/*! 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 26 After all students have left the... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 26

After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. At the beginning of the next lecture, the professor distributes the four books at random to the four students \((1,2,3\), and 4\()\) who claim to have left books. One possible outcome is that 1 receives 2's book, 2 receives 4's book, 3 receives his or her own book, and 4 receives l's book. This outcome can be abbreviated \((2,4,3,1)\). a. List the 23 other possible outcomes. b. Which outcomes are contained in the event that exactly two of the books are returned to their correct owners? Assuming equally likely outcomes, what is the probability of this event? c. What is the probability that exactly one of the four students receives his or her own book? d. What is the probability that exactly three receive their own books? e. What is the probability that at least two of the four students receive their own books?

Short Answer

Expert verified
a. There are 23 other outcomes, which we can find by generating all permutations of 4 distinct items. b. There are 12 outcomes where exactly 2 students receive their books and the probability is \(\frac{1}{2}\). c. The probability that exactly one student receive their book is \(\frac{1}{3}\). d. The probability that 3 students receive their own books is 0, because it implies all students get their book which violates the condition. e. The probability that at least 2 students receive their own books is \(\frac{13}{24}\).

Step by step solution

01

List all potential outcomes

As there are 4 textbooks, we have 4x3x2x1 = 24 possible arrangements, including the given one (2,4,3,1). We list the remaining 23 as required. Fortunately, there are tools available online that can generate permutations for you.
02

Find outcomes with exactly two books returned correctly

In this scenario, 2 out of the 4 students must get their own books, while the other 2 cannot. There are \({4\choose2}=6\) ways to choose which two students get their books, and then 2 ways to arrange the remaining students' books. This gives us a total of \(6*2=12\) favorable outcomes. The probability is therefore \(\frac{12}{24}=\frac{1}{2}\).
03

Find probability of exactly one student getting their book

We use a similar approach. We have \({4\choose1}=4\) ways of choosing which student gets their book. The remaining students forms a cyclic permutation, which has \((3-1)!=2\) possibilities. Hence, the number of favorable outcomes is \(4*2=8\), and the probability is \(\frac{8}{24}=\frac{1}{3}\).
04

Find probability of exactly three students getting their books

If 3 students receive their own books, it implies that the 4th one also gets their own book which goes against our condition. Hence, this situation is not possible and the probability is 0.
05

Find probability of at least two of the students getting their books

If we want at least two students to get their books, we are looking for scenarios where 2, 3, or 4 students get their books. As found in step 2 and 4, there are 12 cases where exactly 2 students get their books and 1 case where all 4 get their own books. Thus, there are \(12+1=13\) favorable outcomes. The probability is \(\frac{13}{24}\).

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.

Permutations
Permutations refer to the different ways in which a set of items can be arranged in order. In our exercise, permutation plays a crucial role as it determines the number of ways the professor can distribute the four textbooks among the students. Since each textbook must go to one student and there are four textbooks, we calculate the total number of permutations for these four items using the factorial of 4, denoted as \(4!\). This is computed as follows:

\[4! = 4 \times 3 \times 2 \times 1 = 24\]

Thus, there are 24 possible ways to distribute the books randomly. Understanding permutations allows us to organize and count these arrangements efficiently.
Combinatorics
Combinatorics is the field of mathematics that deals with counting, both as a means and an end in obtaining results. In the problem at hand, we use combinatorics to solve for various probability scenarios.

**List Possible Outcomes**
To solve part of the exercise, we first need to acknowledge the total number of possible ways to distribute the books, which involves listing or considering all permutations of the textbooks among students.

**Choosing Outcomes**
For specific questions such as when exactly two students receive their correct book, combinatorics helps us calculate this using combinations. We select two students from a group of four who will definitely receive their own books. This is indicated by the combination \({4 \choose 2}\), representing how many subsets of 2 we can form, which is calculated as 6. Then, we account for the arrangement of the other unmapped students' books, leading to a total of \(6 \times 2 = 12\) favorable outcomes.

Combinatorics provides the tools needed for organizing and counting these outcomes to identify probabilities.
Equally Likely Outcomes
In probability theory, equally likely outcomes refer to scenarios where each outcome in the sample space has the same chance of occurring. This assumption is crucial for calculating probabilities.

**Equally Likely in Textbook Distribution**
In our exercise, every possible permutation of the textbooks is considered equally likely. This implies that since there are 24 permutations, the probability of any specific permutation occurring, such as a student receiving their own book, is \(\frac{1}{24}\).

