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Section 11.1 The Fundamental Counting Principle

Section 11.1 The Fundamental Counting Principle. Objective 1. Use the Fundamental Counting Principle to determine the number of possible outcomes in a given situation. Fundamental Counting Principle. Definition

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Section 11.1 The Fundamental Counting Principle

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  1. Section 11.1 The Fundamental Counting Principle Objective 1. Use the Fundamental Counting Principle to determine the number of possible outcomes in a given situation. Section 11.1

  2. Fundamental Counting Principle • Definition If you can choose one item from a group of M items and a second item from a group of N items, then the total number of two-item choices is M∙N. • Tree Diagram A representation of all possible choices. This tree diagram shows that there are 2·3 = 6 different Outfits from 2 pairs of jeans and three T-shirts. Section 11.1

  3. Example 1Applying the Fundamental Counting Principle The Greasy Spoon Restaurant offers 6 appetizers and 14 main courses. In how many ways can a person order a two-course meal? Solution: Choosing from one of 6 appetizers and one of 14 main courses, the total number of two-course meals is: 6 ∙14 = 84 Section 11.1

  4. The Fundamental Counting Principle with More than Two Groups of Items • Definition The number of ways in which a series of successive things can occur is found by multiplying the number of ways in which each thing can occur. • The number of possible outfits from 2 pair of jeans, 3 T-shirts and 2 pairs of sneakers are: 2·3·2 = 12 Section 11.1

  5. Example 2 Applying the Fundamental Principle of Counting with More Than Two Groups of Items Options in Planning a Course Schedule Next semester, you are planning to take three courses – math, English and humanities. There are 8 sections of math, 5 of English, and 4 of humanities that you find suitable. Assuming no scheduling conflicts, how many different three- course schedules are possible? Solution: This situation involves making choices with three groups of items. Math English Humanities {8 choices} {5 choices} {4 choices} There are 8∙5∙4 = 160 different three-course schedules. Section 11.1

  6. Example 3 A Multiple-Choice Test • You are taking a multiple-choice test that has ten questions. Each of the questions has four answer choices, with one correct answer per question. If you select one of these four choices for each question and leave nothing blank, in how man ways can you answer the questions? • Solution: This situation involves making choices with ten questions: Question 1 Question 2 Question 3 ∙ ∙ ∙ Question 9 Question 10 {4 choices} {4 choices} {4 choices} {4 choices} {4 choices} The number of different ways you can answer the questions is: 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 = 410 = 1,048,576 Section 11.1

  7. Example 4Telephone Numbers in the United States • Telephone numbers in the United States begin with three- digit area codes followed by seven-digit local telephone numbers. Area codes and local telephone numbers cannot begin with 0 or 1. How many different telephone numbers are possible? • Solution: This situation involves making choices with ten groups of items. Here are the choices for each of the ten groups of items:Area Code Local Telephone Number 8 10 10 8 10 10 10 10 10 10 The total number of different telephone numbers is: 8 ∙ 10 ∙ 10 ∙ 8 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 = 6,400,000,000 Section 11.1

  8. Section 11.2Permutations Objectives Use the Fundamental Counting Principle to count permutations. Evaluate factorial expressions. Use the permutations formula. Find the number of permutations of duplicate items. Section 11.2

  9. Permutations • Permutation is an ordered arrangement of items that occurs when: • No item is used more than once. • The order of arrangement makes a difference. Section 11.2

  10. Example 1Counting Permutations • You are in charge of planning a concert tour with U2, ‘N Sync, Aerosmith and the Rolling Stones. You decide that the Rolling Stones will perform last. How many different ways can you put together the concert? • Solution: You can now choose any one of the three groups, U2, ‘N sync or Aerosmith, as the opening act. That leaves two groups left to choose from for the second group and only one group left for the third group. Using the Fundamental Counting Principle there are: 3∙2∙1∙1 = 6 different ways to arrange the concert. Section 11.2

