CHAPTER 4 PROBABILITY And COUNTING RULES 1 4

CHAPTER 4 PROBABILITY And COUNTING RULES 1

4. 1 SAMPLE SPACE AND PROBABILITY Definition q. Experiment is a process or activity such that when performed, results in one and only one of many observations. q. The observations are also called outcomes of the experiment. q. A sample space, S, is the collection of all the outcomes. It can be written as S = {List of all outcomes} The elements of a sample space are called sample points. Example #1: John tosses a coin once. Describe the experiment, outcome, and sample space. 2

Experiments, Outcomes, and Sample Spaces Venn Diagram and Tree Diagram q. Venn diagram depicts all the possible outcomes of an experiment. q. The diagram is a close geometric shape such as rectangle, square, or circle. For example, S = {1, 2, 3, 4, 5, 6} 1 3 5 2 4 6 q. A tree diagram depicts each outcome as a branch of the tree. For example, S = {1, 2, 3, 4, 5, 6}. 3

Experiments, Outcomes, and Sample Spaces Example #2: A box contains a certain number of computer parts, a few of which are defective. Two parts are selected at random from this box and inspected to determine if they are good or defective. How many total outcomes are possible? Draw a tree diagram for this experiment. Solution: Let “G” represent good part and “D” represent defective part. Therefore, for each experiment, the outcome is either the part is “D” or “G”. Hence, there are four possible outcomes, {DD, DG, GD} 1 st Selection D 2 nd Final Selection Outcomes DD D G G DG GD GG 4

Experiments, Outcomes, and Sample Spaces Example #3: Draw a tree diagram for three tosses of a coin. List all outcomes for this experiment in a sample space S. Solution: Let “H” represent head and “T” represent tail. Therefore, for each experiment, the outcome is either a “H” or “T”. 3 rd 2 nd 1 st Selection HHH H HHT T H HTH H HTT S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} T THH T H H THT T H TTH H T TTT 5 T T

4. 1. 1 Simple and Compound Events a. b. Event is a collection of one or more of the outcomes of an experiment. An event could be the entire or portion of a sample space. Therefore, an event can be classified as: q Simple event q Compound event A simple event, Ei, consists of one and only one of the final outcomes of an experiment. In Example #3, E 1=(HHH), E 2=(HHT), E 3=(HTH), E 4=(HTT), E 5=(THH), E 6=(THT), E 7=(TTH), and E 8=(TTT) c. A compound event consists of more than one outcome of an experiment. It is represented by A, B, C, D, . . . , or A 1, A 2, A 3, . . . , B 1, B 2, B 3, , . . . Reconsider Example #3, let A be the event that two of the three tosses will result in heads. Then, event A is given by A = { HHT, HTH, THH} 6

Simple and Compound Events Example #4: A box contains a certain number of computer parts, a few of which are defective. Two parts are selected at random from this box and inspected to determine if they are good or defective. List all the outcomes included in each of the following events. Indicate which are simple and which are compound events. a)At least one part is good. b)Exactly one part is defective. c)The first part is good and the second is defective. d)At most one part is good. Solution: Let, D = a defective part G = a good part The experiment has the following outcomes: DD = both parts are defective DG = the 1 st part is defective and the 2 nd is good GG = both parts are good GD = the 1 st part is good and the 2 nd is defective a)At least one part is good = {DG, GD} compound event b)Exactly one part is defective = {DG, GD} compound event c)The 1 st is good and 2 nd defective = {GD} simple event d)At most one part is good = {DD, DG, GD} compound event. 7

CALCULATING PROBABLITY Probability is a numerical measure that tells us the likelihood that a specific event will occur. It is donated by P(Ei) for a probability that a simple event will occur. P(A) for a probability that a compound event will occur. Properties of Probability 1. Probability can take a numerical value from 0 to 1 for a simple or compound event. 2. The sum of probabilities of all simple events (or final outcome) for an experiment is always 1. 8

4. 2. 1 Three Conceptual Approaches to Probability The study of probability can be approached through • Classical probability • Relative frequency concept of probability • Subjective probability q. Classical Probability • Equally likely outcomes is defined as two or more events or outcomes of an experiment that have the same probability of occurrence. • By classical rule, the probability of equally likely outcome is: 9

