Mustafa Jarrar: Lecture Notes in Discrete Mamatics. irzeit University Palestine 2015 Counting 9.1 asics of Probability and Counting 9.2 Possibility Trees and Multiplication Rule 9.3 Counting Elements of Disjoint Sets: ddition Rule 9.5 Counting Subsets of a Set: Combinations 9.6 r-combinations with Repetition llowed mjarrar 2015 1 Watch this lecture and download slides Course Page: http://www.jarrar.info/courses/dmath/ More Online Courses at: http://www.jarrar.info cknowledgement: This lecture is based on (but not limited to) to chapter 9 in Discrete Mamatics with pplications by Susanna S. Epp (3 rd Edition). 2 1
Mustafa Jarrar: Lecture Notes in Discrete Mamatics. irzeit University Palestine 2015 Counting 9.1 asics of Probability and Counting In this lecture: q Part 1: Probability and Sample Space qpart 2: Counting in Sub lists 3 Tossing Coins Tossing two coins and observing wher 0 1 or 2 are. What are chances of having 012? 2
9.1 Introduction Imagine tossing two coins and observing wher 0 1 or 2 are. It would be natural to guess that each of se events occurs about one-third of time but in fact this is not case. Table 9.1.1 below shows actual data from tossing two quarters 50 times. Table 9.1.1 Experimental Data Obtained from Tossing Two Quarters 50 Times Event Tally 2 0 Relative Frequency (Fraction of times event occurred) 11 22% Tossing Coins 1 head Frequency (Number of times event occurred) 27 5% 12 2% s you can see relative frequency of obtaining exactly 1 head was roughly twice Tossing coins and observing wher 1 or 2 as great as thattwo of obtaining 2 or 0. It turns out that 0 mamatical ory of probability can be used to predict that a result like this will almost always occur. 9.1 Introduction. To see howare call two coins and and suppose that each is perfectly balanced. Then each has an equal chance of coming up or tails and when two are tossed What are chances of having 012? toger four pictured in Figure 9.1.2 are all equally likely. 517 9.1 Introduction Imagine tossing two coins and observing wher 0 1 or 2 are. It would be natural to guess that each of se events occurs about one-third of time but in fact this is not case. Table 9.1.1 below shows actual data from tossing two quarters 50 times. 2 Table 9.1.1 1 head from Tossing Two 0Quarters Experimental Data Obtained 50 Times Figure 9.1.2 Equally Likely Outcomes from Tossing Two alanced Coins Frequency Relative Frequency Figure 9.1.2 shows that re is a 1 in chance of obtaining two and a 1 of in times (Number of times (Fraction chance of obtaining no. The chance of obtaining one head however is 2 in Event Tally event occurred) event occurred) because could come up and tails or could come up and tails. 2 11 22% So if you repeatedly toss two balanced coins and record number of you should 1 head similar to those shown in Table27 5% expect relative frequencies 9.1.1. formalize this analysis 0 To and extend it to more complex 12 situations we introduce 2% notions of random process sample space event and probability. To say that a process 5 is random means that when it takes place one outcome from some set of is s can see frequency obtaining 1 head that waswill roughly sureyou to occur but it is relative impossible to predictof with certainty exactly which outcome be. twice Forgreat instance if an person performs tossing ordinary coin as as that ofordinary obtaining 2 orexperiment 0. Itofturns outanthat mamatical into ofair and allowing to used fall flat on ground it can like be predicted certainty ory probability canitbe to predict that a result this willwith almost always occur. that coin will land up or tails up (so set of can be denotedbalanced. To see how call two coins and and suppose that each is perfectly { tails}) but it is not known for sure wher or tails will occur. We restricted Then each has an equal chance of coming up or tails and when two are tossed this experiment to ordinary people because a skilled magician can toss a coin in a way toger random four in Figure 9.1.2with are first-rate all equally likely. devices that appears but is not pictured and a physicist equipped measuring may be able to analyze all forces on coin and correctly predict its landing position. Just a few of many examples of random processes or experiments are choosing winners lotteriesselecting respondents polls and choosing subjects in state in public opinion to 518 Chapter 9 Counting and Probability receive treatments or serve as controls in medical experiments. The set of that can result from a random process or experiment is called a sample space. Sample Space Definition 2 space is set of all possible 1 head of a random process 0 sample or experiment. 