  11. Factorial Notation • The product 7∙6∙5∙4∙3∙2∙1 is called factorial and is written 7! • Definition: If n is a positive integer, the notation n! (read “n factorial”) is the product of all positive integers from n down through 1. n! = n(n - 1)(n - 2)···(3)(2)(1) 0! (zero factorial), by definition, is 1. 0! = 1 Section 11.2

  12. Example 2Using Factorial Notation • Evaluate without using your calculator: ∕ ∕ ∕ ∕ Section 11.2

  13. A Formula for Permutations • Permutations of n things taken r at a time. Section 11.2

  14. Example 3Using the Formula for Permutations • You and 19 of your friends have decided to form a business. The group needs to choose three officers– a CEO, an operating manager, and a treasure. In how many ways can those offices be filled? • Solution: Your group is choosing r = 3 officers from a group of n = 20 people. The order matters because each officer has different responsibilities: ∕ ∕ Section 11.2

  15. A Formula for Permutations of Duplicate Items • Permutations of Duplicate Items The number of permutations of n items, where p items are identical, q items are identical, r items are identical, and so on, is given by: Section 11.2

  16. Example 4Using the Formula for Permutations of Duplicate Items • In how many distinct ways can the letters of the word MISSISSIPPI be arranged? • Solution: The word contains 11 letters (n = 11) where four Is are identical (p = 4), four Ss are identical (q = 4) and 2 Ps are identical (r = 2). The number of distinct permutations are: Section 11.2

  17. Section 11.3Combinations Objectives • Distinguish between permutation and combination problems. • Solve problems involving combinations using the combinations formula. Section 11.3

  18. Combinations A combination of items occurs when • The items are selected from the same group. • No item is used more than once. • The order of items makes no difference. Note: • Permutation problems involve situations in which order matters. • Combination problems involve situations in which the order of items makes no difference. Section 11.3

  19. Comparing Combinations and Permutations Given the letters: A,B,C,D: We can compare how many permutations and how many combinations are possible if we chose 3 letters at a time: Section 11.3

  20. Example 1Distinguishing Between Permutations and Combinations Determine which involve permutations and which involve combinations. • Six student are running for student government president, vice-president and treasure. The student with the greatest number of votes becomes the president, the second highest vote-getter becomes vice-president, and the student who gets the third largest number of votes will be treasurer. How many different outcomes are possible? • Solution: Order matters since the number of the three highest votes determine the office. Permutations Section 11.3

  21. Example 1Distinguishing Between Permutations and Combinations (continued) • Six people are on the board of supervisors for your neighborhood part. A three-person committee is needed to study the possibility of expanding the park. How many different committees could be formed from the six people? • Solution: The order in which the three people are selected does not matter since they are not filling different roles. Combinations Section 11.3

  22. A Formula for Combinations The number of possible combinations if r items are taken from n items is: Section 11.3

  23. Example 2Using the Formula for Combinations • How many three-person committees could be formed from 8 people? • Solution: We are selecting 3 people ( r = 3) from 8 people ( n = 8) ∕ ∕ Section 11.3

  24. Example 3Using the Formula for Combinations and the Fundamental Counting Principle • The U.S Senate of the 104th Congress consisted of 54 Republicans and 46 Democrats. How many committees can be formed if each committee must have 3 Republicans and 2 Democrats? • Solution: The order in which members are selected does not matter so this is a problem of combinations. Section 11.3

  25. Example 3 continued First, select 3 Republicans out of 54 Republicans: Secondly, select 2 Democrats out of 46 Democrats: Use the Fundamental Counting Principle to find the number of committees that can be formed. Section 11.3

  26. Section 11.4Fundamentals of Probability Objectives Compute theoretical probability. Compute empirical probability. Section 11.4

  27. Probability • Probabilities are assigned values from 0 to 1. • The closer the probability of a given event is to 1, the more likely it is that the event will occur. • The closer the probability of a given event is to 0, the less likely that the event will occur. Section 11.4

