Baixe Probability Demystified Bluman e outras Manuais, Projetos, Pesquisas em PDF para Economia, somente na Docsity! orobahilit A SELF-TEACHING GUIDE Common devices illustrate the BASIC CONCEPTS for EASY understanding + 4 NO high-powered math knowledge needed — A perfect INTRODUCTION to actuarial sciences Es Chapter-ending QUIZZES and a final EXAM EM Allan G. Bluman PROBABILITY DEMYSTIFIED Copyright © 2005 by The McGraw-Hill Companies, Inc. All rights reserved. Manufactured in the United States of America. Except as permitted under the United States Copyright Act of 1976, no part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written permission of the publisher. 0-07-146999-0 The material in this eBook also appears in the print version of this title: 0-07-144549-8. All trademarks are trademarks of their respective owners. Rather than put a trademark symbol after every occurrence of a trademarked name, we use names in an editorial fashion only, and to the benefit of the trademark owner, with no intention of infringement of the trademark. Where such designations appear in this book, they have been printed with initial caps. McGraw-Hill eBooks are available at special quantity discounts to use as premiums and sales promotions, or for use in corporate training programs. For more information, please contact George Hoare, Special Sales, at george_hoare@mcgraw-hill.com or (212) 904-4069. TERMS OF USE This is a copyrighted work and The McGraw-Hill Companies, Inc. (“McGraw-Hill”) and its licensors reserve all rights in and to the work. Use of this work is subject to these terms. Except as permitted under the Copyright Act of 1976 and the right to store and retrieve one copy of the work, you may not decompile, disassemble, reverse engineer, reproduce, modify, create derivative works based upon, transmit, distribute, disseminate, sell, publish or sublicense the work or any part of it without McGraw- Hill’s prior consent. You may use the work for your own noncommercial and personal use; any other use of the work is strictly prohibited. Your right to use the work may be terminated if you fail to comply with these terms. THE WORK IS PROVIDED “AS IS.” McGRAW-HILL AND ITS LICENSORS MAKE NO GUARANTEES OR WARRANTIES AS TO THE ACCURACY, ADEQUACY OR COMPLETENESS OF OR RESULTS TO BE OBTAINED FROM USING THE WORK, INCLUDING ANY INFORMATION THAT CAN BE ACCESSED THROUGH THE WORK VIA HYPERLINK OR OTHERWISE, AND EXPRESSLY DISCLAIM ANY WARRANTY, EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO IMPLIED WARRANTIES OF MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. McGraw-Hill and its licensors do not warrant or guarantee that the functions contained in the work will meet your requirements or that its operation will be uninterrupted or error free. Neither McGraw-Hill nor its licensors shall be liable to you or anyone else for any inaccuracy, error or omission, regardless of cause, in the work or for any damages resulting therefrom. McGraw-Hill has no responsibility for the content of any information accessed through the work. Under no circumstances shall McGraw-Hill and/or its licensors be liable for any indirect, incidental, special, punitive, consequential or similar damages that result from the use of or inability to use the work, even if any of them has been advised of the possibility of such damages. This limitation of liability shall apply to any claim or cause whatsoever whether such claim or cause arises in contract, tort or otherwise. DOI: 10.1036/0071469990 �� Want to learn more? We hope you enjoy this McGraw-Hill eBook! If you’d like more information about this book, its author, or related books and websites, please click here. To all of my teachers, whose examples instilled in me my love of mathematics and teaching. . This book is not a book on how to win at gambling! This book presents the basic concepts of probability in a simple, straightforward, easy-to-understand way. It does require, however, a knowledge of arithmetic (fractions, decimals, and percents) and a knowledge of basic algebra (formulas, exponents, order of operations, etc.). If you need a review of these concepts, you can consult another of my books in this series entitled Pre-Algebra Demystified. This book can be used to gain a knowledge of the basic concepts of probability theory, either as a self-study guide or as a supplementary textbook for those who are taking a course in probability or a course in statistics that has a section on probability. The basic concepts of probability are explained in the first two chapters. Then the addition and multiplication rules are explained. Following that, the concepts of odds and expectation are explained. The counting rules are explained in Chapter 6, and they are needed for the binomial and other probability distributions found in Chapters 7 and 8. The relationship between probability and the normal distribution is presented in Chapter 9. Finally, a recent development, the Monte Carlo method of simulation, is explained in Chapter 10. Chapter 11 explains how probability can be used in game theory and Chapter 12 explains how probability is used in actuarial science. Special material on Bayes’ Theorem is presented in the Appendix because this concept is somewhat more difficult than the other concepts presented in this book. In addition to addressing the concepts of probability, each chapter ends with what is called a ‘‘Probability Sidelight.’’ These sections cover some of the historical aspects of the development of probability theory or some commentary on how probability theory is used in gambling and everyday life. I have spent my entire career teaching mathematics at a level that most students can understand and appreciate. I have written this book with the same objective in mind. Mathematical precision, in some cases, has been sacrificed in the interest of presenting probability theory in a simplified way. Good luck! Allan G. Bluman PREFACEx ACKNOWLEDGMENTS I would like to thank my wife, Betty Claire, for helping me with the prepara- tion of this book and my editor, Judy Bass, for her assistance in its pub- lication. I would also like to thank Carrie Green for her error checking and helpful suggestions. xi Copyright © 2005 by The McGraw-Hill Companies, Inc. Click here for terms of use. CHAPTER 1 Basic Concepts Introduction Probability can be defined as the mathematics of chance. Most people are familiar with some aspects of probability by observing or playing gambling games such as lotteries, slot machines, black jack, or roulette. However, probability theory is used in many other areas such as business, insurance, weather forecasting, and in everyday life. In this chapter, you will learn about the basic concepts of probability using various devices such as coins, cards, and dice. These devices are not used as examples in order to make you an astute gambler, but they are used because they will help you understand the concepts of probability. 1 Copyright © 2005 by The McGraw-Hill Companies, Inc. Click here for terms of use. EXAMPLE: Two coins are tossed; find the probability that both coins land heads up. SOLUTION: The sample space for tossing two coins is HH, HT, TH, and TT. Since there are 4 events in the sample space, and only one way to get two heads (HH), the answer is PðHHÞ ¼ 1 4 EXAMPLE: A die is tossed; find the probability of each event: a. Getting a two b. Getting an even number c. Getting a number less than 5 SOLUTION: The sample space is 1, 2, 3, 4, 5, 6, so there are six outcomes in the sample space. a. P(2) ¼ 1 6 , since there is only one way to obtain a 2. b. P(even number) ¼ 3 6 ¼ 1 2 , since there are three ways to get an odd number, 1, 3, or 5. c. P(number less than 5Þ ¼ 4 6 ¼ 2 3 , since there are four numbers in the sample space less than 5. EXAMPLE: A dish contains 8 red jellybeans, 5 yellow jellybeans, 3 black jellybeans, and 4 pink jellybeans. If a jellybean is selected at random, find the probability that it is a. A red jellybean b. A black or pink jellybean c. Not yellow d. An orange jellybean CHAPTER 1 Basic Concepts4 SOLUTION: There are 8+5+3+4=20 outcomes in the sample space. a. PðredÞ ¼ 8 20 ¼ 2 5 b. Pðblack or pinkÞ ¼ 3þ 4 20 ¼ 7 20 c. P(not yellow)=P(red or black or pink) ¼ 8þ 3þ 4 20 ¼ 15 20 ¼ 3 4 d. P(orange)= 0 20 ¼ 0, since there are no orange jellybeans. Probabilities can be expressed as reduced fractions, decimals, or percents. For example, if a coin is tossed, the probability of getting heads up is 12 or 0.5 or 50%. (Note: Some mathematicians feel that probabilities should be expressed only as fractions or decimals. However, probabilities are often given as percents in everyday life. For example, one often hears, ‘‘There is a 50% chance that it will rain tomorrow.’’) Probability problems use a certain language. For example, suppose a die is tossed. An event that is specified as ‘‘getting at least a 3’’ means getting a 3, 4, 5, or 6. An event that is specified as ‘‘getting at most a 3’’ means getting a 1, 2, or 3. Probability Rules There are certain rules that apply to classical probability theory. They are presented next. Rule 1: The probability of any event will always be a number from zero to one. This can be denoted mathematically as 0P(E) 1. What this means is that all answers to probability problems will be numbers ranging from zero to one. Probabilities cannot be negative nor can they be greater than one. Also, when the probability of an event is close to zero, the occurrence of the event is relatively unlikely. For example, if the chances that you will win a certain lottery are 0.00l or one in one thousand, you probably won’t win, unless of course, you are very ‘‘lucky.’’ When the probability of an event is 0.5 or 12, there is a 50–50 chance that the event will happen—the same CHAPTER 1 Basic Concepts 5 probability of the two outcomes when flipping a coin. When the probability of an event is close to one, the event is almost sure to occur. For example, if the chance of it snowing tomorrow is 90%, more than likely, you’ll see some snow. See Figure 1-1. Rule 2: When an event cannot occur, the probability will be zero. EXAMPLE: A die is rolled; find the probability of getting a 7. SOLUTION: Since the sample space is 1, 2, 3, 4, 5, and 6, and there is no way to get a 7, P(7)¼ 0. The event in this case has no outcomes when the sample space is considered. Rule 3: When an event is certain to occur, the probability is 1. EXAMPLE: A die is rolled; find the probability of getting a number less than 7. SOLUTION: Since all outcomes in the sample space are less than 7, the probability is 6 6 ¼1. Rule 4: The sum of the probabilities of all of the outcomes in the sample space is 1. Referring to the sample space for tossing two coins (HH, HT, TH, TT), each outcome has a probability of 14 and the sum of the probabilities of all of the outcomes is 1 4 þ 1 4 þ 1 4 þ 1 4 ¼ 4 4 ¼ 1: Fig. 1-1. CHAPTER 1 Basic Concepts6 8. On a roulette wheel there are 38 sectors. Of these sectors, 18 are red, 18 are black, and 2 are green. When the wheel is spun, find the probability that the ball will land on a. Red. b. Green. 9. A person has a penny, a nickel, a dime, a quarter, and a half-dollar in his pocket. If a coin is selected at random, find the probability that the coin is a. A quarter. b. A coin whose amount is greater than five cents. c. A coin whose denomination ends in a zero. 10. Six women and threemen are employed in a real estate office. If a person is selected at random to get lunch for the group, find the probability that the person is a man. ANSWERS 1. The sample space is $1, $2, $5, $10, $20. a. P($10)= 1 5 . b. P(bill greater than $2)= 3 5 , since $5, $10, and $20 are greater than $2. c. P($50)= 0 5 ¼ 0, since there is no $50 bill. d. P(bill is odd)= 2 5 , since $1 and $5 are odd denominational bills. e. P(number is divisible by 5)= 3 5 , since $5, $10, and $20 are divisible by 5. 2. The sample space is 1, 2, 3, 4, 5, 6. a. P(2)= 1 6 , since there is only one 2 in the sample space. b. P(number greater than 2)= 4 6 ¼ 2 3 , since there are 4 numbers in the sample space greater than 2. CHAPTER 1 Basic Concepts 9 c. P(number less than 1)= 0 6 ¼ 0, since there are no numbers in the sample space less than 1. d. P(number is an odd number)= 3 6 ¼ 1 2 , since 1, 3, and 5 are odd numbers. 3. The sample space is 1, 2, 3, 4, 5, 6, 7, 8, 9. a. P(number divisible by 3)= 3 9 ¼ 1 3 , since 3, 6, and 9 are divisible by 3. b. P(number greater than 7)= 2 9 , since 8 and 9 are greater than 7. c. P(even number)= 4 9 , since 2, 4, 6, and 8 are even numbers. 4. The sample space is HH, HT, TH, TT. a. P(TT)= 1 4 , since there is only one way to get two tails. b. P(at least one head)= 3 4 , since there are three ways (HT, TH, HH) to get at least one head. c. P(HH)= 1 4 , since there is only one way to get two heads. 5. The sample space is A˘, 2^, 3¨, 4˘, 5¯, 6¨. a. P(4˘)= 1 6 . b. P(red card)= 3 6 ¼ 1 2 , since there are three red cards. c. P(club)= 2 6 ¼ 1 3 , since there are two clubs. 6. The sample space is red, blue, green, and white. a. P(blue)= 1 4 , since there is only one blue ball. b. P(red or blue)= 2 4 = 1 2 , since there are two outcomes in the event. c. P(pink)= 0 6 ¼ 0, since there is no pink ball. 7. The sample space consists of the letters in ‘‘computer.’’ a. P(t)= 1 8 . b. P(o or m)= 2 8 ¼ 1 4 . c. P(x)= 0 8 ¼ 0, since there are no ‘‘x’’s in the word. d. P(vowel)= 3 8 , since o, u, and e are the vowels in the word. CHAPTER 1 Basic Concepts10 8. There are 38 outcomes: a. P(red)= 18 38 ¼ 9 19 . b. P(green)= 2 38 ¼ 1 19 . 9. The sample space is 1c= , 5c= , 10c= , 25c= , 50c= . a. P(25c= )¼ 1 5 . b. P(greater than 5c= )¼ 3 5 . c. P(denomination ends in zero)¼ 2 5 . 10. The sample space consists of six women and three men. PðmanÞ ¼ 3 9 ¼ 1 3 : Empirical Probability Probabilities can be computed for situations that do not use sample spaces. In such cases, frequency distributions are used and the probability is called empirical probability. For example, suppose a class of students consists of 4 freshmen, 8 sophomores, 6 juniors, and 7 seniors. The information can be summarized in a frequency distribution as follows: Rank Frequency Freshmen 4 Sophomores 8 Juniors 6 Seniors 7 TOTAL 25 From a frequency distribution, probabilities can be computed using the following formula. CHAPTER 1 Basic Concepts 11 PRACTICE 1. A recent survey found that the ages of workers in a factory is distrib- uted as follows: Age Number 20–29 18 30–39 27 40–49 36 50–59 16 60 or older 3 Total 100 If a person is selected at random, find the probability that the person is a. 40 or older. b. Under 40 years old. c. Between 30 and 39 years old. d. Under 60 but over 39 years old. 2. In a sample of 50 people, 19 had type O blood, 22 had type A blood, 7 had type B blood, and 2 had type AB blood. If a person is selected at random, find the probability that the person a. Has type A blood. b. Has type B or type AB blood. c. Does not have type O blood. d. Has neither type A nor type O blood. 3. In a recent survey of 356 children aged 19–24 months, it was found that 89 ate French fries. If a child is selected at random, find the probability that he or she eats French fries. 4. In a classroom of 36 students, 8 were liberal arts majors and 7 were history majors. If a student is selected at random, find the probability that the student is neither a liberal arts nor a history major. 5. A recent survey found that 74% of those questioned get some of the news from the Internet. If a person is selected at random, find the probability that the person does not get any news from the Internet. CHAPTER 1 Basic Concepts14 ANSWERS 1. a. P(40 or older)¼ 36þ 16þ 3 100 ¼ 55 100 ¼ 11 20 b. P(under 40)¼ 18þ 27 100 ¼ 45 100 ¼ 9 20 c. P(between 30 and 39)¼ 27 100 d. P(under 60 but over 39)¼ 36þ 16 100 ¼ 52 100 ¼ 13 25 2. The total number of outcomes in this sample space is 50. a. PðAÞ ¼ 22 50 ¼ 11 25 b. PðB or ABÞ ¼ 7þ 2 50 ¼ 9 50 c. P(not O)¼ 1 19 50 ¼ 31 50 d. P(neither A nor O)¼P(AB or B)¼ 2þ 7 50 ¼ 9 50 3. P(French fries)¼ 89 356 ¼ 1 4 4. P(neither liberal arts nor history)¼ 1 8þ 7 36 ¼ 1 15 36 ¼ 21 36 ¼ 7 12 5. P(does not get any news from the Internet)¼ 1 0.74¼ 0.26 Law of Large Numbers We know from classical probability that if a coin is tossed one time, we cannot predict the outcome, but the probability of getting a head is 12 and the probability of getting a tail is 12 if everything is fair. But what happens if we toss the coin 100 times? Will we get 50 heads? Common sense tells us that CHAPTER 1 Basic Concepts 15 most of the time, we will not get exactly 50 heads, but we should get close to 50 heads. What will happen if we toss a coin 1000 times? Will we get exactly 500 heads? Probably not. However, as the number of tosses increases, the ratio of the number of heads to the total number of tosses will get closer to 12. This phenomenon is known as the law of large numbers. This law holds for any type of gambling game such as rolling dice, playing roulette, etc. Subjective Probability A third type of probability is called subjective probability. Subjective probability is based upon an educated guess, estimate, opinion, or inexact information. For example, a sports writer may say that there is a 30% probability that the Pittsburgh Steelers will be in the Super Bowl next year. Here the sports writer is basing his opinion on subjective information such as the relative strength of the Steelers, their opponents, their coach, etc. Subjective probabilities are used in everyday life; however, they are beyond the scope of this book. Summary Probability is the mathematics of chance. There are three types of probability: classical probability, empirical probability, and