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# Chapter 4 260110 044531

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• ### Chapter 4 260110 044531

1. 1. Chapter 4 Probability & Counting Rules Reference: Allan G. Bluman (2004) Elementary Statistics: A Step-by Step Approach . New York : McGraw Hill
2. 2. Objectives <ul><li>Determine Sample Spaces and find the probability of an event using classical probability or empirical probability. </li></ul><ul><li>Find the probability of compound events using the addition rules and the multiplication rules. </li></ul><ul><li>Find the conditional probability of an event. </li></ul><ul><li>Find the total number of outcomes in a sequence of events, using the fundamental counting rule. </li></ul><ul><li>Find the number of ways r objects can be selected from n objects using the permutation rule. </li></ul><ul><li>Find the number of ways r objects can be selected from n objects without regard to order using the combination rule. </li></ul><ul><li>Find the probability of an event, using the counting rule. </li></ul>
3. 3. Sample Spaces and Probability <ul><li>A probability experiment is a process that leads to well-defined results called outcomes. </li></ul><ul><li>An outcome is the result of a single trial of a probability experiment. </li></ul><ul><li>An event consists of a set of outcomes of a probability experiment. </li></ul><ul><li>NOTE: A tree diagram can be used as a systematic way to find all possible outcomes of a probability experiment. </li></ul>
4. 4. Tree Diagram for Tossing Two Coins First Toss H T H T H T Second Toss
5. 5. Sample Spaces - Examples
6. 6. Formula for Classical Probability <ul><li>Classical probability assumes that all outcomes in the sample space are equally likely to occur. </li></ul><ul><li>That is, equally likely events are events that have the same probability of occurring. </li></ul>
7. 7. Formula for Classical Probability
8. 8. Classical Probability - Examples <ul><li>For a card drawn from an ordinary deck, find the probability of getting (a) a queen (b) a 6 of clubs (c) a 3 or a diamond. </li></ul><ul><li>Solution: </li></ul><ul><li>(a) Since there are 4 queens and 52 cards, P(queen) = 4/52 = 1/13 . </li></ul><ul><li>(b) Since there is only one 6 of clubs, then P(6 of clubs) = 1/52 . </li></ul>
9. 9. Classical Probability - Examples <ul><li>(c) There are four 3s and 13 diamonds, but the 3 of diamonds is counted twice in the listing. Hence there are only 16 possibilities of drawing a 3 or a diamond, thus P(3 or diamond) = 16/52 = 4/13 . </li></ul>
10. 10. Classical Probability - Examples <ul><li>When a single dice is rolled, find the probability of getting a 9. </li></ul><ul><li>Solution: Since the sample space is 1, 2, 3, 4, 5, and 6, it is impossible to get a 9. </li></ul><ul><li>Hence, P(9) = 0/6 = 0 . </li></ul><ul><li>NOTE: The sum of the probabilities of all outcomes in a sample space is one. </li></ul>
11. 11. Complement of an Event E
12. 12. Complement of an Event - Example <ul><li>Find the complement of each event. </li></ul><ul><li>Rolling a dice and getting a 4. </li></ul><ul><li>Solution: Getting a 1, 2, 3, 5, or 6. </li></ul><ul><li>Selecting a letter of the alphabet and getting a vowel. </li></ul><ul><li>Solution: Getting a consonant. </li></ul>
13. 13. Complement of an Event - Example <ul><li>Selecting a day of the week and getting a weekday. </li></ul><ul><li>Solution: Getting Saturday or Sunday. </li></ul><ul><li>Selecting a one-child family and getting a boy. </li></ul><ul><li>Solution: Getting a girl. </li></ul>
14. 14. Rule for Complementary Event P E P E or P E P E or P E P E ( ) ( )   1 1 1 ( ) =  ( ) ( ) + ( ) = .