  28. Theoretical Probability • Experiment is any occurrence for which the outcome is uncertain. • Sample space is the set of all possible outcomes of an experiment , denoted by S. • Event, denoted by E is any subset of a sample space. • Sum of the theoretical probabilities of all possible outcomes is 1. Section 11.4

  29. Computing Theoretical Probability • If an event E has n(E) equally likely outcomes and its sample space S has n(S) equally-likely outcomes, the theoretical probability of event E, denoted by P(E), is: • P(E) = number of outcomes in event E = n(E) total number of possible outcomes n(S) Section 11.4

  30. Example 1Computing Theoretical Probability • A die is rolled once. The sample space is S = {1,2,3,4,5,6} with n(S) = 6. Find the probability of rolling • a 3 b. an even number • Solution to a: There is only one way to roll a 3 so n(E) = 1. P(3) = number of outcomes that result in 3 = n(E) = 1 total number of possible outcomes n(S) 6 • Solution to b: Rolling an even number describes the event E = {2,4,6}. This event can occur in 3 ways: n(E) = 3. P(even number) = number of outcomes that result in even number = n(E)= 3 =1 total number of possible outcomes n(S) 6 2 Section 11.4

  31. Example 2Probability and a Deck of 52 Cards • You are dealt one card from a standard 52-card deck. Find the probability of being dealt a • King Solution: P(3) = number of outcomes that result in a king = n(E) = 4 =1 total number of possible outcomes n(S) 52 13 • Heart Solution: P(heart) = number of outcomes that result in a heart = n(E) = 13=1 total number of possible outcomes n(S) 52 4 Section 11.4

  32. Empirical Probability • Applies to situations in which we observe how frequently an event occurs. • Computing Empirical Probability The empirical probability of event E is: • P(E) = observed number of times E occurs = n(E) total number of observed occurrences n(S) Section 11.4

  33. Example 4Computing Empirical Probability If one person is randomly selected from the population described above, find the probability that the person is female. Solution. The probability of selecting a female is the observed number of females, 110.1(million), divided by the total number of U.S. adults, 212.5 (million). P (selecting a female) = number females_ = 110.1≈ 0.52 total number of adults 212.5 Section 11.4

  34. Section 11.5Probability with the Fundamental counting Principle, Permutations, and Combinations Objectives Compute probabilities with permutations. Compute probabilities with combinations, Section 11.5

  35. Example 1Probability with Permutations. • Five groups agree to randomly select the order of performance by picking cards out of a hat, one at a time. What is the probability of the Rolling Stones performing fourth and the Beatles last? • Solution: P (Rolling Stones fourth, Beatles last) = Number of permutations with Rolling Stones fourth, Beatles last Total number of possible permutations • Use the Fundamental Counting Principle to find the total number of possible permutations. • There are 5·4·3·2·1 =120 possible permutations. Section 11.5

  36. Example 1 continued • Use the Fundamental Counting Principle to find the number of permutations with the Rolling Stones performing fourth and the Beatles performing last. • There are 3·2·1·1·1 = 6 possible permutations. • Putting both values in the original equation. P (Rolling Stones fourth, Beatles last) = 6 = 1 120 20 Section 11.5

  37. Example 2Probability and Combinations: Winning the Lottery • Florida’s lottery game, LOTTO, is set up so that each player chooses six different numbers from 1 to 53. With one LOTTO ticket, what is the probability of winning this prize? • Solution: Because the order of the six numbers does not matter, this situation involves combinations. P (winning) = number of ways of winning total number of possible combinations Section 11.5

  38. Example 2 continued P (winning) = number of ways of winning total number of possible combinations Using the combinations formula, r = 6 and n = 53. If a person buys only one ticket, then that person has selected only one combination, thus P (winning) = number of ways of winning ___ = __1___ total number of possible combinations 22,957,480 ≈ 0.0000000436 Section 11.5

  39. Example 3Probability and Combinations • A club consists of five men and seven women. Three members are selected at random to attend a conference. Find the probability that the selected group consists of 3 men. • Solution: Order of selection does not matter, so this is a problem involving combinations. P(3 men) = number of ways of selecting 3 men total number of possible combinations Section 11.5