subjective probability. Classical probability uses sample spaces. A sample space is the set of outcomes of a probability experiment. The range of probability is from 0 to 1. If an event cannot occur, its probability is 0. If an event is certain to occur, its probability is 1. Classical probability is defined as the number of ways (outcomes) the event can occur divided by the total number of outcomes in the sample space. Empirical probability uses frequency distributions, and it is defined as the frequency of an event divided by the total number of frequencies. Subjective probability is made by a person’s knowledge of the situation and is basically an educated guess as to the chances of an event occurring. CHAPTER 1 Basic Concepts16 12. If a letter is selected at random from the word ‘‘Mississippi,’’ find the probability that it is an ‘‘s.’’ a. 1 8 b. 1 2 c. 3 11 d. 4 11 13. When a die is rolled, the probability of getting an 8 is a. 1 6 b. 0 c. 1 d. 1 1 2 14. In a survey of 180 people, 74 were over the age of 64. If a person is selected at random, what is the probability that the person is over 64? a. 16 45 b. 32 37 c. 37 90 d. 53 90 15. In a classroom of 24 students, there were 20 freshmen. If a student is selected at random, what is the probability that the student is not a freshman? a. 2 3 CHAPTER 1 Basic Concepts 19 b. 5 6 c. 1 3 d. 1 6 (The answers to the quizzes are found on pages 242–245.) Probability Sidelight BRIEF HISTORY OF PROBABILITY The concepts of probability are as old as humans. Paintings in tombs excavated in Egypt showed that people played games based on chance as early as 1800 B.C.E. One game was called ‘‘Hounds and Jackals’’ and is similar to the present-day game of ‘‘Snakes and Ladders.’’ Ancient Greeks and Romans made crude dice from various items such as animal bones, stones, and ivory. When some of these items were tested recently, they were found to be quite accurate. These crude dice were also used in fortune telling and divination. Emperor Claudius (10 BCE–54 CE) is said to have written a book entitled How To Win at Dice. He liked playing dice so much that he had a special dice board in his carriage. No formal study of probability was done until the 16th century when Girolamo Cardano (1501–1576) wrote a book on probability entitled The Book on Chance and Games. Cardano was a philosopher, astrologer, physician, mathematician, and gambler. In his book, he also included techniques on how to cheat and how to catch others who are cheating. He is believed to be the first mathematician to formulate a definition of classical probability. During the mid-1600s, a professional gambler named Chevalier de Mere made a considerable amount of money on a gambling game. He would bet unsuspecting patrons that in four rolls of a die, he could obtain at least one 6. He was so successful at winning that word got around, and people refused to play. He decided to invent a new game in order to keep winning. He would bet patrons that if he rolled two dice 24 times, he would get at least one double 6. However, to his dismay, he began to lose more often than he would win and lost money. CHAPTER 1 Basic Concepts20 Unable to figure out why he was losing, he asked a renowned mathematician, Blaise Pascal (1623–1662) to study the game. Pascal was a child prodigy when it came to mathematics. At the age of 14, he participated in weekly meetings of the mathematicians of the French Academy. At the age of 16, he invented a mechanical adding machine. Becauseof thediceproblem,Pascalbecame interested in studyingprobability and began a correspondence with a French government official and fellow mathematician, Pierre de Fermat (1601–1665). Together the two were able to solve deMere’s dilemma and formulate the beginnings of probability theory. In 1657, a Dutch mathematician named Christian Huygens wrote a treatise on the Pascal–Fermat correspondence and introduced the idea of mathemat- ical expectation. (See Chapter 5.) Abraham de Moivre (1667–1754) wrote a book on probability entitled Doctrine of Chances in 1718. He published a second edition in 1738. Pierre Simon Laplace (1749–1827) wrote a book and a series of supplements on probability from 1812 to 1825. His purpose was to acquaint readers with the theory of probability and its applications, using everyday language. He also stated that the probability that the sun will rise tomorrow is 1,826,214 1,826,215 . Simeon-Denis Poisson (1781–1840) developed the concept of the Poisson distribution. (See Chapter 8.) Also during the 1800s a mathematician named Carl Friedrich Gauss (1777–1855) developed the concepts of the normal distribution. Earlier work on the normal distribution was also done by de Moivre and Laplace, unknown to Gauss. (See Chapter 9.) In 1895, the Fey Manufacturing Company of San Francisco invented the first automatic slot machine. These machines consisted of three wheels that were spun when a handle on the side of the machine was pulled. Each wheel contained 20 symbols; however, the number of each type of symbols was not the same on each wheel. For example, the first wheel may have 6 oranges, while the second wheel has 3 oranges, and the third wheel has only one. When a person gets two oranges, the person may think that he has almost won by getting 2 out of 3 equitable symbols, while the real probability of winning is much smaller. In the late 1940s, two mathematicians, Jon von Neumann and Stanislaw Ulam used a computer to simulate probability experiments. This method is called the Monte Carlo method. (See Chapter 10.) Today probability theory is used in insurance, gambling, war gaming, the stock market, weather forecasting, and many other areas. CHAPTER 1 Basic Concepts 21 EXAMPLE: A coin is tossed and a die is rolled. Find the probability of getting a. A head on the coin and a 3 on the die. b. A head on the coin. c. A 4 on the die. SOLUTION: a. Since there are 12 outcomes in the sample space and only one way to get a head on the coin and a three on the die, PðH3Þ ¼ 1 12 b. Since there are six ways to get a head on the coin, namely H1, H2, H3, H4, H5, and H6, Pðhead on the coin) ¼ 6 12 ¼ 1 2 c. Since there are two ways to get a 4 on the die, namely H4 and T4, Pð4 on the die) ¼ 2 12 ¼ 1 6 EXAMPLE: Three coins are tossed. Draw a tree diagram and find the sample space. SOLUTION: Each coin can land either heads up (H) or tails up (T); therefore, the tree diagram will consist of three parts and each part will have two branches. See Figure 2-2. Fig. 2-2. CHAPTER 2 Sample Spaces24 Hence the sample space is HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. Once the sample space is found, probabilities can be computed. EXAMPLE: Three coins are tossed. Find the probability of getting a. Two heads and a tail in any order. b. Three heads. c. No heads. d. At least two tails. e. At most two tails. SOLUTION: a. There are eight outcomes in the sample space, and there are three ways to get two heads and a tail in any order. They are HHT, HTH, and THH; hence, P(2 heads and a tail) ¼ 3 8 b. Three heads can occur in only one way; hence PðHHHÞ ¼ 1 8 c. The event of getting no heads can occur in only one way—namely TTT; hence, PðTTTÞ ¼ 1 8 d. The event of at least two tails means two tails and one head or three tails. There are four outcomes in this event—namely TTH, THT, HTT, and TTT; hence, P(at least two tails) ¼ 4 8 ¼ 1 2 e. The event of getting at most two tails means zero tails, one tail, or two tails. There are seven outcomes in this event—HHH, THH, HTH, HHT, TTH, THT, and HTT; hence, P(at most two tails) ¼ 7 8 When selecting more than one object from a group of objects, it is important to know whether or not the object selected is replaced before drawing the second object. Consider the next two examples. CHAPTER 2 Sample Spaces 25 EXAMPLE: A box contains a red ball (R), a blue ball (B), and a yellow ball (Y). Two balls are selected at random in succession. Draw a tree diagram and find the sample space if the first ball is replaced before the second ball is selected. SOLUTION: There are three ways to select the first ball. They are a red ball, a blue ball, or a yellow ball. Since the first ball is replaced before the second one is selected, there are three ways to select the second ball. They are a red ball, a blue ball, or a yellow ball. The tree diagram is shown in Figure 2-3. The sample space consists of nine outcomes. They are RR, RB, RY, BR, BB, BY, YR, YB, YY. Each outcome has a probability of 19: Now what happens if the first ball is not replaced before the second ball is selected? EXAMPLE: A box contains a red ball (R), a blue ball (B), and a yellow ball (Y). Two balls are selected at random in succession. Draw a tree diagram and find the sample space if the first ball is not replaced before the second ball is selected. SOLUTION: There are three outcomes for the first ball. They are a red ball, a blue ball, or a yellow ball. Since the first ball is not replaced before the second ball is drawn, there are only two outcomes for the second ball, and these outcomes depend on the color of the first ball selected. If the first ball selected is blue, then the second ball can be either red or yellow, etc. The tree diagram is shown in Figure 2-4. Fig. 2-3. CHAPTER 2 Sample Spaces26 3. There are nine outcomes in the sample space. a. Pð$6Þ ¼ 2 9 since $1+$5, and $5+$1 equal $6. b. Pðgreater than $10) ¼ 5 9 since there are five ways to get a sum greater than $10. c. Pðless than $15) ¼ 6 9 ¼ 2 3 since there are six ways to get a sum lesser than $15. 