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Chapter 4: Problem 29

Three students are playing a card game. They decide to choose the first person to play by each selecting a card from the 52 -card deck and looking for the highest card in value and suit. They rank the suits from lowest to highest: clubs, diamonds, hearts, and spades. a. If the card is replaced in the deck after each student chooses, how many possible configurations of the three choices are possible? b. How many configurations are there in which each student picks a different card? c. What is the probability that all three students pick exactly the same card? d. What is the probability that all three students pick different cards?

Short Answer

Expert verified
Answer: There are 132,600 configurations in which each student picks a different card.

Step by step solution

01

1. Calculate the number of possible configurations with replacement

Since there are 52 cards in the deck and each of the 3 students can choose any card, there are \(52^3\) possible configurations. Therefore, there are \(52^3 = 140608\) possible configurations. For the second question:
02

2. Calculate the number of configurations with different cards

If each student picks a different card, this is a permutation because the order in which the cards are chosen matters. The total number of possible permutations for the students picking different cards is \(P(52,3) = \frac{52!}{(52-3)!} = 52 \times 51 \times 50 = 132600\). For the third question:
03

3. Calculate the probability of all students picking the same card

If all three students pick the same card, there are 52 possible cards that can be picked. Since all these events are mutually exclusive, we can calculate the probability as follows: \(P(\text{same card}) = \frac{\text{Number of successful outcomes}}{\text{Total possible outcomes}} = \frac{52}{52^3} = \frac{1}{(52)^2} = \frac{1}{2704}\). For the fourth question:
04

4. Calculate the probability of all students picking different cards

Since we've already calculated the total number of configurations with different cards (132600), we can now compute the probability of all students picking different cards: \(P(\text{different cards}) = \frac{\text{Number of successful outcomes}}{\text{Total possible outcomes}} = \frac{132600}{140608} = \frac{75}{76}\).

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