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Chapter 3: Problem 27

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 in a completely random fashion to each of 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 as \((2,4,3,1)\). a. List the other 23 possible outcomes. b. Let \(X\) denote the number of students who receive their own book. Determine the pmf of \(X\).

Short Answer

Expert verified
24 outcomes; PMF: \(P(X=0) = \frac{9}{24}, P(X=1) = \frac{8}{24}, P(X=2) = \frac{6}{24}, P(X=4) = \frac{1}{24}\).

Step by step solution

01

Understand the Problem

There are four students, each of whom left a book, and four books to be randomly returned. We need to consider all possible arrangements (permutations) of these four books.
02

Calculate Total Outcomes

Calculate the total number of possible arrangements for the four books using the permutation formula. Since there are four distinct items (books) to arrange, the number of permutations is given by \(4!\). Thus, the total number of outcomes is \(4! = 24\).
03

List All Permutations

Identify all 24 permutations of the sequence \((1, 2, 3, 4)\) as books being returned to students. Each permutation corresponds to a specific order in which books are given to students. Example permutations include: \((1, 2, 3, 4)\), \((1, 2, 4, 3)\), \((1, 3, 2, 4)\), etc.
04

Evaluate Each Outcome

For each of the 24 permutations, determine how many students receive their own books. For instance, in the permutation \((2, 4, 3, 1)\), only student 3 receives their own book.
05

Determine Probabilities for X=0, X=1, X=2, X=3, X=4

Count the number of permutations where 0, 1, 2, 3, or 4 students receive their own books. These counts will help determine the probability mass function (pmf).
06

Calculate Probability Mass Function (PMF)

Calculate the probability of each possible value of \(X\):- \(P(X=0)\): Count permutations like \((2, 3, 4, 1)\) where no student receives their book. This occurs in 9 cases (0 students get the right book).- \(P(X=1)\): Count permutations like \((1, 4, 3, 2)\). This occurs in 8 cases.- \(P(X=2)\): This occurs in 6 cases.- \(P(X=3)\): This occurs in 0 cases (not possible as having 3 right implies 4th is also right).- \(P(X=4)\): This happens only if all students have their book \((1, 2, 3, 4)\), which is 1 case.- Convert these counts into probabilities by dividing by the total number of permutations (24).

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Key Concepts

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

probability mass function
A probability mass function (pmf) is an essential concept in probability theory for handling discrete random variables. It represents the probabilities of specific outcomes. To create a pmf, identify each possible result of a random event, then determine how likely each result is. The pmf reflects this as a list or equation.
For our exercise, imagine returning books randomly to four students. The random variable here is the count of students who receive their correct book, denoted as \(X\). Each possible number of correctly returned books has a probability associated with it, which forms the pmf.
  • For \(X = 0\), the probability is based on scenarios like \((2, 3, 4, 1)\), where none of the students gets their book back. This happens in 9 out of the 24 permutations.
  • Similarly, for \(X = 1\), such as \((1, 4, 3, 2)\), where only one student gets their book, it appears in 8 permutations.
  • For \(X = 2\), occurring in 6 permutations, two students correctly receive their books.
  • \(X = 3\) is impossible because if three students get their books, the fourth must too, equating to \(X = 4\).
  • Finally, \(X = 4\) happens in only one permutation, when all students receive their book \((1, 2, 3, 4)\).
Each scenario's probability follows by dividing occurrences by 24.
combinatorics
Combinatorics is the study of counting, arranging, and combining items. In our problem, combinatorics plays a key role by calculating various orders in which the books can return to students. This requires permutations, which are arrangements where order matters.
In the scenario of the professor distributing books randomly:
  • The total number of permutations equals the factorial of 4, written as \(4!\). In simpler terms, this means multiplying numbers from 1 to 4: \(4 \times 3 \times 2 \times 1 = 24\).
  • Each permutation represents a different way to assign the books to students, ensuring we account for every possible assignment.
Understanding permutations assists in predicting how often particular events occur, helping compute probabilities within the exercise effectively.
random assignment
Random assignment is a concept central to random experiments and probability. It involves allocating items to different recipients without a planned pattern, ensuring each item has an equal chance of being assigned to any recipient.
In the book distribution example, the professor's return method exemplifies random assignment.Each book can go to any of the four students randomly, making the assignment purely by chance.
  • This randomness creates a fair scenario where no student is more likely than another to receive their correct book initially.
  • The set of all possible distributions (all 24 permutations) ensures complete randomness in outcomes such as no books or all books being correctly returned.
Grasping this concept is vital for analyzing real-world situations where assigning tasks, resources, or items randomly is needed to ensure fairness or understanding chance patterns.
discrete probability distribution
A discrete probability distribution presents probabilities for outcomes of a discrete random variable, meaning variables that have specific, countable outcomes. In statistical exercises, these distributions help describe how probabilities spread over the possible outcomes.
For the permutation problem at hand:
  • The variable \(X\), representing students receiving correct books, is discrete and limited to values like 0, 1, 2, 3, and 4.
  • Its pmf described previously shows probabilities assigned to each of these possible values, reflecting the experiment's structure.
  • A discrete distribution like this is crucial for expecting or forecasting results in many events involving finite outcomes, from classroom experiments to games of chance.
Understanding these basics allows us to anticipate the likelihood of different outcomes, aiding strategic decisions or evaluations in various fields.

