/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Soccer Team Problem 1: Eleven gi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 10

Soccer Team Problem 1: Eleven girls try out for the 11 positions on a soccer team In how many different ways could the 11 positions be filled if there are no restrictions on who plays which position? b. In how many different ways could the positions be filled if Mabel must be the goalkeeper? c. If the positions are selected at random, what is the probability that Mabel will be the goalkeeper? d. What is the probability in part c expressed as a percent?

Short Answer

Expert verified
a. 11! different ways b. 10! different ways c. The probability is 1/11 d. The probability expressed as a percent is approximately 9.09%.

Step by step solution

01

Calculating Total Ways without Restrictions

To calculate the number of ways the 11 positions can be filled without restrictions, use the permutation formula since the order in which the positions are filled matters. This is the same as finding 11 factorial (11!).
02

Calculating Ways with Mabel as Goalkeeper

Since Mabel must be the goalkeeper, there are now 10 positions to be filled by 10 players. This is similar to finding 10 factorial (10!).
03

Calculating the Probability of Mabel Being Goalkeeper

The probability that Mabel will be the goalkeeper is the number of favorable outcomes over the number of possible outcomes, which is 1/11 because there are 11 players and 1 desired outcome for Mabel to be the goalkeeper.
04

Converting Probability to Percent

To convert the probability into percent, multiply the probability by 100. Therefore, the percent chance that Mabel will be the goalkeeper is (1/11) * 100.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Factorial Calculation
Understanding factorial calculation is crucial in solving problems related to permutations and combinations in probability. A factorial is represented by an exclamation mark (!) and is defined as the product of all positive integers up to a given number. For example, the factorial of 3, written as 3!, is calculated as follows:

3! = 3 脳 2 脳 1 = 6

In the context of our soccer team problem, when calculating the number of ways 11 positions on a soccer team can be filled, 11! or '11 factorial' represents the product of all numbers from 1 to 11. Using factorial calculations simplifies the representation of large permutations, which might otherwise require extensive multiplication.
Permutation Formula
The permutation formula is used to find the number of ways to arrange a set number of items. In problems like team selection, where order matters, we apply the permutation formula. Generally, if we have n items, the number of permutations is n!.

Using this principle, for our soccer team problem, there are 11 positions to fill with 11 different players, resulting in:

11! = 11 脳 10 脳 ... 脳 2 脳 1

This gives us the total number of ways to fill all positions when there are no restrictions. When we place a restriction, such as having a specific player in a specific position, we reduce the number of items (players) by 1, hence for 10 remaining positions, we have 10! permutations.
Probability to Percent Conversion
Converting probabilities to percent is a way of expressing the likelihood of an event in terms that are often easier to understand. To convert a probability to a percentage, we multiply the probability by 100%. For example, suppose the probability of an event is 0.25; to convert this to a percent, we do:

0.25 脳 100% = 25%

In the case of determining the probability that Mabel is the goalkeeper, we first calculate the probability as a fraction, 1/11, and then convert it to a percentage by multiplying by 100%, yielding approximately 9.09%. This translates the abstract idea of probability into a more tangible form, typically used in everyday language.
Favorable Outcomes in Probability
When determining the probability of a specific outcome, like Mabel being chosen as the goalkeeper, we refer to this as a favorable outcome. Probability is calculated by dividing the number of favorable outcomes by the total number of possible outcomes.

For Mabel to be the goalkeeper, the number of favorable outcomes is 1 (Mabel being the goalkeeper), and the total number of possible outcomes is 11 (the total number of players who could be goalkeeper). This gives us:
\( P(\text{Mabel as goalkeeper}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} = \frac{1}{11} \)

Understanding how to calculate favorable outcomes is key to answering probability questions accurately.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Uranium Fission Problem: When a uranium atom splits ("fissions"), it releases 0,1,2,3 or 4 neutrons. Let \(P(x)\) be the probability that \(x\) neutrons are released. Assume the probability distribution is $$\begin{array}{ll}x & P(x) \\\\\hline 0 & 0.05 \\\1 & 0.2 \\\2 & 0.25 \\\3 & 0.4 \\\4 & 0.1\end{array}$$ a. What is the mathematically expected number of neutrons released per fission? b. The number of neutrons released in any one fission must be an integer. How do you explain the fact that the mathematically expected number in part a is not an integer?

Evaluate the number of combinations or permutations two ways: a. Using factorials, as in the examples of this section b. Directly, using the built-in features of your grapher $$27 C_{19}$$

First Girl Problem: Eva and Paul want to have a baby girl. They know that the probability of having a girl on any single birth is 0.5 (A). Let \(x\) be the number of babies they have, and let \(P(x)\) be the probability that the \(x\) th baby is the first girl. Then \(P(1)=0.5 . P(2)\) is the probability that the first baby is not a girl and that the second baby is a girl. Calculate \(P(2), P(3),\) and \(P(4)\) (B). Plot the graph of \(P\). Sketch the graph, showing what happens as \(x\) becomes large. (C). Besides being called a probability distribution, what other special kind of function is this? (D). Show that the sum of the values of \(P(x)\) approaches 1 as \(x\) becomes very large.

Traffic Light Problem 1: Two traffic lights on Broadway operate independently. Your probability of being stopped at the first light is \(40 \% .\) Your probability of being stopped at the second one is \(70 \% .\) Find the probability of being stopped at a. Both lights b. Neither light c. The first light but not the second d. The second light but not the first e. Exactly one of the lights

Permutations with Repeated Elements-Problems 15 and 16: The word CARRIER has seven letters. But there are fewer than \(7 !\) permutations, because in any arrangement of these seven letters the three \(R\) 's are interchangeable. If these \(R\) 's were distinguishable, there would be \(3 !\), or \(6,\) ways of arranging them. This implies that only \(\frac{1}{6}\) (that is, \(\frac{1}{3 !}\) ) of the \(7 !\) permutations are actually different. So the number of permutations is $$ \frac{7 !}{3 !}=840 $$ There are four \(I^{\prime}\) 's, four \(S\) 's, and two \(P\) 's in the word MISSISSIPPI, so the number of different permutations of its letters is $$ \frac{111}{4 ! 4 ! 2 !}=34,650 $$ Nine pennies are lying on a table. Five show heads and four show tails. In how many different ways, such as "HHTHTTHHT," could the coins be lined up if you consider all the heads to be identical and all the tails to be identical?
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.