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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 stu- dents \((1,2,3\), 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 1 '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? As- suming 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?

Step by step solution

01

Understand the nature of the problem

There are four books and four students. We are to find the different ways in which the books can be distributed to students. For each book, there are four possibilities (it could go to any of the four students). Hence total possible outcomes are \(4^4 = 256\)

02

Solve for event 'a'

We have already counted one possible outcome \((2,4,3,1)\). A simple way to find others will be to generate all permutations of the numbers 1 through 4, and eliminate the situation where a student gets their own book. However, manually generating all such permutations can be time-consuming and error-prone.

03

Solve for event 'b'

A student can either have their own book or not. If exactly two have their own books, the others must have someone else's. There are six ways (or \( \binom{4}{2} = 6\)) to choose which two students have their own books. Considering the students that don’t get their own book, there are two ways (or 2!) for them to exchange books, leading to \(6*2!=12\) total outcomes which satisfy this criterion. Hence the required probability is \( \frac{12}{256} = 0.046875 \)

04

Solve for event 'c'

If exactly one student gets their own book, there are \( \binom{4}{1} = 4\) ways to select that student. For the remaining three students, considering the rule that nobody receives their own book, there are 2 possible ways (or 2!) for them to exchange books, leading to \(4*2!=8\) total outcomes. Hence the required probability is \( \frac{8}{256} = 0.03125 \)

05

Solve for event 'd'

If exactly three students receive their own books, it essentially means that all students get their books back because the fourth student also needs to get their own book. The probability of this event happening is 0, since it contradicts the nature of the initial problem where at least one student must not have their own book.

06

Solve for event 'e'

The probability of at least two students getting their own books is the sum of the probabilities of exactly two students and exactly three students getting their own books. But as we found out in the last step, the probability that exactly three students receive their own books is 0. Hence the probability that at least two of the students receive their own books equals the probability that exactly two students receive their own books i.e. 0.046875

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Key Concepts

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

Permutations

In mathematics, a permutation refers to the arrangement of all the members of a set into some sequence or order. Permutations are crucial when considering problems where the arrangement, or order, of the elements matters.

For example, if you have a set of 4 books, there are several permutations in which you can distribute these books to 4 students. A permutation takes into account both the order of placing each book and the identity of the student receiving it.

Permutations are typically characterized by factorial notation. The number of ways to arrange 4 items is given by the factorial of 4, noted as 4! which equals 24. Thus, there are 24 permutations of 4 books when each student receives one book.

• Example: Considering the sequence (2,4,3,1), where each number represents the student receiving the specific book.
• This implies that student 1 receives book 2, student 2 receives book 4, and so on, highlighting the importance of order.

Combinatorics

Combinatorics is a field of mathematics focused on counting, arrangement, and combination of set elements. It provides tools and principles to solve problems related to selecting and arranging objects.

In our exercise, combinatorics plays a vital role in determining how books can be matched with students, ensuring all possible arrangements are accounted for.

Combinatorics helps identify patterns and possible outcomes: from counting possibilities when distributing books, to calculating probability events.

• For instance, when calculating the number of ways students can receive exactly two correct books, combinatorics helps us use combinations to determine possible outcomes.
• This process involves selecting 2 students from 4 using the combination formula \( \binom{4}{2} \).

Permutations and Combinations

Permutations and combinations are both fundamental concepts in combinatorics, but they differ in one critical aspect: the significance of order.

While permutations concern arrangements where the order matters, combinations focus on selections where order does not play a role. This distinction is crucial in solving problems involving distribution of items without replacement.

In our exercise, we use these concepts to calculate the probability of certain book distributions.

• In one part, we examine scenarios where exactly one student gets their own book. This utilizes combinations to choose which one out of the four students receives the correct book, while permutations determine the swap for others.
• To solve for probabilities, understanding both permutations and combinations gives a comprehensive approach to tackling such allocation problems.

Combinations disregard order and are used here to determine probabilities where the book return to the correct student does not depend on order, whereas permutations take precedence when computing the exact sequence of returns.