Example – Classical Probability Example #5: Suppose a randomly selected passenger is about to go through the metal detector at JFK Airport in New York City. Consider the following two outcomes. The passenger sets off the metal detector, or the passenger does not set off the metal detector. Are the two outcomes equally likely? Explain why or why not. If you are to find the probability of these two outcomes, would you use the classical approach or another approach? Explain why. Solution: a. The two outcomes, “passenger sets off the metal detector” and “passenger does not set of the metal detector”, are not equally likely because if they were, 50% of the passengers would set off the detector. This would be a daunting task for the Transportation Authority Administration (TSA). b. Classical approach will not be appropriate for determining the probability of these two outcomes. Therefore, another approach is needed. This other approach will require obtaining a random sample of passengers going through NY JFK airport and collecting information whether they set off the detector or not. 10

Example – Classical Probability Example #6: A box contains 40 marbles. Of them, 18 are red and 22 are green. If one marble is randomly selected out of this box, what is the probability that this marble is: a. red? b. Solution: green? Solution: Example #7 A multiple-choice question on a test has five answers. If Dianne chooses one answer based on “pure guess” , what is the probability that her answer is: a. correct? b. wrong? Do these two probabilities add up to 1? If yes why? Yes. The experiment has two and only two outcomes and according to the 2 nd property of probability, the sum of the probability must be 1. 11

4. 2. 1 Three Conceptual Approaches to Probability q Relative Frequency Concept of Probability • To determine the probability, we will have to perform each of the experiments for a large number of times. • If the experiment is repeated n times and an event “A” occurred f times, then according to the relative frequency concept of probability, the probability of event “A” is given by, 12

4. 2. 1 Three Conceptual Approaches to Probability • • • From the formula, it is clear that the relative frequency of each event is used as an approximation for the probability of the event. The probability may change because relative frequencies are determined every time the experiment is performed. Relative frequencies are not exact probabilities but are approximate probabilities unless if they are based on census. However, if the experiment is carried out for a large number of times, the relative frequency approaches the actual probability of the outcome. This is called the law of large number. 13

4. 2. 1 Three Conceptual Approaches to Probability Example #8: n Thirty-two persons have applied for a security guard position with a company. Of them, 7 have previous experience and 25 do not. Suppose one applicant is selected at random. If you are to find the probability of these two outcomes, would you use the classical approach or the relative frequency approach? Explain why. n Solution: We can use the classical approach because each applicant is equally likely to be selected 14

Example – Relative Frequency Concept of Probability Example #9: A sample of 820 adults showed that 80 of them had no credit cards, 116 had one card each, 94 had two cards each, 77 had three cards each, 43 had four cards each, and 410 had five or more cards each. Write the frequency distribution table for the number of credit cards an adult possesses. Calculate the relative frequencies for all categories. Suppose one adult is randomly selected from these 820 adults. Find the probability that this adult has (a) three credit cards (b) five or more cards Solution: # of Credit Cards 0 1 2 3 4 >=5 Frequenc y (f) 80 116 94 77 43 410 ∑f = 820 a. P(3 cards) = 0. 0939 b. P(>=5) = 0. 5000 Relative Frequency (rf) 80/820 = 0. 0976 116/820 = 0. 1415 94/820 = 0. 1146 77/820 = 0. 0939 43/820 = 0. 0524 410/820 = 0. 5000 ∑rf = 1. 0000 15

4. 2. 1 Three Conceptual Approaches to Probability q Subjective Probability • Neither the classical rule nor relative frequency approach can be used to calculate the probability of an event or outcome that is neither equally likely nor can be performed various times to generate • So, the probability of such event is called subjective probability because it is based on ones experience, information, and belief. • For example, what is the probability that New York Giant will win the Super Bowl next season? The two outcomes, “NY Giant will win” and “NY Giant will not win” are not equally likely to occur. It cannot be repeated like rolling a dice. Therefore, one has to use past Super Bowl outcome to determine the probability that NY Giant will or not win. 16

MARGINAL AND CONDITIONAL PROBABILITIES The following table gives the responses of 2, 000 randomly selected adults who were asked whether or not they have shopped on internet. Have shopped Have never shopped Male 500 700 Female 300 500 Discussion 1. The table is a two-way classification of 2, 000 adults. 2. The table is called contingency table and each box with a numeric entry is called a cell. 3. Each cell gives the frequency of two characteristics: a) Gender (male or female) and b) Opinion (have shopped or have never shopped). 17