By understanding that each outcome has an equal chance, we simplify the calculation of probabilities. For example, the probability that exactly two students receive their own books is obtained by dividing the number of favorable outcomes (12) by the total number of outcomes (24), resulting in \(\frac{1}{2}\).

This concept is foundational in probability and helps ensure fairness and accuracy in our calculations when each arrangement or case is assumed to be equally probable.

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

The manager of a music store has kept records of the number of CDs bought in a single transaction by customers who make a purchase at the store. The accompanying table gives six possible outcomes and the estimated probability associated with each of these outcomes for the chance experiment that consists of observing the number of CDs purchased by the next customer at the store. $$ \begin{aligned} &\begin{array}{l} \text { Number of CDs } \\ \text { purchased } \end{array} & 1 & 2 & 3 & 4 & 5 & 6 \text { or more } \\ &\begin{array}{c} \text { Estimated } \\ \text { probability } \end{array} & .45 & .25 & .10 & .10 & .07 & .03 \end{aligned} $$ a. What is the estimated probability that the next customer purchases three or fewer CDs? b. What is the estimated probability that the next customer purchases at most three CDs? How does this compare to the probability computed in Part (a)? c. What is the estimated probability that the next customer purchases five or more CDs? d. What is the estimated probability that the next customer purchases one or two CDs? e. What is the estimated probability that the next customer purchases more than two CDs? Show two different ways to compute this probability that use the probability rules of this section.

A library has five copies of a certain textbook on reserve of which two copies ( 1 and 2 ) are first printings and the other three \((3,4\), and 5\()\) are second printings. \(\mathrm{A}\) student examines these books in random order, stopping only when a second printing has been selected. a. Display the possible outcomes in a tree diagram. b. What outcomes are contained in the event \(A\), that exactly one book is examined before the chance experiment terminates? c. What outcomes are contained in the event \(C\), that the chance experiment terminates with the examination of book 5 ?

Refer to the following information on births in the United States over a given period of time: $$ \begin{array}{lr} \text { Type of Birth } & \text { Number of Births } \\ \hline \text { Single birth } & 41,500,000 \\ \text { Twins } & 500,000 \\ \text { Triplets } & 5,000 \\ \text { Quadruplets } & 100 \\ \hline \end{array} $$ Use this information to approximate the probability that a randomly selected pregnant woman who reaches full term a. Delivers twins b. Delivers quadruplets c. Gives birth to more than a single child

Medical insurance status-covered (C) or not covered (N) - is determined for each individual arriving for treatment at a hospital's emergency room. Consider the chance experiment in which this determination is made for two randomly selected patients. The simple events are \(O_{1}=(\mathrm{C}, \mathrm{C})\), meaning that the first patient selected was covered and the second patient selected was also covered, \(O_{2}=(\mathrm{C}, \mathrm{N}), O_{3}=(\mathrm{N}, \mathrm{C})\), and \(O_{4}=\) \((\mathrm{N}, \mathrm{N}) .\) Suppose that probabilities are \(P\left(O_{1}\right)=.81\), \(P\left(O_{2}\right)=.09, P\left(O_{3}\right)=.09\), and \(P\left(O_{4}\right)=.01\). a. What outcomes are contained in \(A\), the event that at most one patient is covered, and what is \(P(A)\) ? b. What outcomes are contained in \(B\), the event that the two patients have the same status with respect to coverage, and what is \(P(B)\) ?

Consider a system consisting of four components, as pictured in the following diagram: Components 1 and 2 form a series subsystem, as do Components 3 and 4 . The two subsystems are connected in parallel. Suppose that \(P(1\) works \()=.9, P(2\) works \()=\) \(.9, P(3\) works \()=.9\), and \(P(4\) works \()=.9\) and that the four components work independently of one another. a. The 1-2 subsystem works only if both components work. What is the probability of this happening? b. What is the probability that the \(1-2\) subsystem doesn't work? that the \(3-4\) subsystem doesn't work? c. The system won't work if the \(1-2\) subsystem doesn't work and if the \(3-4\) subsystem also doesn't work. What is the probability that the system won't work? that it will work? d. How would the probability of the system working change if a \(5-6\) subsystem were added in parallel with the other two subsystems? e. How would the probability that the system works change if there were three components in series in each of the two subsystems?
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

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.