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Five distinct numbers are randomly distributed to players numbered 1 through \(5 .\) Whenever two players compare their numbers, the one with the higher one is declared the winner. Initially, players 1 and 2 compare their numbers; the winner then compares her number with that of player \(3,\) and so on. Let \(X\) denote the number of times player 1 is a winner. Find \(P\\{X=i\\}, i=0,1,2,3,4\)

Short Answer

Expert verified
The probabilities for different values of X are: P{X=0} = \(\frac{1}{5}\) P{X=1} = \(\frac{1}{20}\) P{X=2} = \(\frac{1}{120}\) P{X=3} = 0 P{X=4} = 0

Step by step solution

01

Total number of scenarios

The total number of scenarios is 5 factorial, or 5! = 5 × 4 × 3 × 2 × 1 = 120. Next, we will calculate the probability of Player 1 winning i times by finding the number of scenarios in which this happens, dividing it by the total number of scenarios, and then expressing the result as a simplified fraction.
02

Case 1: Player 1 wins 0 times (X=0)

If Player 1 does not win any games, then Player 2 has a higher number than Player 1. To create this arrangement, Player 1's number can be anything, but Player 2's number must be chosen from the 4 remaining numbers that are greater than Player 1's. After this, the 3 remaining numbers can be placed in any order (3! ways). So, there are 4 × 3! = 24 scenarios in which Player 1 wins 0 times. The probability is 24/120 = \(\frac{1}{5}\).
03

Case 2: Player 1 wins 1 time (X=1)

If Player 1 wins 1 game, then Player 3's number must be higher than both Player 1's and Player 2's. There are 3 ways to choose the number for Player 3 (it has to be higher than Player 2's and Player 1's), and then the remaining two numbers can be placed in any order (2! ways), giving 3 × 2! = 6 scenarios in which Player 1 wins 1 time. The probability is 6/120 = \(\frac{1}{20}\).
04

Case 3: Player 1 wins 2 times (X=2)

If Player 1 wins 2 games, then Player 4's number must be higher than Player 1's, Player 2's, and Player 3's, leaving only one possible number for Player 4. The remaining number will be assigned to Player 5. There is only 1! = 1 scenario in which Player 1 wins 2 times. The probability is 1/120.
05

Case 4: Player 1 wins 3 times (X=3) and Case 5: Player 1 wins 4 times (X=4)

These cases are not possible, as Player 1 can win at most 2 times. Thus, the probability for X=3 and X=4 is 0. In summary, the probabilities for different values of X are as follows: P{X=0} = \(\frac{1}{5}\) P{X=1} = \(\frac{1}{20}\) P{X=2} = \(\frac{1}{120}\) P{X=3} = 0 P{X=4} = 0

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorial
Factorial is a mathematical concept that is crucial when dealing with permutations and combinations. It helps calculate how many ways you can arrange or order a set of items. When you see the term "factorial," it's often denoted by an exclamation mark, as in "5!", which is read as "five factorial."

The factorial of a number is the product of all positive integers less than or equal to that number. For instance, to find 5!, you multiply 5 by all the smaller positive integers: 5! = 5 × 4 × 3 × 2 × 1 = 120. This means there are 120 different ways to arrange five different numbers.

Factorials are particularly useful in probability calculations, especially when determining the number of possible arrangements that meet certain criteria. So when asked to find how many ways something can happen, think of factorial as an invaluable tool for counting possibilities neatly and efficiently.
Probability Calculation
Probability calculations are integral to understanding the likelihood of various outcomes in random events. When you encounter a probability problem, you often want to find the chance or probability of a specific event happening.

To calculate probability, you use the formula \( P(A) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \). For example, in the exercise, to find the probability that Player 1 wins 0 games, you count how many ways this can happen and divide it by the total number of ways the game can play out. You found 24 scenarios where Player 1 wins none, out of a total of 120 possible scenarios, so the probability is \(\frac{24}{120} = \frac{1}{5}\).

By consistently applying this method, you can determine the probabilities for Player 1 winning 1 or 2 games, ensuring your calculations consider only the viable scenarios within the total possible outcomes.
Combinatorial Analysis
Combinatorial analysis is the mathematical study of counting, arranging, and grouping objects. It comes into play when you need to figure out how many different ways something can happen, especially when constraints are involved.

In problems like the one discussed, combinatorial analysis helps determine the number of ways different outcomes can occur based on selected conditions. The exercise demonstrates this by analyzing cases like Player 1 winning a certain number of times, where the key is determining how the numbers can be ordered or chosen given specific rules.

Combinatorial analysis involves breaking down the problem into manageable parts. You first consider each potential outcome (how many times Player 1 wins), analyze the conditions necessary for these outcomes to occur, and then use concepts like factorials to count possibilities. By combining these techniques, you arrive at the probabilities needed to fully understand and predict the different outcomes in structured random events.

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Most popular questions from this chapter

Suppose that the number of accidents occurring on a highway each day is a Poisson random variable with parameter \(\lambda=3\) (a) Find the probability that 3 or more accidents occur today. (b) Repeat part (a) under the assumption that at least 1 accident occurs today.

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