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Chapter 4: Problem 8

Express all probabilities as fractions. In soccer, a tie at the end of regulation time leads to a shootout by three members from each team. How many ways can 3 players be selected from 11 players available? For 3 selected players, how many ways can they be designated as first, second, and third?

Short Answer

Expert verified
There are 990 ways to select and order 3 players from 11.

Step by step solution

01

Calculate the number of ways to choose 3 players from 11

The number of ways to choose 3 players from 11 can be calculated using the combination formula. The combination formula is given by: \[ C(n, r) = \frac{n!}{r!(n-r)!} \] where \( n \) is the total number of players available and \( r \) is the number of players to choose. Plugging in the values, we get: \[ C(11, 3) = \frac{11!}{3!(11-3)!} = \frac{11!}{3! \times 8!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165 \] So, there are 165 ways to choose 3 players from 11.
02

Calculate the number of ways to order the 3 selected players

After choosing 3 players, we need to determine how many ways they can be designated as first, second, and third. This is a permutation problem where the number of ways to order \( r \) items is given by \( r! \). For 3 players, the number of ways to order them is: \[ 3! = 3 \times 2 \times 1 = 6 \] So, there are 6 ways to designate the 3 players as first, second, and third.
03

Combine the results

To find the total number of possible ways to choose and order the players, multiply the number of ways to choose the players by the number of ways to order them: \[ 165 \times 6 = 990 \] Therefore, there are 990 ways to select 3 players from 11 and designate them as first, second, and third.

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Key Concepts

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

combinations
When selecting a group of items where the order does not matter, you are dealing with combinations. In the soccer player selection example, we need to choose 3 players out of 11 without caring about the order in which they are selected. This is a classic combination problem.
The formula used for combinations is: \ C(n, r) = \frac{n!}{r!(n-r)!} \ where:
  • \( n \) is the total number of items
  • \( r \) is the number of items to choose
. Applying this to selecting 3 players out of 11: \[ C(11, 3) = \frac{11!}{3!(11-3)!} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165 \] . So, there are 165 ways to choose 3 players from 11.
permutations
After selecting 3 players, we need to decide the order in which they will play. This involves permutations, where the order does matter. For permutations, we use the factorial of the number of items.
The formula for permutations of \( r \) items is: \[ P(r) = r! \]
In our example, we have 3 players, so the number of ways to order them is: \[ 3! = 3 \times 2 \times 1 = 6 \] Therefore, there are 6 permutations for 3 selected players, meaning 6 different ways to arrange them as first, second, and third.
factorials
A factorial, denoted by an exclamation point (\(!\)), represents the product of all positive integers up to a given number. Factorials are fundamental in combinatorics for calculating combinations and permutations.
For example: \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). The factorial grows very quickly with larger numbers.
In our exercise, factorials helped calculate:
  • \( 11! \) in determining combinations
  • \( 3! \) in determining permutations
probability calculations
Probability calculations often use combinations and permutations. In our soccer example, we used these concepts to count possible ways of arranging players. Probability can be seen as a ratio of favorable outcomes to total outcomes, often expressed as a fraction.
For instance, if we know there are 990 possible ways to arrange players and wonder the probability of a specific arrangement, it would be \( \frac{1}{990} \). By breaking down problems into combinations and permutations, we simplify complex probability questions greatly.

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Most popular questions from this chapter

Find the probability and answer the questions.Men have XY (or YX) chromosomes and women have XX chromosomes. X-linked recessive genetic diseases (such as juvenile retinoschisis) occur when there is a defective \(\mathrm{X}\) chromosome that occurs without a paired \(\mathrm{X}\) chromosome that is not defective. In the following, represent a defective \(X\) chromosome with lowercase \(x,\) so a child with the \(x\) Y or Yx pair of chromosomes will have the disease and a child with \(X X\) or XY or YX or \(x\) X or \(X x\) will not have the disease. Each parent contributes one of the chromosomes to the child.a. If a father has the defective \(x\) chromosome and the mother has good XX chromosomes, what is the probability that a son will inherit the disease? b. If a father has the defective \(x\) chromosome and the mother has good \(X X\) chromosomes, what is the probability that a daughter will inherit the disease? c. If a mother has one defective \(x\) chromosome and one good \(X\) chromosome and the father has good XY chromosomes, what is the probability that a son will inherit the disease? d. If a mother has one defective \(x\) chromosome and one good \(X\) chromosome and the father has good XY chromosomes, what is the probability that a daughter will inherit the disease?

Find the probability.At Least One. In Exercises \(5-12,\) find the probability. If you make random guesses for 10 multiple choice SAT test questions (each with five possible answers), what is the probability of getting at least 1 correct? If these questions are part of a practice test and an instructor says that you must get at least one correct answer before continuing, is there a good chance you will continue?

Nasonex Treatment Analysis Nasonex is a nasal spray used to treat allergies. In clinical trials, 1671 subjects were given a placebo, and 2 of them developed upper respiratory tract infections. Another 2103 patients were treated with Nasonex and 6 of them developed upper respiratory tract infections. Assume that Nasonex has no effect on upper respiratory tract infections so that the rate of those infections also applies to Nasonex users. Using the placebo rate of \(2 / 1671,\) simulate groups of 2103 subjects given the Nasonex treatment, and determine whether a result of 6 upper respiratory tract infections could easily occur. What does that suggest about Nasonex as a cause of upper respiratory tract infections?

Composite Water Samples The Fairfield County Department of Public Health tests water for the presence of \(E\). coli (Escherichia coli) bacteria. To reduce laboratory costs, water samples from 10 public swimming areas are combined for one test, and further testing is done only if the combined sample tests positive. Based on past results, there is a 0.005 probability of finding \(E\) coli bacteria in a public swimming area. Find the probability that a combined sample from 10 public swimming areas will reveal the presence of \(E\). coli bacteria. Is that probability low enough so that further testing of the individual samples is rarely necessary?

Find the probability.At Least One. In Exercises \(5-12,\) find the probability. In China, where many couples were allowed to have only one child, the probability of a baby being a boy was 0.545. Among six randomly selected births in China, what is the probability that at least one of them is a girl? Could this system continue to work indefinitely? (Phasing out of this policy was begun in \(2015 .\) )
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