9.1.2 of Equally Likely Outcomes from Tossing Two alanced Coins n eventfigure is a subset a sample space. Copyright 2010 Cengage Learning. ll Rights Reserved. May not be copied scanned or duplicated in whole or in part. Due to electronic rights some third party content may be suppressed from eook and/or echapter(s). Editorial review has deemed that any suppressed content does not materially affect overall learning experience. Cengage Learning reserves right to remove additional content at any time if subsequent rights restrictions require it. Figure 9.1.2 shows that re is a 1 in chance of obtaining two and a 1 in Inchance of experiment obtaining nohas. Themany chance of obtaining one head however is 2 inlikely case an finitely and all are equally could come (set and of tails or could come upratio ofand tails. to because occur probability of anupevent ) is just number if you repeatedly toss to two balanced coins and number youthis should of So in event total number of record. Strictlyofspeaking result expect relative frequencies similar to those shown in Table 9.1.1. can be deduced from a set of axioms for probability formulated in 1933 by Russian To formalize this analysis and extend it to more complex situations we introduce mamatician. N. Kolmogorov. In Section 9.8 we discuss axioms and show how to notions of random process sample space event and probability. To say that a process derive ir consequences formally. t present we take a naïve approach to probability is random means that when it takes place one outcome from some set of is andsure simply state but itresult as a principle. to occur is impossible to predict with certainty which outcome that will be. For instance if an ordinary person performs experiment of tossing an ordinary coin into air and allowing it to fall flat on ground it can be predicted with certainty Equally Likely Formula that coin will land up or Probability tails up (so set of can be denoted { tails}) but it is not known for sure wher or tails will occur. If S is a finite sample space in which all are equally likely We andrestricted E is an this experiment to ordinary people because a skilled magician can toss a coin in a way 6 event in S n probability of E denoted P(E) is that appears random but is not and a physicist equipped with first-rate measuring devices on number ofand E may be able to analyze all forces coin correctlyinpredict its landing position.. P(E) = Just a few of many examples of processes experiments random total number of or in Sare choosing winners in state lotteries selecting respondents in public opinion polls and choosing subjects to receive treatments or serve as controls in medical experiments. The set of that can result from a random process or experiment is called a sample space. Notation For any finite set N ( ) denotes number of elements in. Copyright 2010 Cengage Learning. ll Rights Reserved. May not be copied scanned or duplicated in whole or in part. Due to electronic rights some third party content may be suppressed from eook and/or echapter(s). Editorial review has deemed that any suppressed content does not materially affect overall learning experience. Cengage Learning reserves right to remove additional content at any time if subsequent rights restrictions require it. With this notation equally likely probability formula becomes 3
Sample Space Equally Likely Probability Formula If S is a finite sample space in which all are equally likely and E is an event in S nprobability of E denotedp(e) is Notation P(E) = number of in E total number of in S. For any finite set N() denotes number of elements in. P(E) = N(E) N(S). 7 52 cards Probabilities for a Deck of Cards a. What is sample space of? è 52 cards in deck. of diamonds cards co( ) suit hearts contains ( ) 1 into four s rds clubs of ( ) fo spades 2 3 ( ) 5 8 9 10 J b. What is event that chosen card is a black face card? è E = {J Q K J Q K } c. What is probability that chosen card is a black face card? 8
Solution a. The in sample space S are 52 cards in deck. b. Let E be event that a black face card is chosen. The in E are jack queen and king of clubs and jack queen and king of spades. Symbolically E ={J Q K J Q K }. 9.1 Introduction 519 9.1 Introduction 519 c. y part (b) N(E) = 6 and according to description of situation all 52 out- in Rolling sample space a Pair are of equally Dicelikely. Therefore by equally likely proba- Solution Solutioncomes a. Thea. The bility in formula insample sample space probability space S are S are that 52 52cards in chosen in deck. card is a black face card is b. Let b. E Let be E beevent event that that a blacka black face face card card is is chosen. in E are jack queen and king of clubs and jack P(E) = N(E) The N(S) = 6 52 in E are jack queen and king of clubs and jack queen and king of spades. Symbolically = 11.5%. Example 9.1.2 Rolling a Pair of Dice E ={J E ={J Q Q K K J Q K }. c. y c. part y(b) part N(E) (b) N(E) = 6 = and 6 and according according to to description of situation all all 52 52 in sample space are equally likely. Therefore by equally likely proba- in sample space are equally likely. Therefore by equally likely probability formula probability that bility formula probability that chosen card is a black face card is P(E) = N(E) chosen card N(S) = 6 is a black face card is 52 = 11.5%. P(E) = N(E) Example 9.1.2 Rolling a Pair of Dice N(S) = 6 52 = 11.5%. dieisoneofapairofdice.itisacubewithsixsideseachcontainingfromonetosix dots called pips. Supposeabluedieandagraydiearerolledtoger andnumbers of dots that occur face up on each are recorded. The possible can be listed as follows where in each case die on left is blue and one on right is gray. Example 9.1.2 Rolling a Pair of Dice dieisoneofapairofdice.itisacubewithsixsideseachcontainingfromonetosix dieisoneofapairofdice.itisacubewithsixsideseachcontainingfromonetosix dots called