1. Slide 12-1 Copyright © 2005 Pearson Education, Inc. SEVENTH EDITION and EXPANDED SEVENTH EDITION

2. Copyright © 2005 Pearson Education, Inc. Chapter 12 Probability

3. Copyright © 2005 Pearson Education, Inc. 12.1 The Nature of Probability

4. Slide 12-4 Copyright © 2005 Pearson Education, Inc. Definitions An experiment is a controlled operation that yields a set of results. The possible results of an experiment are called its outcomes. An event is a subcollection of the outcomes of an experiment.

5. Slide 12-5 Copyright © 2005 Pearson Education, Inc. Definitions continued Empirical probability is the relative frequency of occurrence of an event and is determined by actual observations of an experiment. Theoretical probability is determined through a study of the possible outcome that can occur for the given experiment.

6. Slide 12-6 Copyright © 2005 Pearson Education, Inc. Empirical Probability Example: In 100 tosses of a fair die, 19 landed showing a 3. Find the empirical probability of the die landing showing a 3. Let E be the event of the die landing showing a 3.

7. Slide 12-7 Copyright © 2005 Pearson Education, Inc. The Law of Large Numbers The law of large numbers states that probability statements apply in practice to a large number of trials, not to a single trial. It is the relative frequency over the long run that is accurately predictable, not individual events or precise totals.

8. Copyright © 2005 Pearson Education, Inc. 12.2 Theoretical Probability

9. Slide 12-9 Copyright © 2005 Pearson Education, Inc. Equally likely outcomes If each outcome of an experiment has the same chance of occurring as any other outcome, they are said to be equally likely outcomes.

10. Slide 12-10 Copyright © 2005 Pearson Education, Inc. Example A die is rolled. Find the probability of rolling a) a 3. b) an odd number c) a number less than 4 d) a 8. e) a number less than 9.

11. Slide 12-11 Copyright © 2005 Pearson Education, Inc. Solutions: There are six equally likely outcomes: 1, 2, 3, 4, 5, and 6. a) b) There are three ways an odd number can occur 1, 3 or 5. c) Three numbers are less than 4.

12. Slide 12-12 Copyright © 2005 Pearson Education, Inc. Solutions: There are six equally likely outcomes: 1, 2, 3, 4, 5, and 6. continued d) There are no outcomes that will result in an 8. e) All outcomes are less than 10. The event must occur and the probability is 1.

13. Slide 12-13 Copyright © 2005 Pearson Education, Inc. Important Facts The probability of an event that cannot occur is 0. The probability of an event that must occur is 1. Every probability is a number between 0 and 1 inclusive; that is, 0  P(E)  1. The sum of the probabilities of all possible outcomes of an experiment is 1.

14. Slide 12-14 Copyright © 2005 Pearson Education, Inc. Example A standard deck of cards is well shuffled. Find the probability that the card is selected. a) a 10. b) not a 10. c) a heart. d) a ace, one or 2. e) diamond and spade. f) a card greater than 4 and less than 7.

15. Slide 12-15 Copyright © 2005 Pearson Education, Inc. Example continued a) a 10 There are four 10’s in a deck of 52 cards. b) not a 10

16. Slide 12-16 Copyright © 2005 Pearson Education, Inc. Example continued c) a heart There are 13 hearts in the deck. d) an ace, 1 or 2 There are 4 aces, 4 ones and 4 twos, or a total of 12 cards.

17. Slide 12-17 Copyright © 2005 Pearson Education, Inc. Example continued d) diamond and spade The word and means both events must occur. This is not possible. e) a card greater than 4 and less than 7 The cards greater than 4 and less than 7 are 5’s, and 6’s.

18. Copyright © 2005 Pearson Education, Inc. 12.3 Odds

19. Slide 12-19 Copyright © 2005 Pearson Education, Inc. Odds Against Example: Find the odds against rolling a 5 on one roll of a die. The odds against rolling a 5 are 5:1.

20. Slide 12-20 Copyright © 2005 Pearson Education, Inc. Odds in Favor

21. Slide 12-21 Copyright © 2005 Pearson Education, Inc. Example Find the odds in favor of landing on blue in one spin of the spinner. The odds in favor of spinning blue are 3:5.