15. 15. Empirical Probability <ul><li>The difference between classical and empirical probability is that classical probability assumes that certain outcomes are equally likely while empirical probability relies on actual experience to determine the probability of an outcome. </li></ul>
16. 16. Formula for Empirical Probability
17. 17. Empirical Probability - Example <ul><li>In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had AB blood. Set up a frequency distribution. </li></ul>
18. 18. Empirical Probability - Example Type Frequency A B AB O 22 5 2 21 50 = n
19. 19. Empirical Probability - Example <ul><li>Find the following probabilities for the previous example. </li></ul><ul><li>A person has type O blood. </li></ul><ul><li>Solution: P (O) = f / n = 21/50. </li></ul><ul><li>A person has type A or type B blood. </li></ul><ul><li>Solution: P (A or B) = 22/50+ 5/50 = 27/50. </li></ul>
20. 20. The Addition Rules for Probability <ul><li>Two events are mutually exclusive if they cannot occur at the same time (i.e. they have no outcomes in common). </li></ul>
21. 21. The Addition Rules for Probability A B A and B are mutually exclusive
22. 22. Addition Rule 1 When two events A and B are mutually exclusive, the probability that A or B will occur is P A or B P A P B ( ) ( ) ( )  
23. 23. Addition Rule 1- Example <ul><li>At a political rally, there are 20 Republicans (R), 13 Democrats (D), and 6 Independents (I). If a person is selected, find the probability that he or she is either a Democrat or an Independent. </li></ul><ul><li>Solution: P (D or I) = P (D) + P (I) = 13/39 + 6/39 = 19/39. </li></ul>
24. 24. Addition Rule 1- Example <ul><li>A day of the week is selected at random. Find the probability that it is a weekend. </li></ul><ul><li>Solution: P (Saturday or Sunday) = P (Saturday) + P (Sunday) = 1/7 + 1/7 = 2/7. </li></ul>
25. 25. Addition Rule 2 When two events A and B are not mutually exclusive, the probability y that A or B will occur is P A or B P A P B P A and B ( ) ( ) ( ) ( )   
26. 26. Addition Rule 2 A B A and B (common portion)
27. 27. Addition Rule 2- Example <ul><li>In a hospital unit there are eight nurses and five physicians. Seven nurses and three physicians are females. If a staff person is selected, find the probability that the subject is a nurse or a male. </li></ul><ul><li>The next slide has the data. </li></ul>
28. 28. Addition Rule 2 - Example
29. 29. Addition Rule 2 - Example <ul><li>Solution: P (nurse or male) = P (nurse) + P (male) – P (male nurse) = 8/13 + 3/13 – 1/13 = 10/13. </li></ul>
30. 30. Addition Rule 2 - Example <ul><li>On New Year’s Eve, the probability that a person driving while intoxicated is 0.32, the probability of a person having a driving accident is 0.09, and the probability of a person having a driving accident while intoxicated is 0.06. What is the probability of a person driving while intoxicated or having a driving accident? </li></ul>
31. 31. Addition Rule 2 - Example <ul><li>Solution: </li></ul><ul><li>P (intoxicated or accident) = P (intoxicated) + P (accident) – P(intoxicated and accident) </li></ul><ul><li>= 0.32 + 0.09 – 0.06 = 0.35. </li></ul>
32. 32. <ul><li>Two events A and B are independent if the fact that A occurs does not affect the probability of B occurring. </li></ul><ul><li>Example: Rolling a dice and getting a 6, and then rolling another dice and getting a 3 are independent events. </li></ul>The Multiplication Rules and Conditional Probability
33. 33. Multiplication Rule 1
34. 34. Multiplication Rule 1 - Example <ul><li>A card is drawn from a deck and replaced; then a second card is drawn. Find the probability of getting a queen and then an ace. </li></ul><ul><li>Solution: Because these two events are independent (why?), P (queen and ace) = (4/52)  (4/52) = 16/2704 = 1/169. </li></ul>
35. 35. Multiplication Rule 1 - Example <ul><li>A Harris pole found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer stress at least once a week. </li></ul><ul><li>Solution: Let S denote stress. Then P ( S and S and S ) = (0.46) 3 = 0.097. </li></ul>
36. 36. Multiplication Rule 1 - Example <ul><li>The probability that a specific medical test will show positive is 0.32. If four people are tested, find the probability that all four will show positive. </li></ul><ul><li>Solution: Let T denote a positive test result. Then P ( T and T and T and T) = (0.32) 4 = 0.010. </li></ul>
37. 37. <ul><li>When the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed, the events are said to be dependent. </li></ul><ul><li>Example: Having high grades and getting a scholarship are dependent events. </li></ul>The Multiplication Rules and Conditional Probability
38. 38. <ul><li>The conditional probability of an event B in relationship to an event A is the probability that an event B occurs after event A has already occurred. </li></ul><ul><li>The notation for the conditional probability of B given A is P ( B | A ). </li></ul><ul><li>NOTE: This does not mean B  A . </li></ul>The Multiplication Rules and Conditional Probability
