Ibe Fundamentals of Applied Probability and Random Processes

2. Auflage 2014
ISBN: 978-0-12-801035-8
Verlag: Elsevier Science & Techn.
Format: EPUB
Kopierschutz: Adobe DRM (»Systemvoraussetzungen)

E-Book, Englisch, 456 Seiten

The long-awaited revision of Fundamentals of Applied Probability and Random Processes expands on the central components that made the first edition a classic. The title is based on the premise that engineers use probability as a modeling tool, and that probability can be applied to the solution of engineering problems. Engineers and students studying probability and random processes also need to analyze data, and thus need some knowledge of statistics. This book is designed to provide students with a thorough grounding in probability and stochastic processes, demonstrate their applicability to real-world problems, and introduce the basics of statistics. The book's clear writing style and homework problems make it ideal for the classroom or for self-study. - Demonstrates concepts with more than 100 illustrations, including 2 dozen new drawings - Expands readers' understanding of disruptive statistics in a new chapter (chapter 8) - Provides new chapter on Introduction to Random Processes with 14 new illustrations and tables explaining key concepts. - Includes two chapters devoted to the two branches of statistics, namely descriptive statistics (chapter 8) and inferential (or inductive) statistics (chapter 9).

Dr Ibe has been teaching at U Mass since 2003. He also has more than 20 years of experience in the corporate world, most recently as Chief Technology Officer at Sineria Networks and Director of Network Architecture for Spike Broadband Corp.

Autoren/Hrsg.

Ibe, Oliver

Leseproben

1.10 Basic Combinatorial Analysis
Combinatorial analysis deals with counting the number of different ways in which an event of interest can occur. Two basic aspects of combinatorial analysis that are used in probability theory are permutation and combination. 1.10.1 Permutations
Sometimes we are interested in how the outcomes of an experiment can be arranged; that is, we are interested in the order of the outcomes of an experiment. For example, if the possible outcomes are A, B and C, we can think of six possible arrangements of these outcomes: ABC, ACB, BAC, BCA, CAB, and CBA. Each of these arrangements is called a permutation. Thus, there are six permutations of a set of three distinct objects. This number can be derived as follows: There are 3 ways of choosing the first object; after the first object has been chosen, there are two ways of choosing the second object; and after the first two objects have been chosen, there is 1 way to choose the third object. This means that there are 3 × 2 × 1 = 6 permutations. For a system of n distinct objects we can apply a similar reasoning to obtain the following number of permutations: ×n-1×n-2×…×3×2×1=n! where n! is read as “n factorial.” By convention, 0! = 1. Assume that we want to arrange the n objects taking r at a time. The problem now becomes that of finding how many possible sequences of r objects we can get from n objects, where r = n. This number is denoted by P(n, r) and defined as follows: nr=n!n-r!=n×n-1×n-2×…×n-r+1,r=1,2,…,n   (1.20) The number P(n, r) represents the number of permutations (or sequences) of r objects taken at a time from n objects when the arrangement of the objects within a given sequence is important. Note that when n = r, we obtain: nn=n!n-n!=n!0!=n! Example 1.19 A little girl has six building blocks and is required to select four of them at a time to build a model. If the order of the blocks in each model is important, how many models can she build? Solution: Since the order of objects is important, this is a permutation problem. Therefore, the number of models is given by: 64=6!6-4!=6!2!=6×5×4×3×2×12×1=360 Note that if the little girl were to select three blocks at a time, the number of permutations decreases to 120; and if she were to select two blocks at a time, the number of permutations will be 30. Example 1.20 How many words can be formed from the word SAMPLE? Assume that a formed word does not have to be an actual English word, but it may contain at most as many instances of a letter as there are in the original word (for example, “maa” is not acceptable since “a” does not appear twice in SAMPLE, but “mas” is allowed). Solution: The words can be single-letter words, two-letter words, three-letter words, four-letter words, five-letter words or six-letter words. Since the letters of the word SAMPLE are all unique, there are P(6, k) ways of forming k-letter words, k = 1, 2, …, 6. Thus, the number of words that can be formed is =P61+P(6,2)+P(6,3)+P(6,4)+P(6,5)+P(6,6)=6+30+120+360+720+720=1956 We present the following theorem without proof: Theorem Given a population of n elements, let n1, n2, …, nk be positive integers such that n1 + n2 + … + nk = n. Then there are =n!n1!×n2!×…×nk!   (1.21) ways to partition the population into k subgroups of sizes n1, n2, …, nk, respectively. Example 1.21 Five identical red blocks, two identical white blocks and three identical blue blocks are arranged in a row. How many different arrangements are possible? Solution: In this example, n1 = 5, n2 = 2, n3 = 3 and n = 5 + 2 + 3 = 10. Thus, the number of possible arrangements is given by: =10!5!×2!×3!=2520 Example 1.22 How many words can be formed by using all the letters of the word MISSISSIPPI? Solution: The word contains 11 letters consisting of 1 M, 4 S’s, 4 I’s, and 2 P’s. Thus, the number of words that can be formed is =11!1!×4!×4!×2!=34650 1.10.2 Circular Arrangement
Consider the problem of seating n people in a circle. Assume that the positions are labeled 1, 2, …, n. Then, if after one arrangement everyone moves one place either to the left or to the right, that will also be an arrangement because each person is occupying a new location. However, each person’s previous neighbors to the left and right are still his/her neighbors in the new arrangement. This means that such a move does not lead to a new valid arrangement. To solve this problem, one person must remain fixed while the others move. Thus, the number of people being arranged is n - 1, which means that the number of possible arrangements is (n - 1)! For example, the number of ways that 10 people can be seated in a circle is (10 - 1)! = 9 ! = 362880. 1.10.3 Applications of Permutations in Probability
Consider a system that contains n distinct objects labeled a1, a2, …, an. Assume that we choose r of these objects in the following manner. We choose the first object, record its type and put it back into the “population.” We then choose the second object, record its type and put it back into the population. We continue this process until we have chosen a total of r objects. This gives an “ordered sample” consisting of r of the n objects. The question is to determine the number of distinct ordered samples that can be obtained, where two ordered samples are said to be distinct if they differ in at least one entry in a particular position within the samples. Since the number of ways of choosing an object in each round is n, the total number of distinct samples is n × n × … × n = nr. Assume now that the sampling is done without replacement. That is, after an object has been chosen, it is not put back into the population. Then the next object from the remainder of the population is chosen and not replaced, and so on until all the r objects have been chosen. The total number of possible ways of making this sampling can be obtained by noting that there are n ways to choose the first object, n - 1 ways to choose the second object, n - 2 ways to choose the third object, and so on, and finally there are n - r + 1 ways to choose the rth object. Thus, the total number of distinct samples is n × (n - 1) × (n - 2) × … × (n - r + 1). Example 1.23 A subway train consists of n cars. The number of passengers waiting to board the train is k < n and each passenger enters a car at random. What is the probability that all the k passengers end up in different cars of the train? Solution: The probability p of an event of this nature where a constraint or restriction is imposed on the participants is given by =NumberofRestrictedorConstrainedArrangementsNumberofUnrestrictedorUnconstrainedArrangements Now, without any restriction on the occupancy of the cars, each of the k passengers can enter any one of the n cars. Thus, the number of distinct, unrestricted arrangements of the passengers in the cars is N = n × n × … × n = nk. If the passengers enter the cars in such a way that there is no more than one passenger in a car, then the first passenger can...