pips. Supposeablue andagray arerolledtoger andnumbers dots called of dotspips. that Suppose occur faceaup blue on each dieand are recorded. agraydiethe arepossible rolledtoger and can be listed numbers as of dots follows that occur where face in each up on case each die are on recorded. left is The blue possible and one on right can is be gray. listed as follows where in each case die on left is blue and one on right is gray. 9 Rolling a Pair of Dice morecompactnotationidentifiessay with notation 2 with 53 and so forth. more compact notation identifiessay with notation 2 with 53 a. Use compact notation to write sample space S of possible. moreand compact so forth. notationidentifiessay with notation 2 with 53 b. Use set notation to write event E that numbers showing face up have a sum of and so forth. 6andfindprobabilityofthisevent. a. a. Write Use compact sample notation space S toof write possible sample. space S of possible. a. Solution Use compact notation to write sample space S of possible. b. Use set notation to write event E that numbers showing face up have a sum of 6andfindprobabilityofthisevent. b. Usea. set S notation ={11 12 to13 write 1 15 16 event 21 E22 that 23 2 25 numbers 26 31 showing 32 33 3 face 35 36 up have 1 2 a 3 sum of 6andfindprobabilityofthisevent. 5 6 51 52 53 5 55 56 61 62 63 6 65 66}. Solution b. E ={15 2 33 2 51}. Solution b. The write probability event that sum E that of numbers numbers is 6 = P(E) showing = N(E) face up have N(S) = 5 a. S ={11 12 13 1 15 16 21 22 23 2 25 26 31 32 33 3 35 36. 36 1 2 3 a. a sum S 5 ={11 6 of 51 612 and 52 13 53 find 5 1 55 15 56 16 probability 61 21 62 22 63 6 23 of 65 2 this 66}. 25 event. 26 31 32 33 3 35 36 1 2 3 The next example is called Monty Hall problem after host of an old game 5 6 51 52 53 5 55 56 61 62 63 6 65 66}. b. E ={15 show Let s 2 33 Make 2 51}. Deal. When it was originally publicized in a newspaper column and on a radio show it created tremendous controversy. Many highly educated people The probability b. E ={15 2 33 2 51}. even somethat with Ph.D. s sum submitted of numbers incorrect is solutions 6 = P(E) or argued = N(E) N(S) vociferously = 5 36. against correct solution. efore you read answer think about what your own response to situation The probability would be. that sum of numbers is 6 = P(E) = N(E) N(S) = 5 The next example is called Monty Hall problem after host of an old game 36. show Let s Make Deal. When it was originally publicized in a newspaper column 10 and on a radio Theshow nextitexample created tremendous is calledcontroversy. MontyMany Hallhighly problem educated after people host of an old game even some with Ph.D. s submitted incorrect solutions or argued vociferously against show Let s Make Deal. When it was originally publicized in a newspaper column correct solution. efore you read answer think about what your own response to situation and would on be. a radio show it created tremendous controversy. Many highly educated people even some with Ph.D. s submitted incorrect solutions or argued vociferously against correct solution. efore you read answer think about what your own response to5 situation would be. ge Learning. ll Rights Reserved. May not be copied scanned or duplicated in whole or in part. Due to electronic rights some third party content may be suppressed from eook and/or echapter(s). emed that any suppressed content does not materially affect overall learning experience. Cengage Learning reserves right to remove additional content at any time if subsequent rights restrictions require it. Copyright 2010 Cengage Learning. ll Rights Reserved. May not be copied scanned or duplicated in whole or in part. Due to electronic rights some third party content may be suppressed from eook and/or echapter(s). Editorial review has deemed that any suppressed content does not materially affect overall learning experience. Cengage Learning reserves right to remove additional content at any time if subsequent rights restrictions require it. 10 Cengage Learning. ll Rights Reserved. May not be copied scanned or duplicated in whole or in part. Due to electronic rights some third party content may be suppressed from eook and/or echapter(s). ew has deemed that any suppressed content does not materially affect overall learning experience. Cengage Learning reserves right to remove additional content at any time if subsequent rights restrictions require it.
Mustafa Jarrar: Lecture Notes in Discrete Mamatics. irzeit University Palestine 2015 Counting 9.1 asics of Probability and Counting In this lecture: q Part 1: Probability and Sample Space qpart 2: Counting in Sub lists 11 Counting Elements of a List list: 5 6 7 8 9 10 11 12 count: 1 2 3 5 + 6 7 8 + + list: m(= m + 0) m + 1 m + 2... n (= m + (n m)) count: 1 2 3... (n m) + 1 + + + + = + + Theorem 9.1.1 The Number of Elements in a List If m and n are integers and m n n re are n m + 1integersfromm to n inclusive. 12 6
Counting Elements of a Sublist a. How many three-digit integers (integers from 100 to 999 inclusive) are divisible by 5? 100 101 102 103 10 105 106 107 108 109 110 99 995 996 997 998 999 5 20 5 21 5 22 5 199 b. What is probability that a randomly chosen three-digit integer is divisible by 5? 999 100 + 1 = 900. 180/900 = 1/5. 13 7