Marginal and Conditional Probabilities Discussion 3. By adding the row totals and column totals, we obtain a new table. 4. Have shopped Have never shopped Total Male 500 700 1200 Female 300 500 800 Total 800 1200 2000 If only one characteristic, “have shopped”, “have never shopped”, “male”, or “female”, is being considered at a time, the probability of each event is called marginal probability or simple probability. 18

Marginal and Conditional Probabilities Have shopped Have never shopped Total Male 500 700 1200 Female 300 500 800 Total 800 1200 2000 The marginal probability or simple probability is a probability of a single event without consideration of any other event. From the table, the marginal probabilities of the characteristics are as follows: 19

Marginal and Conditional Probabilities Now suppose we want to find the probability that the randomly selected adult has shopped on the internet, assuming that the adult is female. In other words, the event that the adult is female has already occurred. This probability is called conditional probability, and it is written, P(has shopped | female) and is read as The probability that the selected adult has shopped on the internet given that the event “female” has already occurred. 20

Marginal and Conditional Probabilities General Statement Suppose A and B are two events, then the conditional probability of A given B is written as , P(A|B). Again from the contingent table, 21

MUTUALLY EXCLUSIVE EVENTS Definition q Mutually exclusive events are events that do not have any outcome in common. q q q Ex. Events for rolling a die: A = an even number is observed B = an odd number is observe C = a number less than five is observed Mutually Exclusive Mutually nonexclusive event Most importantly, the occurrence of one event prevents the occurrence of the other mutually exclusive events. 23

Mutually Exclusive Events p For example, the outcomes of tossing a coin are mutually exclusive because both Head and Tail outcomes could not occur at the same time. The occurrence of Head prevents occurrence of Tail to occur. H T HH HT H H T T TH TT 24

Mutually Exclusive Events Example #12 (4. 59) There are 160 practicing physicians in a city. Of them, 75 are female and 25 are pediatricians. Of the 75 female, 20 are pediatricians. Are the events “female” and “pediatrician” mutually exclusive? Explain why or why not. Solution Ped Non-Ped Totals Male 5 80 85 Female 20 55 75 The events “female” and “pediatrician” are not mutually exclusive because a physician could be a female and a pediatrician as shown above. Example #13 Define the following two events for two tosses of a coin: A = at least one head B = both tails are obtained Are A and B mutually exclusive? Explain why or why not. Solution: The experiment involves tossing a coin twice. The sample space S = {HH, HT, TH} where H = Head and T = Tail. The events are: A = {HH, HT, TH} & B = {TT} A and B are mutually exclusive. They do not have any outcome in common. 25

INDEPENDENT VERSUS DEPENDENT EVENTS Definition q Two events are said to be independent if the occurrence of one does not affect the probability of the occurrence of the other. In other words, A and B are independent events if either P(A | B) = P(A) or P(B | A) = P(B). q q q If P(A | B) = P(A) is true, then P(B | A) = P(B) is also true. If P(A | B) = P(A) is false, then P(B | A) = P(B) is also false. If the occurrence of one event affects the probability of the other, then we say that the events are dependent. In other words, two events are dependent if either P(A | B) ≠ P(A) or P(B | A) ≠ P(B). 26

Independent Versus Dependent Events General Statement 1. Two events are either mutually exclusive or independent. a. Mutually exclusive events are dependent b. Independent events are never mutually exclusive 2. Dependent events may or may not be mutually exclusive. Example #14 Solution There are 160 practicing physicians in a city. Of them, 75 are female and 25 are pediatricians. Of the 75 female, 20 are pediatricians. Are the events “female” and “pediatrician” independent? Explain why or why not. Ped Non-Ped Totals Male 5 80 85 Female 20 55 75 Totals 25 135 160 Since P(female | Pediatrician) ≠ P(female), then the events are not independent. 27