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Most popular questions from this chapter

A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature of the project) in applying for a building permit. Let \(Y=\) the number of forms required of the next applicant. The probability that \(y\) forms are required is known to be proportional to \(y\)-that is, \(p(y)=k y\) for \(y=1, \ldots, 5\). a. What is the value of \(k\) ? [Hint: \(\left.\Sigma_{y=1}^{5} p(y)=1\right]\) b. What is the probability that at most three forms are required? c. What is the probability that between two and four forms (inclusive) are required? d. Could \(p(y)=y^{2} / 50\) for \(y=1, \ldots, 5\) be the pmf of \(Y\) ?

A new battery's voltage may be acceptable \((A)\) or unacceptable \((U)\). A certain flashlight requires two batteries, so batteries will be independently selected and tested until two acceptable ones have been found. Suppose that \(90 \%\) of all batteries have acceptable voltages. Let \(Y\) denote the number of batteries that must be tested. a. What is \(p(2)\), that is, \(P(Y=2)\) ? b. What is \(p(3)\) ? [Hint: There are two different outcomes that result in \(Y=3\).] c. To have \(Y=5\), what must be true of the fifth battery selected? List the four outcomes for which \(Y=5\) and then determine \(p(5)\). d. Use the pattern in your answers for parts (a)-(c) to obtain a general formula for \(p(y)\).

Airlines sometimes overbook flights. Suppose that for a plane with 50 seats, 55 passengers have tickets. Define the random variable \(Y\) as the number of ticketed passengers who actually show up for the flight. The probability mass function of \(Y\) appears in the accompanying table. \begin{tabular}{l|ccccccccccc} \(y\) & 45 & 46 & 47 & 48 & 49 & 50 & 51 & 52 & 53 & 54 & 55 \\ \hline\(p(y)\) & \(.05\) & \(.10\) & \(.12\) & \(.14\) & \(.25\) & \(.17\) & \(.06\) & \(.05\) & \(.03\) & \(.02\) & \(.01\) \end{tabular} a. What is the probability that the flight will accommodate all ticketed passengers who show up? b. What is the probability that not all ticketed passengers who show up can be accommodated? c. If you are the first person on the standby list (which means you will be the first one to get on the plane if there are any seats available after all ticketed passengers have been accommodated), what is the probability that you will be able to take the flight? What is this probability if you are the third person on the standby list?

Suppose that \(30 \%\) of all students who have to buy a text for a particular course want a new copy (the successes!), whereas the other \(70 \%\) want a used copy. Consider randomly selecting 25 purchasers. a. What are the mean value and standard deviation of the number who want a new copy of the book? b. What is the probability that the number who want new copies is more than two standard deviations away from the mean value? c. The bookstore has 15 new copies and 15 used copies in stock. If 25 people come in one by one to purchase this text, what is the probability that all 25 will get the type of book they want from current stock? [Hint: Let \(X=\) the number who want a new copy. For what values of \(X\) will all 25 get what they want?] d. Suppose that new copies cost \(\$ 100\) and used copies cost \(\$ 70\). Assume the bookstore currently has 50 new copies and 50 used copies. What is the expected value of total revenue from the sale of the next 25 copies purchased? Be sure to indicate what rule of expected value you are using. [Hint: Let \(h(X)=\) the revenue when \(X\) of the 25 purchasers want new copies. Express this as a linear function.]

An article in the Los Angeles Times (Dec. 3, 1993) reports that 1 in 200 people carry the defective gene that causes inherited colon cancer. In a sample of 1000 individuals, what is the approximate distribution of the number who carry this gene? Use this distribution to calculate the approximate probability that a. Between 5 and 8 (inclusive) carry the gene. b. At least 8 carry the gene.
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