39. 39. Multiplication Rule 2
40. 40. <ul><li>In a shipment of 25 microwave ovens, two are defective. If two ovens are randomly selected and tested, find the probability that both are defective if the first one is not replaced after it has been tested. </li></ul><ul><li>Solution: See next slide. </li></ul>The Multiplication Rules and Conditional Probability - Example
41. 41. <ul><li>Solution: Since the events are dependent, </li></ul><ul><li>P ( D 1 and D 2 ) = P ( D 1 )  P ( D 2 | D 1 ) </li></ul><ul><li> = (2/25)(1/24) </li></ul><ul><li>= 2/600 </li></ul><ul><li> = 1/300. </li></ul>The Multiplication Rules and Conditional Probability - Example
42. 42. <ul><li>The WW Insurance Company found that 53% of the residents of a city had homeowner’s insurance with its company. Of these clients, 27% also had automobile insurance with the company. If a resident is selected at random, find the probability that the resident has both homeowner’s and automobile insurance. </li></ul>The Multiplication Rules and Conditional Probability - Example
43. 43. <ul><li>Solution: Since the events are dependent, </li></ul><ul><li>P ( H and A ) = P ( H )  P ( A | H ) = (0.53)(0.27) = 0.1431. </li></ul>The Multiplication Rules and Conditional Probability - Example
44. 44. <ul><li>Box 1 contains two red balls and one blue ball. Box 2 contains three blue balls and one red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn. Find the probability of selecting a red ball. </li></ul>The Multiplication Rules and Conditional Probability - Example
45. 45. Tree Diagram for Example P ( B 1 ) 1/2 Red Red Blue Blue Box 1 P ( B 2 ) 1/2 Box 2 P ( R | B 1 ) 2/3 P ( B | B 1 ) 1/3 P ( R | B 2 ) 1/4 P ( B | B 2 ) 3/4 (1/2)(2/3) (1/2)(1/3) (1/2)(1/4) (1/2)(3/4)
46. 46. <ul><li>Solution: P (red) = (1/2)(2/3) + (1/2)(1/4) = 2/6 + 1/8 = 8/24 + 3/24 = 11/24. </li></ul>The Multiplication Rules and Conditional Probability - Example
47. 47. Conditional Probability - Formula
48. 48. <ul><li>The probability that Sam parks in a no-parking zone and gets a parking ticket is 0.06, and the probability that Sam cannot find a legal parking space and has to park in the no-parking zone is 0.2. On Tuesday, Sam arrives at school and has to park in a no-parking zone. Find the probability that he will get a ticket. </li></ul>Conditional Probability - Example
49. 49. <ul><li>Solution: Let N = parking in a no-parking zone and T = getting a ticket. </li></ul><ul><li>Then P ( T | N ) = [ P ( N and T ) ]/ P ( N ) = 0.06/0.2 = 0.30. </li></ul>Conditional Probability - Example
50. 50. <ul><li>A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat. The results are shown in the table on the next slide. </li></ul>Conditional Probability - Example
51. 51. Conditional Probability - Example
52. 52. <ul><li>Find the probability that the respondent answered “yes” given that the respondent was a female. </li></ul><ul><li>Solution: Let M = respondent was a male; </li></ul><ul><li> F = respondent was a female; </li></ul><ul><li>Y = respondent answered “yes”; N = respondent answered “no”. </li></ul>Conditional Probability - Example
53. 53. <ul><li>P ( Y | F ) = [ P ( F and Y ) ]/ P ( F ) = [8/100]/[50/100] = 4/25. </li></ul><ul><li>Find the probability that the respondent was a male, given that the respondent answered “no”. </li></ul><ul><li>Solution: P ( M | N ) = [ P ( N and M )]/ P ( N ) = [18/100]/[60/100] = 3/10. </li></ul>Conditional Probability - Example
54. 54. Tree Diagrams <ul><li>A tree diagram is a device used to list all possibilities of a sequence of events in a systematic way. </li></ul>
55. 55. Tree Diagrams - Example <ul><li>Suppose a sales person can travel from New York to Pittsburgh by plane, train, or bus, and from Pittsburgh to Cincinnati by bus, boat, or automobile. Display the information using a tree diagram. </li></ul>
56. 56. Tree Diagrams - Example Cincinnati Bus New York Pittsburgh Plane Train Bus Boat Auto Bus Boat Boat Bus Auto Auto Plane, Bus Plane, boat Plane, auto Train, bus Train, boat Train, auto Bus, bus Bus, boat Bus, auto
57. 57. The Multiplication Rule for Counting <ul><li>Multiplication Rule : In a sequence of n events in which the first one has k 1 possibilities and the second event has k 2 and the third has k 3 , and so forth, the total possibilities of the sequence will be k 1  k 2  k 3  k n . </li></ul>
58. 58. The Multiplication Rule for Counting - Example <ul><li>A nurse has three patients to visit. How many different ways can she make her rounds if she visits each patient only once? </li></ul>
59. 59. The Multiplication Rule for Counting - Example <ul><li>She can choose from three patients for the first visit and choose from two patients for the second visit, since there are two left. On the third visit, she will see the one patient who is left. Hence, the total number of different possible outcomes is 3  2  1  = 6. </li></ul>