Independent Versus Dependent Events Solution Example #15: Two donut bakers baked 1000 donut holes. Baker A baked 600 donuts, of which 450 were sold and the remaining were discarded. Baker B baked the remaining donuts, of which 100 were discarded. The events are “Baker A”, “Baker B”, “sold donuts”, and “discarded donuts”. Prepare a contingent table for this experiment. Are the events “Baker A” and “sold donuts” independent? Explain why or why not. A S 0 00 1 / 0 75 D 250/1000 0 /75 0 45 B 300/750 A 0 /25 0 15 B 100/250 S|A S|B D|A D|B Sold Donut Discarded Donuts Totals Baker A 450 150 600 Baker B 300 100 400 Totals 750 250 1000 Since P(Baker A | sold donut) = P(Baker A), then the events are independent because the occurrence of event “sold donuts” does affect the probability of event “Baker A”. 28

Independent Versus Dependent Events Example #15. 1: A statistical experiment has 10 equally likely outcomes that are denoted by 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. Let event A = { 10, 12, 14, 16} and event B = {11, 13, 15} a. Are events A and B mutually exclusive? yes b. Are event A and B independent events? Because the two probabilities are not the same, the two events are not independent. Also, we know that mutually exclusive events are always dependent. 29

COMPLEMENTARY EVENTS Definition 1. The complement of event A, denoted by Ā and is read as “A bar” or “A complement, ” is the event that includes all the outcomes for an experiment that are not in A. 2. Therefore, complementary events are always mutually exclusive. 3. Two complementary events, combined together, includes all the outcomes of the experiment. 30

Complementary Events Example #15 – Problem 4. 63: Let A be the event that a number less than 3 is obtained if we roll a die once. What is the probability of A? What is the complementary event of A, and what is its probability? Solution 31

Complementary Events Example #15. 1: A statistical experiment has 10 equally likely outcomes that are denoted by 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. Let event A = { 10, 12, 14, 16} and event B = {11, 13, 15} What are the complements of event A and B, respectively, and their probabilities? Solution 32

4. 3 INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE Suppose an experiment resulted in a sample space described as, S = {1, 2, 3, 4, 5, 6, 7, 8} Also, three events from the experiment are define as consisting of the following outcomes: A = {1, 2, 3, 4} A and B B = {3, 4, 5, 6} A B, or simply AB C = {5, 6, 7, 8} 1. 2. 3. You can see that the Events A and B are not mutually exclusive because they have two common outcomes, 3 and 4. Likewise, Events B and C are not mutually exclusive because they have two outcomes, 5 and 6, in common. Given two Events, A and B, we can say that the intersection of Events A and B is the collection of all outcomes that are common to both A and B. It can be written as, 33

4. 8. 2 Multiplication Rule The probability of the intersection of Events A and B is called the joint probability and is define as the product of the marginal and conditional probabilities. Joint probability is written as, P(A B) = P(A) P(B|A) or P(A B) = P(B) P(A|B) From the two formulas, we can calculate the conditional probabilities as, 34

Multiplication Rule for Independent Events Recall, P(A B) = P(A) P(B|A) or P(A B) = P(B) P(A|B) This is true for dependent events if the probability of one is affected by the occurrence of the other event. In other words, P(A|B) ≠ P(A) and P(B|A)≠ P(B). For independent events, the occurrence of one event does not affect the probability of the other. Therefore, P(A|B) = P(A) and P(B|A) = P(B). Hence, we can rewrite the formula for calculating probability of intersection of two independent events as, P(A B) = P(A) P(B) Note: You can extend the multiplication rule to calculate the joint probability of as many events as you want. 35

Joint Probability of Mutually Exclusive Events We know from discussion of mutually exclusive events that mutually exclusive events have no common outcomes. Therefore, they do not have an intersection. In this case, we write the intersection of two or more mutually exclusive events as P(A B) = 0 Solution Example #17 Find the joint probability of A and B for the a. following: b. a. P(B) = 0. 59 and P(A|B) = 0. 77 b. P(A) = 0. 28 and P(B|A) = 0. 35 Example #18 Find the joint probability for the following three independent events: a. P(A) = 0. 49, P(B) = 0. 67, P(C) =. 75 b. P(A) = 0. 71, P(B) = 0. 34, P(C) = 0. 45 P(A B)= P(B) P(A|B) = (0. 59)(0. 77) = 0. 4543 P(A B)= P(A) P(B|A) = (0. 28)(0. 35) = 0. 098 Solution a. b. P(A B C)= P(A)P(B)P(C) = (. 49)(. 67)(. 75) = 0. 2462 P(A B C)= P(A)P(B)P(C) = (. 71)(. 34)(. 45) = 0. 1086 36