4. There are eight outcomes in the sample space. a. Pð3 girls) ¼ 1 8 since three girls is GGG. b. P(2 boys and one girl in any order) ¼ 3 8 since there are threeways to get two boys and one girl in any order. They are BBG, BGB, and GBB. Fig. 2-8. Fig. 2-7. CHAPTER 2 Sample Spaces 29 c. P(at least 2 boys) ¼ 4 8 ¼ 1 2 since at least two boys means two or three boys. The outcomes are BBG, BGB, GBB, and BBB. 5. The probability that one marble is white is 4 6 ¼ 2 3 since the outcomes are WB, WG, BW, and GW. Tables Another way to find a sample space is to use a table. EXAMPLE: Find the sample space for selecting a card from a standard deck of 52 cards. SOLUTION: There are four suits—hearts and diamonds, which are red, and spades and clubs, which are black. Each suit consists of 13 cards—ace through king. Hence, the sample space can be shown using a table. See Figure 2-10. Face cards are kings, queens, and jacks. Once the sample space is found, probabilities for events can be computed. Fig. 2-9. Fig. 2-10. CHAPTER 2 Sample Spaces30 EXAMPLE: A single card is drawn at random from a standard deck of cards. Find the probability that it is a. The 4 of diamonds. b. A queen. c. A 5 or a heart. SOLUTION: a. The sample space consists of 52 outcomes and only one outcome is the four of diamonds; hence, Pð4^Þ ¼ 1 52 b. Since there are four queens (one of each suit), PðQÞ ¼ 4 52 ¼ 1 13 c. In this case, there are 13 hearts and 4 fives; however, the 5˘ has been counted twice, so the number of ways to get a 5 or a heart is 13þ 4 1¼ 16. Hence, Pð5 or ˘Þ ¼ 16 52 ¼ 4 13 : A table can be used for the sample space when two dice are rolled. Since the first die can land in 6 ways and the second die can land in 6 ways, there are 6 6 or 36 outcomes in the sample space. It does not matter whether the two dice are of the same color or different color. The sample space is shown in Figure 2-11. Fig. 2-11. CHAPTER 2 Sample Spaces 31 ANSWERS 1. There are 52 outcomes in the sample space. a. There are four 9s, so Pð9Þ ¼ 4 52 ¼ 1 13 b. There is only one ace of diamonds, so PðA^Þ ¼ 1 52 c. There are 13 clubs, so Pð¨Þ ¼ 13 52 ¼ 1 4 2. There are 52 outcomes in the sample space. a. There are 26 black cards: they are 13 clubs and 13 spades, so Pðblack card) ¼ 26 52 ¼ 1 2 b. There are two red queens: they are the queen of diamonds and the queen of hearts, so Pðred queen) = 2 52 ¼ 1 26 c. There are 13 hearts and 13 spades, so Pð˘ or ¨Þ ¼ 26 52 ¼ 1 2 3. There are 52 outcomes in the sample space. a. There are 13 diamonds and 12 face cards, but the jack, queen, and king of diamonds have been counted twice, so P(diamond or face card) ¼ 13þ 12 4 52 ¼ 21 52 b. There are 13 clubs and four 8s, but the 8 of clubs has been counted twice, so Pð¨ or an 8Þ ¼ 13þ 4 1 52 ¼ 16 52 ¼ 4 13 CHAPTER 2 Sample Spaces34 c. There are 26 red cards and four 6s, but the 6 of hearts and the 6 of diamonds have been counted twice, so Pðred card or 6) ¼ 26þ 4 2 52 ¼ 28 52 ¼ 7 13 4. There are 36 outcomes in the sample space. a. There are six ways to get a sum of seven. They are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1); hence, P(sum of 7) ¼ 6 36 ¼ 1 6 b. A sum greater than 8 means a sum of 9, 10, 11, 12, so P(sum greater than 8) ¼ 10 36 ¼ 5 18 c. A sum less than or equal to five means a sum of five, four, three or two. There are ten ways to get a sum less than or equal to five; hence, Pðsum less than or equal to fiveÞ ¼ 10 36 ¼ 5 18 5. There are 36 outcomes in the sample space. a. There are 11 ways to get a 5 on one or both dice. They are (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (5, 6), (5, 4), (5, 3) (5, 2), and (5, 1); hence, P(5 on one or both dice) ¼ 11 36 b. There are 0 ways to get a sum greater than 12; hence, P(sum greater than 12) ¼ 0 36 ¼ 0 The event is impossible. c. Since all sums are less than 13 when two dice are rolled, there are 36 ways to get a sum less than 13; hence, P(sum less than 13) ¼ 36 36 ¼ 1 The event is certain. CHAPTER 2 Sample Spaces 35 Summary Two devices can be used to represent sample spaces. They are tree diagrams and tables. A tree diagram can be used to determine the outcome of a probability experiment. A tree diagram consists of branches corresponding to the outcomes of two or more probability experiments that are done in sequence. Sample spaces can also be represented by using tables. For example, the outcomes when selecting a card from an ordinary deck can be represented by a table. When two dice are rolled, the 36 outcomes can be represented by using a table. Once a sample space is found, probabilities can be computed for specific events. CHAPTER QUIZ 1. When a coin is tossed and then a die is rolled, the probability of getting a tail on the coin and an odd number on the die is a. 1 2 b. 1 4 c. 3 4 d. 1 12 2. When a coin is tossed and a die is rolled, the probability of getting a head and a number less than 5 on the die is a. 1 3 b. 2 3 c. 1 2 d. 5 6 CHAPTER 2 Sample Spaces36 9. A card is drawn from an ordinary deck of 52 cards. The probability that it is a spade is a. 1 4 b. 1 13 c. 1 52 d. 1 26 10. A card is drawn from an ordinary deck of 52 cards. The probability that it is a 9 or a club is a. 17 52 b. 5 8 c. 4 13 d. 3 4 11. A card is drawn from an ordinary deck of 52 cards. The probability that it is a face card is a. 3 52 b. 1 4 c. 9 13 d. 3 13 CHAPTER 2 Sample Spaces 39 12. Two dice are rolled. The probability that the sum of the spots on the faces will be nine is a. 1 9 b. 5 36 c. 1 6 d. 3 13 13. Two dice are rolled. The probability that the sum of the spots on the faces is greater than seven is a. 2 3 b. 7 36 c. 3 4 d. 5 12 14. Two dice are rolled. The probability that one or both numbers on the faces will be 4 is a. 1 3 b. 4 13 c. 11 36 d. 1 6 CHAPTER 2 Sample Spaces40 15. Two dice are rolled. The probability that the sum of the spots on the faces will be even is a. 3 4 b. 5 6 c. 1 2 d. 1 6 Probability Sidelight HISTORY OF DICE AND CARDS Dice are one of the earliest known gambling devices used by humans. They have been found in ancient Egyptian tombs and in the prehistoric caves of people in Europe and America. The first dice were made from animal bones—namely the astragalus or the heel bone of a hoofed animal. These bones are very smooth and easily carved. The astragalus had only four sides as opposed to modern cubical dice that have six sides. The astragalus was used for fortune telling, gambling, and board games. By 3000 B.C.E. the Egyptians had devised many board games. Ancient tomb paintings show pharaohs playing board games, and a game similar to today’s ‘‘Snakes and Ladders’’ was found in an Egyptian tomb dating to 1800 B.C.E. Eventually crude cubic dice evolved from the astragalus. The dice were first made from bones, then clay, wood, and finally polished stones. Dots were used instead of numbers since writing numbers was very complicated at that time. It was thought that the outcomes of rolled dice were controlled by the gods that the people worshipped. As one story goes, the Romans incorrectly reasoned that there were three ways to get a sum of seven when two dice are rolled. They are 6 and 1, 5 and 2, and 4 and 3. They also reasoned incorrectly that there were three ways to get a sum of six: 5 and 1, 4 and 2, and 3 and 3. CHAPTER 2 Sample Spaces 41 Mutually Exclusive Events Many problems in probability involve finding the probability of two or more events. For example, when a card is selected at random from a deck, what is the probability that the card is a king or a queen? In this case, there are two situations to consider. They are: 1. The card selected is a king 2. The card selected is a queen Now consider another example. When a card is selected from a deck, find the probability that the card is a king or a diamond. In this case, there are three situations to consider: 1. The card is a king 2. The card is a diamond 3. The card is a king and a diamond. That is, the card is the king of diamonds. The difference is that in the first example, a card cannot be both a king and a queen at the same time, whereas in the second example, it is possible for the card selected to be a king and a diamond at the same time. In the first example, we