(\binom{48}{12} \cdot \frac{40!}{(13!)^3}\bigr)}{\frac{52!}{(13!)^4}}\) - \( P(E_3) = \frac{\bigl(\binom{48}{12} \cdot \frac{40!}{(13!)^3}\bigr)}{\frac{52!}{(13!)^4}}\) - \( P(E_4) = \frac{\bigl(\binom{48}{12} \cdot \frac{40!}{(13!)^3}\bigr)}{\frac{52!}{(13!)^4}}\) As the numerators are the same for all the events, we can conclude that all 4 events have the same probability. Calculate the probability for any of the events, and that will be true for all other events as well.