60. 60. The Multiplication Rule for Counting - Example <ul><li>Employees of a large corporation are to be issued special coded identification cards. The card consists of 4 letters of the alphabet. Each letter can be used up to 4 times in the code. How many different ID cards can be issued? </li></ul>
61. 61. The Multiplication Rules for Counting - Example <ul><li>Since 4 letters are to be used, there are 4 spaces to fill ( _ _ _ _ ). Since there are 26 different letters to select from and each letter can be used up to 4 times, then the total number of identification cards that can be made is 26  26  26  26  = 456,976. </li></ul>
62. 62. The Multiplication Rule for Counting - Example <ul><li>The digits 0, 1, 2, 3, and 4 are to be used in a 4-digit ID card. How many different cards are possible if repetitions are permitted? </li></ul><ul><li>Solution: Since there are four spaces to fill and five choices for each space, the solution is 5  5  5  5 = 5 4 = 625. </li></ul>
63. 63. The Multiplication Rule for Counting - Example <ul><li>What if the repetitions were not permitted in the previous example? </li></ul><ul><li>Solution: The first digit can be chosen in five ways. But the second digit can be chosen in only four ways, since there are only four digits left; etc. Thus the solution is 5  4  3  2 = 120. </li></ul>
64. 64. Permutations <ul><li>Consider the possible arrangements of the letters a , b , and c . </li></ul><ul><li>The possible arrangements are: abc, acb, bac, bca, cab, cba . </li></ul><ul><li>If the order of the arrangement is important then we say that each arrangement is a permutation of the three letters. Thus there are six permutations of the three letters. </li></ul>
65. 65. Permutations <ul><li>An arrangement of n distinct objects in a specific order is called a permutation of the objects. </li></ul><ul><li>Note: To determine the number of possibilities mathematically, one can use the multiplication rule to get: 3  2  1 = 6 permutations. </li></ul>
66. 66. Permutations <ul><li>Permutation Rule : The arrangement of n objects in a specific order using r objects at a time is called a permutation of n objects taken r objects at a time. It is written as n P r and the formula is given by n P r = n! / (n – r)! . </li></ul>
67. 67. Permutations - Example <ul><li>How many different ways can a chairperson and an assistant chairperson be selected for a research project if there are seven scientists available? </li></ul><ul><li>Solution: Number of ways = 7 P 2 = 7! / (7 – 2)! = 7!/5! = 42 . </li></ul>
68. 68. Permutations - Example <ul><li>How many different ways can four books be arranged on a shelf if they can be selected from nine books? </li></ul><ul><li>Solution: Number of ways = 9 P 4 = 9! / (9 – 4)! = 9!/5! = 3024 . </li></ul>
69. 69. Combinations <ul><li>Consider the possible arrangements of the letters a , b , and c . </li></ul><ul><li>The possible arrangements are: abc, acb, bac, bca, cab, cba . </li></ul><ul><li>If the order of the arrangement is not important then we say that each arrangement is the same. We say there is one combination of the three letters. </li></ul>
70. 70. Combinations <ul><li>Combination Rule : The number of combinations of of r objects from n objects is denoted by n C r and the formula is given b n C r = n! / [(n – r)!r!] . </li></ul>
71. 71. Combinations - Example <ul><li>How many combinations of four objects are there taken two at a time? </li></ul><ul><li>Solution: Number of combinations: 4 C 2 = 4! / [(4 – 2)! 2!] = 4!/[2!2!] = 6 . </li></ul>
72. 72. Combinations - Example <ul><li>In order to survey the opinions of customers at local malls, a researcher decides to select 5 malls from a total of 12 malls in a specific geographic area. How many different ways can the selection be made? </li></ul><ul><li>Solution: Number of combinations: 12 C 5 = 12! / [(12 – 5)! 5!] = 12!/[7!5!] = 792 . </li></ul>
73. 73. Combinations - Example <ul><li>In a club there are 7 women and 5 men. A committee of 3 women and 2 men is to be chosen. How many different possibilities are there? </li></ul><ul><li>Solution: Number of possibilities: (number of ways of selecting 3 women from 7)  (number of ways of selecting 2 men from 5) = 7 C 3  5 C 2 = (35)(10) = 350 . </li></ul>
74. 74. Combinations - Example <ul><li>A committee of 5 people must be selected from 5 men and 8 women. How many ways can the selection be made if there are at least 3 women on the committee? </li></ul>
75. 75. Combinations - Example <ul><li>Solution: The committee can consist of 3 women and 2 men, or 4 women and 1 man, or 5 women. To find the different possibilities, find each separately and then add them: </li></ul><ul><li>8 C 3  5 C 2 + 8 C 4  5 C 1 + 8 C 5  5 C 0 </li></ul><ul><li>= (56)(10) + (70)(5) + (56)(1) </li></ul><ul><li>= 966 . </li></ul>