say the two events are mutually exclusive. In the second example, we say the two events are not mutually exclusive. Two events then are mutually exclusive if they cannot occur at the same time. In other words, the events have no common outcomes. EXAMPLE: Which of these events are mutually exclusive? a. Selecting a card at random from a deck and getting an ace or a club b. Rolling a die and getting an odd number or a number less than 4 c. Rolling two dice and getting a sum of 7 or 11 d. Selecting a student at random who is full-time or part-time e. Selecting a student who is a female or a junior SOLUTION: a. No. The ace of clubs is an outcome of both events. b. No. One and three are common outcomes. c. Yes d. Yes e. No. A female student who is a junior is a common outcome. CHAPTER 3 The Addition Rules44 Addition Rule I The probability of two or more events occurring can be determined by using the addition rules. The first rule is used when the events are mutually exclusive. Addition Rule I: When two events are mutually exclusive, PðA or BÞ ¼ PðAÞ þ PðBÞ EXAMPLE: When a die is rolled, find the probability of getting a 2 or a 3. SOLUTION: As shown in Chapter 1, the problem can be solved by looking at the sample space, which is 1, 2, 3, 4, 5, 6. Since there are 2 favorable outcomes from 6 outcomes, P(2 or 3)¼ 26 ¼ 1 3. Since the events are mutually exclusive, addition rule 1 also can be used: Pð2 or 3Þ ¼ Pð2Þ þ Pð3Þ ¼ 1 6 þ 1 6 ¼ 2 6 ¼ 1 3 EXAMPLE: In a committee meeting, there were 5 freshmen, 6 sophomores, 3 juniors, and 2 seniors. If a student is selected at random to be the chairperson, find the probability that the chairperson is a sophomore or a junior. SOLUTION: There are 6 sophomores and 3 juniors and a total of 16 students. P(sophomore or junior) ¼ PðsophomoreÞ þ Pð juniorÞ ¼ 6 16 þ 3 16 ¼ 9 16 EXAMPLE: A card is selected at random from a deck. Find the probability that the card is an ace or a king. SOLUTION: P(ace or king) ¼ PðaceÞ þ PðkingÞ ¼ 4 52 þ 4 52 ¼ 8 52 ¼ 2 13 The word or is the key word, and it means one event occurs or the other event occurs. CHAPTER 3 The Addition Rules 45 PRACTICE 1. In a box there are 3 red pens, 5 blue pens, and 2 black pens. If a person selects a pen at random, find the probability that the pen is a. A blue or a red pen. b. A red or a black pen. 2. A small automobile dealer has 4 Buicks, 7 Fords, 3 Chryslers, and 6 Chevrolets. If a car is selected at random, find the probability that it is a. A Buick or a Chevrolet. b. A Chrysler or a Chevrolet. 3. In a model railroader club, 23 members model HO scale, 15 members model N scale, 10 members model G scale, and 5 members model O scale. If a member is selected at random, find the probability that the member models a. N or G scale. b. HO or O scale. 4. A package of candy contains 8 red pieces, 6 white pieces, 2 blue pieces, and 4 green pieces. If a piece is selected at random, find the probability that it is a. White or green. b. Blue or red. 5. On a bookshelf in a classroom there are 6 mathematics books, 5 reading books, 4 science books, and 10 history books. If a student selects a book at random, find the probability that the book is a. A history book or a mathematics book. b. A reading book or a science book. ANSWERS 1. a. P(blue or red)¼P(blue)þP(red)¼ 5 10 þ 3 10 ¼ 8 10 ¼ 4 5 b. P(red or black)¼P(red)þP(black)¼ 3 10 þ 2 10 ¼ 5 10 ¼ 1 2 CHAPTER 3 The Addition Rules46 EXAMPLE: At a political rally, there are 8 Democrats and 10 Republicans. Six of the Democrats are females and 5 of the Republicans are females. If a person is selected at random, find the probability that the person is a female or a Democrat. SOLUTION: There are 18 people at the rally. Let PðfemaleÞ ¼ 6þ 518 ¼ 11 18 since there are 11 females, and PðDemocratÞ ¼ 818 since there are 8 Democrats. Pðfemale and DemocratÞ ¼ 618 since 6 of the Democrats are females. Hence, Pðfemale or DemocratÞ ¼ PðfemaleÞ þ PðDemocratÞ Pðfemale and DemocratÞ ¼ 11 18 þ 8 18 6 18 ¼ 13 18 EXAMPLE: The probability that a student owns a computer is 0.92, and the probability that a student owns an automobile is 0.53. If the probability that a student owns both a computer and an automobile is 0.49, find the probability that the student owns a computer or an automobile. SOLUTION: Since P(computer)¼ 0.92, P(automobile)¼ 0.53, and P(computer and auto- mobile)¼ 0.49, P(computer or automobile)¼ 0.92þ 0.53 0.49¼ 0.96. The key word for addition is ‘‘or,’’ and it means that one event or the other occurs. If the events are not mutually exclusive, the probability of the outcomes that the two events have in common must be subtracted from the sum of the probabilities of the two events. For the mathematical purist, only one addition rule is necessary, and that is PðA or BÞ ¼ PðAÞ þ PðBÞ PðA and BÞ The reason is that when the events are mutually exclusive, P(A and B) is equal to zero because mutually exclusive events have no outcomes in common. CHAPTER 3 The Addition Rules 49 PRACTICE 1. When a card is selected at random from a 52-card deck, find the probability that the card is a face card or a spade. 2. A die is rolled. Find the probability that the result is an even number or a number less than 3. 3. Two dice are rolled. Find the probability that a number on one die is a six or the sum of the spots is eight. 4. A coin is tossed and a die is rolled. Find the probability that the coin falls heads up or that there is a 4 on the die. 5. In a psychology class, there are 15 sophomores and 18 juniors. Six of the sophomores are males and 10 of the juniors are males. If a student is selected at random, find the probability that the student is a. A junior or a male. b. A sophomore or a female. c. A junior. ANSWERS 1. P(face card or spade)¼P(face card)þP(spade)P(face card and spade)¼ 12 52 þ 13 52 3 52 ¼ 22 52 ¼ 11 26 2. P(even or less than three)¼P(even)þP(less than three)P(even and less than three)¼ 3 6 þ 2 6 1 6 ¼ 4 6 ¼ 2 3 3. P(6 or a sum of 8)¼P(6)þP(sum of 8)P(6 and sum of 8)¼ 11 36 þ 5 36 2 36 ¼ 14 36 ¼ 7 18 4. P(heads or 4)¼P(heads)þP(4)P(heads and 4)¼ 1 2 þ 1 6 1 12 ¼ 7 12 5. a. P( junior or male)¼P( junior)þP(male)P( junior and male)¼ 18 33 þ 16 33 10 33 ¼ 24 33 ¼ 8 11 b. P(sophomore or female)¼P(sophomore)þP(female) P(sophomore or female) ¼ 15 33 þ 17 33 9 33 ¼ 23 33 c. P( junior)¼ 18 33 ¼ 6 11 CHAPTER 3 The Addition Rules50 Summary Many times in probability, it is necessary to find the probability of two or more events occurring. In these cases, the addition rules are used. When the events are mutually exclusive, addition rule I is used, and when the events are not mutually exclusive, addition rule II is used. If the events are mutually exclusive, they have no outcomes in common. When the two events are not mutually exclusive, they have some common outcomes. The key word in these problems is ‘‘or,’’ and it means to add. CHAPTER QUIZ 1. Which of the two events are not mutually exclusive? a. Rolling a die and getting a 6 or a 3 b. Drawing a card from a deck and getting a club or an ace c. Tossing a coin and getting a head or a tail d. Tossing a coin and getting a head and rolling a die and getting an odd number 2. Which of the two events are mutually exclusive? a. Drawing a card from a deck and getting a king or a club b. Rolling a die and getting an even number or a 6 c. Tossing two coins and getting two heads or two tails d. Rolling two dice and getting doubles or getting a sum of eight 3. In a box there are 6 white marbles, 3 blue marbles, and 1 red marble. If a marble is selected at random, what is the probability that it is red or blue? a. 2 5 b. 1 3 c. 9 10 d. 1 9 CHAPTER 3 The Addition Rules 51 10. Two dice are rolled. What is the probability of getting doubles or a sum of 10? a. 11 18 b. 2 9 c. 1 4 d. 11 36 11. The probability that a family visits Safari Zoo is 0.65, and the prob- ability that a family rides on the Mt. Pleasant Tourist Railroad is 0.55. The probability that a family does both is 0.43. Find the prob- ability that the family visits the zoo or the railroad. a. 0.77 b. 0.22 c. 0.12 d. 0.10 12. If a card is drawn from a deck, what is the probability that it is a king, queen, or an ace? a. 5 13 b. 7 13 c. 6 13 d. 3 13 CHAPTER 3 The Addition Rules54 Probability Sidelight WIN A MILLION OR BE STRUCK BY LIGHTNING? Do you think you are more likely to win a large lottery and become a millionaire or are you more likely to be struck by lightning? Consider each probability. In a recent article, researchers estimated that the chance of winning a million or more dollars in a lottery is about one in 2 million. In a recent Pennsylvania State Lottery, the chances of winning a million dollars were 1 in 9.6 million. The chances of winning a $10 million prize in Publisher’s Clearinghouse Sweepstakes were 1 in 2 million. Now the chances of being struck by lightning are about 1 in 600,000. Thus, a person is at least three times more likely to be struck by lightning than win a million dollars! But wait a minute! Statisticians are critical of these types of comparisons, since winning the lottery is a random occurrence. But being struck by lightning depends on several factors. For example, if a person lives in a region where there are a lot of thunderstorms, his or her chances of being struck increase. Also, where a person is during a thunder storm influences his or her chances of being struck by lightning. If the person is in a safe place such as inside a building or in an automobile, the probability of being struck is relatively small compared to a person standing out in a field or on a golf course during a thunderstorm. So be wary of such comparisons. As the old saying goes, you cannot compare apples and oranges. CHAPTER 3 The Addition Rules 55 4 CHAPTER The Multiplication Rules Introduction The previous chapter showed how the addition rules could be used to solve problems in probability. This chapter will show you how to use the multiplication rules to solve many problems in probability. In addition, the concept of independent and dependent events will be introduced. 