Intersection of Events and Multiplication Rule Solution Example #19 The following table gives two way classification of all basketball players at a state university who began their college careers between 2001 and 2005, based on gender and whether or not they graduate. Graduate Did not Graduate Totals Male 126 55 181 Female 133 32 165 Totals 259 87 346 a. If one of these players is selected at random, find the following probabilities: i. P(female and graduate) ii. P(male and did not graduate) b. Find P(graduate and did not graduate). Is this probability zero? If yes, why? P(graduate and did not graduate) = 0, because these events are mutually exclusive and you could not have someone that is both “a graduate” and “a non graduate”. 37

Intersection of Events and Multiplication Rule Example #20 The following table gives two way classification of the responses based on the education levels of the persons included in the survey and whether they are financially better off, the same as, or worse off than their parents. < High Sch (D) High Sch (E) > High Sch (F) Better off (A) 140 450 420 Same as (B) 60 250 110 Worse off (C) 200 300 70 a. Suppose one adult is selected at random from these 2000 adults. Find the following probabilities: i. P(better off and high school) ii. P(more than high school and worse off) b. Find the joint probability of the events “worse off” and “better off. ” Is this probability zero? Explain why or why not. 38

Intersection of Events and Multiplication Rule Solution < High Sch (D) High Sch (E) > High Sch (F) Totals Better off (A) 140 450 420 1010 Same as (B) 60 250 110 420 Worse off (C) 200 300 70 570 Totals 400 1000 600 2000 39

Intersection of Events and Multiplication Rule Example #21 In a statistics class of 42 students, 28 have volunteered for community service in the past. If two students are selected at random from this class, what is the probability that both of them have volunteered for community service in the past? Draw a tree diagram for this problem. Solution Let A = Student #1 has volunteered for community service. B = Student #1 has never volunteered for community service. C = Student #2 has volunteered for community service. D = Student #2 has never volunteered for community service. 40

Intersection of Events and Multiplication Rule Example #22 The probability that a student graduating from Suburban State University has a student loan to pay off after graduation is 0. 60. If two students are randomly selected from this university, what is the probability that neither of them has loans to pay off after graduation? Solution Let A = Student #1 with loans to pay off after graduation. B = Student #1 without loans to pay off after graduation. C = Student #2 with loans to pay off after graduation. D = Student #2 without loans to pay off after graduation. 41

Intersection of Events and Multiplication Rule Example #23 Five percent of all items sold by a mail order company are returned by customers for a refund. Find the probability that, of two items sold during a given hour by the company, a. both will be returned for a refund. b. neither will be returned for a refund Example #22 - Solution Let A= B= First item will be returned for a refund. First item will not be returned for a refund. Second item will not be returned for a refund. 42

4. 2 UNION OF EVENTS AND THE ADDITION RULE Definition Let a sample space, S, consist of all outcomes in Events A and B. Then the union of the two events is the collection of all outcomes that belong to either A or B or to both A and B. This is denoted by A B or just A or B Addition Rule Addition rule is the method for calculating the probability of the union of events. It is defined as, P(A B) = P(A)+P(B)-P(A B) S A B S Essentially, we calculate the probability of union of events by: 1. Adding the probability of each event and 2. Subtract the probability of the intersection of the events from result in (1). 43

Addition Rule for Mutually Exclusive Events Let re-examine the formula for calculating the probability of union of events. P(A B) = P(A)+P(B)-P(A B) However, we have said that for mutually exclusive events, P(A B) = 0 Then, for mutually exclusive events, the union of two events is, P(A B) = P(A)+P(B) 44

Union of Events and the Addition Rule Example #24 - Solution Example #24 The following table gives two way classification of all basketball players at a state university who began their college careers between 2001 and 2005, based on gender and whether or not they graduate. Graduate (C) Did not Graduate (D) Totals Male (A) 126 55 181 Female (B) 133 32 165 Totals 259 87 346 If one of these players is selected at random, find the following probabilities: a. P(female or did not graduate) b. P(graduate or male) 45

Union of Events and the Addition Rule Example #25 The following table gives two way classification of the responses based on the education levels of the persons included in the survey and whether they are financially better off, the same as, or worse off than their parents. < High Sch (D) High Sch (E) > High Sch (F) Better off (A) 140 450 420 Same as (B) 60 250 110 Worse off (C) 200 300 70 Suppose one adult is selected at random from these 2000 adults. Find the following probabilities: i. P(better off or high school) ii. P(more than high school or worse off) iii. P(better off or worse off) 46