  40. Example 3 continued Consider the denominator: We are selecting r = 3 people from a total group of n = 12. Consider the numerator. We are interested in selecting 3 (r = 3)men from 5 (n = 5) men. Therefore: P(3 men ) = 10 = 1 220 22 Section 11.5

  41. 11.6Events Involving Not and Or; Odds Objectives Find the probability that an event will not occur. Find the probability of one event or a second event occurring. Understand and use odds. Section 11.6

  42. Probability of an Event Not Occurring • Complement of E: If we know P(E), the probability of an event E, we can determine the probability that the event will not occur, denoted by P(not E). • The probability that an event E will not occur is equal to 1 minus the probability that it will occur. P(not E) = 1 – P(E) • The probability that an event E will occur is equal to 1 minus the probability that it will not occur. P(E) = 1 – P(not E) Using set notation, if E’ is the complement of E, then P(E’) = 1 – P(E) and P(E) = 1 – P(E’) Section 11.6

  43. Example 1The Probability of an Event Not Occurring • If you are dealt one card from a standard 52-card deck, find the probability that you are not dealt a queen. • Solution: Because P(not E) = 1 – P(E) then P(not a queen) = 1 – P(queen). • There are 4 queens in a standard deck. • The probability of being dealt a queen is 4 =1 52 13 • P(not a queen) = 1 – P(queen) = 1 – 1 = 12 13 13 Section 11.6

  44. Or Probabilities with Mutually Exclusive Events • Mutually Exclusive Events: Events A and B are mutually exclusive if it is impossible for them to occur simultaneously. • Or Probabilities with Mutually Exclusive Events: • If A and B are mutually exclusive events, then P(A or B) = P(A) + P(B) • Using set notation, P(A U B) = P(A) + P(B) Section 11.6

  45. Example 2The Probability of Either of Two Mutually Exclusive Events Occurring • If one card is randomly selected from a deck of cards, what is the probability of selecting a king or a queen? • Solution: P ( king or queen) = P(king) + P(queen) = 4 + 4 = 8 = 2 52 52 52 13 Section 11.6

  46. Or Probabilities with Events That Are Not Mutually Exclusive • If A and B are not mutually exclusive events then the probability that A or B will occur is determined by adding their individual probabilities and then subtracting the probability that A and B occur simultaneously. • P(A or B) = P(A) + P(B) – P(A and B) • Using set notation, P(A U B) = P(A) + P(B) – P(A ∩B) Section 11.6

  47. Example 3An OR Probability With Events That Are Not Mutually Exclusive • In a group of 25 baboons, 18 enjoy grooming their neighbors, 16 enjoy screeching wildly, while 10 enjoy doing both. If one baboon is selected at random, find the probability that it enjoy grooming its neighbors or screeching wildly. • Solution: Since 10 of the baboons enjoy both grooming their neighbors and screeching wildly, these events are not mutually exclusive. Section 11.6

  48. Example 3 continued Section 11.6

  49. Probability to Odds • If we know the probability of an event, we can determine the odds in favor, or the odds against, the event. Probability to Odds: • The odds in favor of E are found by taking the probability that E will occur and dividing by the probability that E will not occur. Odds in favor of E = P(E) . P(not E) • The odds against E are found by taking the probability that E will not occur and dividing by the probability that E will occur. Odds against E = P(not E) P(E) The odds against E can also be found by reversing the ratio representing the odds in favor of E. Section 11.6

  50. Example 4From Probability to Odds • You roll a single, six-sided die. Find the odds in favor of rolling a 2 Solution: Let E represent the event of rolling a 2 • First, find the probability of E occurring and the probability of E not occurring. S = {1,2,3,4,5,6} E = {2} P(E) =1 and P(not E) = 1 – P(E) = 1 – 1 = 5 6 6 6 1 • Odds in favor of E( rolling a 2 ) = P(E) . = 6 = 1 .6 =1 P(not E) 5 6 5 5 6 Section 11.6

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