56 Copyright © 2005 by The McGraw-Hill Companies, Inc. Click here for terms of use. The previous example can also be solved using classical probability. Recall that the sample space for tossing a coin and rolling a die is H1, H2, H3, H4, H5, H6 T1, T2, T3, T4, T5, T6 Notice that there are 12 outcomes in the sample space and only one outcome is a tail and a 5; hence, P(tail and 5)¼ 112. EXAMPLE: An urn contains 2 red balls, 3 green balls, and 5 blue balls. A ball is selected at random and its color is noted. Then it is replaced and another ball is selected and its color is noted. Find the probability of each of these: a. Selecting 2 blue balls b. Selecting a blue ball and then a red ball c. Selecting a green ball and then a blue ball SOLUTION: Since the first ball is being replaced before the second ball is selected, the events are independent. a. There are 5 blue balls and a total of 10 balls; therefore, the probability of selecting two blue balls with replacement is P(blue and blue) ¼ PðblueÞ PðblueÞ ¼ 5 10 5 10 ¼ 25 100 ¼ 1 4 b. There are 5 blue balls and 2 red balls, so the probability of selecting a blue ball and then a red ball with replacement is Pðblue and redÞ ¼ PðblueÞ PðredÞ ¼ 5 10 2 10 ¼ 10 100 ¼ 1 10 CHAPTER 4 The Multiplication Rules 59 c. There are 3 green balls and 5 blue balls, so the probability of selecting a green ball and then a blue ball with replacement is Pðgreen and blueÞ ¼ PðgreenÞ PðblueÞ ¼ 3 10 5 10 ¼ 15 100 ¼ 3 20 The multiplication rule can be extended to 3 or more events that occur in sequence, as shown in the next example. EXAMPLE: A die is tossed 3 times. Find the probability of getting three 6s. SOLUTION: When a die is tossed, the probability of getting a six is 16; hence, the probabil- ity of getting three 6s is Pð6 and 6 and 6Þ ¼ Pð6Þ Pð6Þ Pð6Þ ¼ 1 6 1 6 1 6 ¼ 1 216 Another situation occurs in probability when subjects are selected from a large population. Even though the subjects are not replaced, the probability changes only slightly, so the change can be ignored. Consider the next example. EXAMPLE: It is known that 66% of the students at a large college favor building a new fitness center. If two students are selected at random, find the probability that all of them favor the building of a new fitness center. SOLUTION: Since the student population at the college is large, selecting a student does not change the 66% probability that the next student selected will favor the building of a new fitness center; hence, the probability of selecting two students who both favor the building of a new fitness center is (0.66)(0.66)¼ 0.4356 or 43.56%. CHAPTER 4 The Multiplication Rules60 PRACTICE 1. A card is drawn from a deck, then replaced, and a second card is drawn. Find the probability that two kings are selected. 2. If 12% of adults are left-handed, find the probability that if three adults are selected at random, all three will be left-handed. 3. If two people are selected at random, find the probability that both were born in August. 4. A coin is tossed 4 times. Find the probability of getting 4 heads. 5. A die is rolled and a card is selected at random from a deck of 52 cards. Find the probability of getting an odd number on the die and a club on the card. ANSWERS 1. The probability that 2 kings are selected is Pðking and kingÞ ¼ PðkingÞ PðkingÞ ¼ 41 52 13 41 52 13 ¼ 1 169 2. The probability of selecting 3 adults who are left-handed is (0.12)(0.12)(0.12)¼ 0.001728. 3. Each person has approximately 1 chance in 12 of being born in August; hence, the probability that both are born in August is 1 12 1 12 ¼ 1 144 : 4. The probability of getting 4 heads is 1 2 1 2 1 2 1 2 ¼ 1 16 . 5. The probability of getting an odd number on the die is 36 ¼ 1 2, and the probability of getting a club is 1352 ¼ 1 4; hence, the P(odd and club)¼PðoddÞ PðclubÞ ¼ 12 1 4 ¼ 1 8. Multiplication Rule II When two sequential events are dependent, a slight variation of the multiplication rule is used to find the probability of both events occurring. For example, when a card is selected from an ordinary deck of 52 cards the CHAPTER 4 The Multiplication Rules 61 PRACTICE 1. In a study, there are 8 guinea pigs; 5 are black and 3 are white. If 2 pigs are selected without replacement, find the probability that both are white. 2. In a classroom there are 8 freshmen and 6 sophomores. If three students are selected at random for a class project, find the probabil- ity that all 3 are freshmen. 3. Three cards are drawn from a deck of 52 cards without replacement. Find the probability of getting 3 diamonds. 4. A box contains 12 calculators of which 5 are defective. If two calcu- lators are selected without replacement, find the probability that both are good. 5. A large flashlight has 6 batteries. Three are dead. If two batteries are selected at random and tested, find the probability that both are dead. ANSWERS 1. P(white and white) ¼ 3 84 21 7 ¼ 3 28 2. P(3 freshmen) ¼ 8 14 2 71 13 61 12 2 ¼ 2 13 3. P(3 diamonds) ¼ 13 1 52 4 12 51 11 50 ¼ 11 850 4. P(2 good) ¼ 7 12 2 61 11 ¼ 7 22 5. P(2 batteries dead) ¼ 3 63 21 5 ¼ 3 15 ¼ 1 5 Conditional Probability Previously, conditional probability was used to find the probability of sequential events occurring when they were dependent. Recall that P(B|A) means the probability of event B occurring given that event A has already occurred. Another situation where conditional probability can be used is when additional information about an event is known. Sometimes it might be CHAPTER 4 The Multiplication Rules64 known that some outcomes in the sample space have occurred or that some outcomes cannot occur. When conditions are imposed or known on events, there is a possibility that the probability of the certain event occurring may change. For example, suppose you want to determine the probability that a house will be destroyed by a hurricane. If you used all houses in the United States as the sample space, the probability would be very small. However, if you used only the houses in the states that border the Atlantic Ocean as the sample space, the probability would be much higher. Consider the following examples. EXAMPLE: A die is rolled; find the probability of getting a 4 if it is known that an even number occurred when the die was rolled. SOLUTION: If it is known that an even number has occurred, the sample space is reduced to 2, 4, or 6. Hence the probability of getting a 4 is 13 since there is one chance in three of getting a 4 if it is known that the result was an even number. EXAMPLE: Two dice are rolled. Find the probability of getting a sum of 3 if it is known that the sum of the spots on the dice was less than six. SOLUTION: There are 2 ways to get a sum of 3. They are (1, 2) and (2, 1), and there are 10 ways to get a sum less than six. They are (1, 1), (1, 2), (2, 1), (3, 1), (2, 2), (1, 3), (1, 4), (2, 3), (3, 2), and (4, 1); hence, P(sum of 3|sum less than 6)¼ 2 10 ¼ 1 5. The two previous examples of conditional probability were solved using classical probability and reduced sample spaces; however, they can be solved by using the following formula for conditional probability. The conditional probability of two events A and B is PðAjBÞ ¼ PðA and BÞ PðBÞ : P(A and B) means the probability of the outcomes that events A and B have in common. The two previous examples will now be solved using the formula for conditional probability. EXAMPLE: A die is rolled; find the probability of getting a 4, if it is known that an even number occurred when the die was rolled. CHAPTER 4 The Multiplication Rules 65 SOLUTION: P(A and B) is the probability of getting a 4 and an even number at the same time. Notice that there is only