Union of Events and the Addition Rule Solution < High Sch (D) High Sch (E) > High Sch (F) Totals Better off (A) 140 450 420 1010 Same as (B) 60 250 110 420 Worse off (C) 200 300 70 570 Totals 400 1000 600 2000 47

Union of Events and the Addition Rule Example #26 The probability of a student getting an A grade in an economics class is 0. 24 and that of getting a B grade is 0. 28. What is the probability that a randomly selected student from this class will get an A or a B in this class? Explain why the probability is not equal to 1. 0. Solution P(A) = 0. 24 P(B) = 0. 28 Then, the probability of getting an A or B is, P(A or B) = P(A) + P(B) = 0. 24 + 0. 28 = 0. 52 The probability is not equal to 1. 0 because the student can get a grade of C, D, or F. 48

4. 4 COUNTING RULE The task of determining final outcomes in an experiment could be daunting, especially if the experiment is repeated a large number of times. For this reason, we must find a better way. Example #10 Now, suppose a dice is rolled 4 times, what is the number of possible outcomes? Solution: The 1 st die has 6 outcomes The 2 nd die has 6 outcomes for each outcome in 1 st die The 3 rd die has 6 outcomes for each outcome in 2 nd die The 4 th die has 6 outcomes for each outcome in 3 rd die Therefore, Total outcomes = 6. 6 = 1296 49

Counting Rule General Statement Let an experiment be performed four times. The 1 st experiment results in n outcomes, the 2 nd results in m outcomes, the 3 rd results in k outcomes, and the 4 th results in l outcomes. Then, Total number of outcomes = n × m × k × l Example #11 A small ice cream shop has 10 flavors of ice cream and 5 kinds of toppings for its sundaes. How many different selections of one flavor of ice cream and one kind of topping are possible? Solution This experiment has two steps: 1. Selection of an ice cream flavor 2. Selection of a sundae topping For ice cream flavor, the Number of outcomes = 10 For topping, the Number of outcome = 5 Therefore, the total number of possible outcomes = 10 × 5 = 50 50

FACTORIALS, COMBINATIONS, AND PERMUTATIONS Factorial is denoted by the symbol “!”. The factorial of a number is calculated by multiplying all integers from the number to 1. Formal Definition The symbol n!, is define as the product of all the integers from n to 1. In other words, n! = n(n - 1)(n – 2)(n – 3) · · · 3 · 2 · 1 Also note that by definition, 0! = 1 Example #26 51 51

Combinations Suppose we want to select two books from 6 books, and the following conditions applies: 1. This type of selection is without replacement. 2. The order in which selection is made is not important. That is, it is the same selection whether we select book 1 first (1, 2) or book 2 first (2, 1). Then, the question is how many ways can we make this selection? Of course, we can do it the old fashion way, write the selection as, (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 3) (2, 4) (2, 5) (2, 6) (3, 4) (3, 5) (3, 6) (4, 5) (4, 6) (5, 6) Therefore, we can select 2 books from 6 books 15 different ways. Note: 52 Each of the possible selection is called a combination.

Combinations Definition Combinations give the number of ways x element can be selected from n distinct elements. The total number of combinations is given by, and is read as “the number of combinations of n elements selected x at a time. ” The formula for the number of combinations for selecting x from n distinct elements is, Note: 53 53

Combinations Example #27 54 54

Permutations Permutation is similar to combination except that the order of selection is important. For example, how many ways can we select two letters from “A”, “B”, and “C” without replacement and if the order does not matter? This selection is a combination and the total number of combinations is These 3 ways are “AB”, “AC, and “BC”. Now suppose the order matters, then the two letter-combination can be selected or arranged in 6 ways: “AB”, “BA”, “AC”, “CA”, “BC” and “CB” These 6 selections are called 6 permutations or arrangements. 55 55

Permutations Definition Permutations give the total selection of x elements from n different elements such that the order of selection matter. The notation and formula is Example #28 56 56

Permutations Example #29 A ski patrol unit has nine members available for duty, and two of them are to be sent to rescue an injured skier. In how many ways can two of these nine members be selected? Now suppose the order of selection is important. How many arrangements are possible in this case? Solution 57 57