one way to get a 4 and an even number—the outcome 4. Hence P(A and B)¼ 16. Also P(B) is the probability of getting an even number which is 36 ¼ 1 2. Now PðAjBÞ ¼ PðA and BÞ PðBÞ ¼ 1 6 1 2 ¼ 1 6 1 2 ¼ 1 63 21 1 ¼ 1 3 Notice that the answer is the same as the answer obtained when classical probability was used. EXAMPLE: Two dice are rolled. Find the probability of getting a sum of 3 if it is known that the sum of the spots on the dice was less than 6. SOLUTION: P(A and B) means the probability of getting a sum of 3 and a sum less than 6. Hence P(A and B)¼ 236 : PðBÞ means getting a sum less than 6 and is 10 36. Hence, PðAjBÞ ¼ PðA and BÞ PðBÞ ¼ 2 36 10 36 ¼ 2 36 10 36 ¼ 21 36 36 10 5 ¼ 1 5 CHAPTER 4 The Multiplication Rules66 3. There are 2 black aces and 26 black cards; hence, P(ace|black card)¼ 226 or 1 13 or P(ace black cardj Þ ¼ P(ace and black card) P(black card) ¼ 2 52 26 52 ¼ 2 52 26 52 ¼ 21 521 521 2613 ¼ 1 13 4. P (Sand CrabjRainbow Gardens) ¼ P(Sand Crab and Rainbow Gardens) P(Rainbow Gardens) ¼ 0:20 0:80 ¼ 0:25 5. There is one way to get 3 twos—(2, 2, 2) and there are 10 ways to get a sum of six—(2, 2, 2), (3, 2, 1), (1, 2, 3), (2, 1, 3), (2, 3, 1), (3, 1, 2), (1, 3, 2), (4, 1, 1), (1, 4, 1,) and (1, 1, 4). Hence, P(3 twosjsum of 6) ¼ P(3 twos and sum of 6) P(sum of 6) ¼ 1 36 10 36 ¼ 1 36 10 36 ¼ 1 361 361 10 ¼ 1 10 CHAPTER 4 The Multiplication Rules 69 Summary When two events occur in sequence, the probability that both events occur can be found by using one of the multiplication rules. When two events are independent, the probability that the first event occurs does not affect or change the probability of the second event occurring. If the events are independent, multiplication rule I is used. When the two events are dependent, the probability of the second event occurring is changed after the first event occurs. If the events are dependent, multiplication rule II is used. The key word for using the multiplication rule is ‘‘and.’’ Conditional probability is used when additional information is known about the probability of an event. CHAPTER QUIZ 1. Which of the following events are dependent? a. Tossing a coin and rolling a die b. Rolling a die and then rolling a second die c. Sitting in the sun all day and getting sunburned d. Drawing a card from a deck and rolling a die 2. Three dice are rolled. What is the probability of getting three 4s? a. 1 216 b. 1 6 c. 1 36 d. 1 18 CHAPTER 4 The Multiplication Rules70 3. What is the probability of selecting 4 spades from a deck of 52 cards if each card is replaced before the next one is selected? a. 4 13 b. 1 2561 c. 1 52 d. 1 13 4. A die is rolled five times. What is the probability of getting 5 twos? a. 1 8 b. 1 4 c. 1 6 d. 1 32 5. A coin is tossed four times; what is the probability of getting 4 heads? a. 1 2 b. 1 16 c. 1 d. 1 4 6. If 25% of U.S. prison inmates are not U.S. citizens, what is the probability of randomly selecting three inmates who will not be U.S. citizens? a. 0.75 b. 0.421875 c. 0.015625 d. 0.225 CHAPTER 4 The Multiplication Rules 71 13. In a certain group of people, it is known that 40% of the people take Vitamins C and E on a daily basis. It is known that 75% take Vitamin C on a daily basis. If a person is selected at random, what is the probability that the person takes Vitamin E given that the person takes Vitamin C? a. 5 13 b. 8 15 c. 3 11 d. 7 12 14. Two dice are tossed; what is the probability that the numbers are the same on both dice if it is known that the sum of the spots is 6? a. 2 3 b. 1 6 c. 4 5 d. 1 5 Probability Sidelight THE LAW OF AVERAGES Suppose I asked you that if you tossed a coin 9 times and got 9 heads, what would you bet that you would get on the tenth toss, heads or tails? Most people would bet on a tail. When asked why they would select a tail, they would probably respond that a tail was ‘‘due’’ according to the ‘‘law of averages.’’ In reality, however, the probability of getting a head on the CHAPTER 4 The Multiplication Rules74 tenth toss is 12, and the probability of getting a tail on the tenth toss is 1 2, so it doesn’t really matter since the probabilities are the same. A coin is an inanimate object. It does not have a memory. It doesn’t know that in the long run, the number of heads and the number of tails should balance out. So does that make the law of averages wrong? No. You see, there’s a big difference between asking the question, ‘‘What is the probability of getting 10 heads if I toss a coin ten times?’’ and ‘‘If I get 9 heads in a row, what is the probability of getting a head on the tenth toss?’’ The answer to the first question is 1 210 ¼ 11024, that is about 1 chance in 1000, and the answer to the second question is 1 2 . This reasoning can be applied to many situations. For example, suppose that a prize is offered for tossing a coin and getting 10 heads in a row. If you played the game, you would have only one chance in 1024 of winning, but if 1024 people played the game, there is a pretty good chance that somebody would win the prize. If 2028 people played the game, there would be a good chance that two people might win. So what does this mean? It means that the probability of winning big in a lottery or on a slot machine is very small, but since there are many, many people playing, somebody will probably win; however, your chances of winning big are very small. A similar situation occurs when couples have children. Suppose a husband and wife have four boys and would like to have a girl. It is incorrect to reason that the chance of having a family of 5 boys is 132, so it is more likely that the next child will be a girl. However, after each child is born, the probability that the next child is a girl (or a boy for that matter) is about 12. The law of averages is not appropriate here. My wife’s aunt had seven girls before the first boy was born. Also, in the Life Science Library’s book entitled Mathematics, there is a photograph of the Landon family of Harrison, Tennessee, that shows Mr. and Mrs. Emery Landon and their 13 boys! Another area where people incorrectly apply the law of averages is in attempting to apply a betting system to gambling games. One such system is doubling your bet when you lose. Consider a game where a coin is tossed. If it lands heads, you win what you bet. If it lands tails, you lose. Now if you bet one dollar on the first toss and get a head, you win one dollar. If you get tails, you lose one dollar and bet two dollars on the next toss. If you win, you are one dollar ahead because you lost one dollar on the first bet but won two dollars on the second bet. If you get a tail on the second toss, you bet four dollars on the third toss. If you win, you start over with a one dollar bet, but if you lose, you bet eight dollars on the next toss. With this system, you win every time you get a head. Sounds pretty good, doesn’t it? CHAPTER 4 The Multiplication Rules 75 This strategy won’t work because if you play long enough, you will eventually run out of money since if you get a series of tails, you must increase your bet substantially each time. So if you lose five times in a row, you have lost $1þ $2þ $4þ $8þ $16 or $31, and your next bet has to be $32. So you are betting $63 to win $1. Runs do occur and when they do, hope that they are in your favor. Now let’s look at some unusual so-called ‘‘runs.’’ In 1950, a person won 28 straight times playing the game of craps (dice) at the Desert Inn in Las Vegas. He lost on the twenty-ninth roll. He did not win big though because after each win he stuffed some bills in his pocket. The event took about one hour and twenty minutes. In 1959 in a casino in Puerto Rico at a roulette game, the number 10 occurred six times in succession. There are 38 numbers on a roulette wheel. At a casino in New York in 1943 the color red occurred in a roulette game 32 times in a row, and at a casino in Monte Carlo an even number occurred in a roulette game 28 times in a row. These incidents have been reported in two books, one entitled Scarne’s Complete Guide to Gambling and the other entitled Lady Luck by Warren Weaver. So what can be concluded? First, rare events (events with a small probability of occurring) can and do occur. Second, the more people who play a game, the more likely someone will win. Finally, the law of averages applies when there is a large number of independent outcomes in which the probability of each outcome occurring does not change. CHAPTER 4 The Multiplication Rules76 EXAMPLE: Two coins are tossed; find the odds in favor of getting two heads SOLUTION: When two coins are tossed, there are four outcomes andPðHHÞ ¼ 14 : PðEÞ ¼ 1 14 ¼ 3 4 ; hence, odds in favor of two heads ¼ PðE Þ 1 PðE Þ ¼ 1 4 1 1 4 ¼ 1 4 3 4 ¼ 1 4 3 4 ¼ 1 41 41 3 ¼ 1 3 The odds are 1 : 3. EXAMPLE: Two dice are rolled; find the odds against getting a sum of 9. SOLUTION: There are 36 outcomes in the sample space and four ways to get a sum of 9. Pðsum of 9Þ ¼ 436 ¼ 1 9 , PðEÞ ¼ 1 1 9 ¼ 8 9 : Hence, odds of not getting a sum of 9 ¼ PðE Þ 1 PðE Þ ¼ 8 9 1 8 9 ¼ 8 9 1 9 ¼ 8 9 1 9 ¼ 8 91 91 1 ¼ 8 1 The odds are 8 : 1. If the odds in favor of an event occurring are A :B, then the odds against the event occurring are B :A. For example, if the odds are 1 : 15 that an event will occur, then the odds against the event occurring are 15 : 1. Odds can also be expressed as odds in favor ¼ number of outcomes in favor of the event number of outcomes not in favor of the event For example, if two coins are tossed, the odds in favor of getting two heads were computed previously as 1 : 3. Notice that there is only one way to get two heads (HH) and three ways of not getting two heads (HT, TH, TT); hence the odds are 1 : 3. CHAPTER 5 Odds and Expectation 79 When the probability of an event occurring is 12, then the odds are 1 : 1. In the realm of gambling, we say the odds are ‘‘even’’ and the chance of the event is ‘‘fifty–fifty.’’ The game is said to be fair. Odds can be other numbers, such as 2 : 5, 7 : 4, etc. PRACTICE 1. When two dice are rolled, find the odds in favor of getting a sum of 12. 2. When a single card is drawn from a deck of 52 cards, find the odds against getting a diamond. 3. When three coins are tossed, find the odds in favor of getting two tails and a head in any order. 4. When a single die is rolled, find the odds in favor of getting an even number. 5. When two dice are rolled, find the odds against getting a sum of 7. ANSWERS 1. There is only one way to get a sum of 12, and that is (6, 6). There are 36 outcomes in the sample space. Hence, P(sum of 12) ¼ 136 : The odds in favor are 1 36 1 1 36 ¼ 1 36 35 36 ¼ 1 36 35 36 ¼ 1 361 361 35 ¼ 1 35 The odds are 1 : 35. 2. There are 13 diamonds in 52 cards; hence, Pð^)¼ 13 52 ¼ 1 4 : Pð^Þ ¼ 1 1 4 ¼ 3 4 : The odds against getting a diamond 3 4 1 3 4 ¼ 3 4 1 4 ¼ 3 4 1 4 ¼ 3 41 41 1 ¼ 3 1 The odds are 3 : 1. CHAPTER 5 Odds and Expectation80 3. When three coins are tossed, there are three ways to get two tails and a head. They are (TTH, THT, HTT), and there are eight outcomes in the sample space. The odds in favor of getting two tails and a head are 3 8 1 3 8 ¼ 3 8 5 8 ¼ 3 8 5 8 ¼ 3 81 81 5 ¼ 3 5 The odds are 3 : 5. 4. There are 3 even numbers out of 6 outcomes; hence, PðevenÞ ¼ 3 6 ¼ 1 2 : The odds in favor of an even number are 1 2 1 1 2 ¼ 1 2 1 2 ¼ 1 2 1 2 ¼ 1 21 21 1 ¼ 1 1 The odds are 1:1. 5. There are six ways to get a sum of 7 and 36 outcomes in the sample space. Hence, P(sum of 7) ¼ 636 ¼ 1 6 and P(not getting a sum of 7) ¼ 1 16 ¼ 5 6 : The odds against getting a sum of 7 are 5 6 1 5 6 ¼ 5 6 1 6 ¼ 5 6 1 6 ¼ 5 61 61 1 ¼ 5 1 The odds are 5 : 1. Previously it was shown that given the probability of an event, the odds in favor of the event occurring or the odds against the event occurring can be found. The opposite is also true. If you know the odds in favor of an event occurring or the odds against an event occurring, you can find the probability of the event occurring. If the odds in favor of an event occurring are A :B, then the probability that the event will occur is PðE Þ ¼ AAþB : If the odds against the event occurring are B :A, the probability that the event will not occur is PðEÞ ¼ BBþA : Note: Recall that PðEÞ is the probability that the event will not occur or the probability of the complement of event E. CHAPTER 5 Odds and Expectation 81 $2, $3, $4, $5, or $6; however, on average, you would win $3.50 on each roll. So if you rolled the die 100 times, you would win on average $3.50 100¼ $350. Now if you had to pay to play this game, you should pay $3.50 for each roll. That would make the game fair. If you paid more to play the game, say $4.00 each time you rolled the die, you would lose on average $0.50 on each roll. If you paid $3.00 to play the game, you would win an average $0.50 per roll. EXAMPLE: When two coins are tossed, find the expected value for the number of heads obtained. SOLUTION: Consider the sample space when two coins are tossed. HH HT TH TT j n = j two heads one head zero heads The probability of getting two heads is 14. The probability of getting one head is 14þ 1 4 ¼ 1 2. The probability of getting no heads is 1 4. The expected value for the number of heads is EðXÞ ¼ 2 14þ 1 1 2þ 0 1 4 ¼ 1: Hence the average number of heads obtained on each toss of 2 coins is 1. In order to find the expected value for a gambling game,multiply the amount you win by the probability of winning that amount, and then multiply the amount you lose by the probability of losing that amount, then add the results. Winning amounts are positive and losses are negative. EXAMPLE: One thousand raffle tickets are sold for a prize of an entertainment center valued at $750. Find the expected value of the game if a person buys one ticket. SOLUTION: The problem can be set up as follows: Gain, (X ) Win Lose $749 $1 Probability, P(X) 1 1000 999 1000 CHAPTER 5 Odds and Expectation84 Since the person who buys a ticket does not get his or her $1 back, the net gain if he or she wins is $750 $1¼ $749. The probability of winning is one chance in 1000 since 1000 tickets are sold. The net loss is $1 denoted as negative and the chances of not winning are 1000 11000 or 999 1000 : Now EðXÞ ¼ $749 11000þ ð$1Þ 999 1000 ¼ $0:25: Here again it is necessary to realize that one cannot lose $0.25 but what this means is that the house makes $0.25 on every ticket sold. If a person purchased one ticket for raffles like this one over a long period of time, the person would lose on average $0.25 each time since he or she would win on average one time in 1000. There is an alternative method that can be used to solve problems when tickets are sold or when people pay to play a game. In this case, multiply the prize value by the probability of winning and subtract the cost of the ticket or the cost of playing the game. Using the information in the previous example, the solution looks like this: EðXÞ ¼ $750 1 1000 $1 ¼ $0:75 $1 ¼ $0:25: When the expected value is zero, the game is said to be fair. That is, there is a fifty–fifty chance of winning. When the expected value of a game is negative, it is in favor of the house (i.e., the person or organization running the game). When the expected value of a game is positive, it is in favor of the player. The last situation rarely ever happens unless the con man is not knowledgeable of probability theory. EXAMPLE: One thousand tickets are sold for $2 each and there are four prizes. They are $500, $250, $100, and $50. Find the expected value if a person purchases 2 tickets. SOLUTION: Find the expected value if a person purchases one ticket. Gain, X $499 $249 $99 $49 $1 Probability P(X) 1 1000 1 1000 1 1000 1 1000 996 1000 EðXÞ ¼ $499 1 1000 þ $249 1 1000 þ $99 1 1000 þ $49 1 1000 $1 996 1000 ¼ $0:10 CHAPTER 5 Odds and Expectation 85 The expected value is $0.10 for one ticket. It is 2($0.10)¼$0.20 for two tickets. Alternate solution EðXÞ ¼ $500 1 1000 þ $250 1 1000 þ $100 1 1000 þ $50 1 1000 $1 ¼ $0:10 2ð$0:10Þ ¼ $0:20 Expectation can be used to determine the average amount of money the house can make on each play of a gambling game. Consider the game called Chuck-a-luck. A player pays $1 and chooses a number from 1 to 6. Then three dice are tossed (usually in a cage). If the player’s number comes up once, the player gets $2. If it comes up twice, the player gets $3, and if it comes up on all three dice, the player wins $4. Con men like to say that the probability of any number coming up is 16 on each die; therefore, each number has a probability of 36 or 1 2 of occurring, and if it occurs more than once, the player wins more money. Hence, the game is in favor of the player. This is not true. The next example shows how to compute the expected value for the game of Chuck-a-luck. EXAMPLE: Find the expected value for the game Chuck-a-luck. SOLUTION: There are 6 6 6¼ 216 outcomes in the sample space for three dice. The probability of winning on each die is 16 and the probability of losing is 5 6: The probability that you win on all three dice is 16 1 6 1 6 ¼ 1 216 : The probability that you lose on all three dice is 56 5 6 5 6 ¼ 125 216 : The probability that you win on two dice is 16 1 6 5 6 ¼ 5 216, but this can occur in three different ways: (i) win on the first and the second dice, and lose on the third die, (ii) win on the first die, lose on the second die, and win on the third die, (iii) lose on the first die, and win on the second and third dice. Therefore, the probability of winning on two out of three dice is 3 5216 ¼ 15 216 : The probability of winning on one die is 16 5 6 5 6 ¼ 25 216, and there are three different ways to win. Hence, the probability of winning on one die is 3 25216 ¼ 75 216 : CHAPTER 5 Odds and Expectation86