22. Slide 12-22 Copyright © 2005 Pearson Education, Inc. Probability from Odds Example: The odds against spinning a blue on a certain spinner are 4:3. Find the probability that a) a blue is spun. b) a blue is not spun.

23. Slide 12-23 Copyright © 2005 Pearson Education, Inc. Solution Since the odds are 4:3 the denominators must be 4 + 3 = 7. The probabilities ratios are:

24. Copyright © 2005 Pearson Education, Inc. 12.4 Expected Value (Expectation)

25. Slide 12-25 Copyright © 2005 Pearson Education, Inc. Expected Value E = x 1 P(x 1 ) + x 2 P(x 2 ) + x 3 P(x 3 ) +…+ x n P(x n ) or E = P(x 1 )x 1 + P(x 2 )x 2 + P(x 3 )x 3 +…+ P(x n )x n (order doesn’t matter) The symbol P 1 represents the probability that the first event will occur, and A 1 represents the net amount won or lost if the first event occurs.

26. Slide 12-26 Copyright © 2005 Pearson Education, Inc. Example Teresa is taking a multiple-choice test in which there are four possible answers for each question. The instructor indicated that she will be awarded 3 points for each correct answer and she will lose 1 point for each incorrect answer and no points will be awarded or subtracted for answers left blank.  If Teresa does not know the correct answer to a question, is it to her advantage or disadvantage to guess?  If she can eliminate one of the possible choices, is it to her advantage or disadvantage to guess at the answer?

27. Slide 12-27 Copyright © 2005 Pearson Education, Inc. Solution Expected value if Teresa guesses.

28. Slide 12-28 Copyright © 2005 Pearson Education, Inc. Solution continued—eliminate a choice

29. Slide 12-29 Copyright © 2005 Pearson Education, Inc. Example: Winning a Prize When Calvin Winters attends a tree farm event, he is given a free ticket for the $75 door prize. A total of 150 tickets will be given out. Determine his expectation of winning the door prize.

30. Slide 12-30 Copyright © 2005 Pearson Education, Inc. Example When Calvin Winters attends a tree farm event, he is given the opportunity to purchase a ticket for the $75 door prize. The cost of the ticket is $3, and 150 tickets will be sold. Determine Calvin’s expectation if he purchases one ticket.

31. Slide 12-31 Copyright © 2005 Pearson Education, Inc. Solution Calvin’s expectation is  $2.50 when he purchases one ticket.

32. Slide 12-32 Copyright © 2005 Pearson Education, Inc. Fair Price Fair price = expected value + cost to play

33. Slide 12-33 Copyright © 2005 Pearson Education, Inc. Example Suppose you are playing a game in which you spin the pointer shown in the figure, and you are awarded the amount shown under the pointer. If is costs $10 to play the game, determine a) the expectation of the person who plays the game. b) the fair price to play the game. $10 $2 $20$15

34. Slide 12-34 Copyright © 2005 Pearson Education, Inc. Solution $0 3/8 $10 $5  $8 Amt. Won/Lost 1/8 3/8Probability $20$15$2Amt. Shown on Wheel

35. Slide 12-35 Copyright © 2005 Pearson Education, Inc. Solution Fair price = expectation + cost to play =  $1.13 + $10 = $8.87 Thus, the fair price is about $8.87.

36. Copyright © 2005 Pearson Education, Inc. 12.5 Tree Diagrams

37. Slide 12-37 Copyright © 2005 Pearson Education, Inc. Counting Principle If a first experiment can be performed in M distinct ways and a second experiment can be performed in N distinct way, then the two experiments in that specific order can be performed in M N ways. (Note: Order is important.)

38. Slide 12-38 Copyright © 2005 Pearson Education, Inc. Definitions Sample space: A list of all possible outcomes of an experiment. Sample point (simple event): Each individual outcome in the sample space.

39. Slide 12-39 Copyright © 2005 Pearson Education, Inc. Example Two balls are to be selected without replacement from a bag that contains one purple, one blue, and one green ball. a) Use the counting principle to determine the number of points in the sample space. b) Construct a tree diagram and list the sample space. c) Find the probability that one blue ball is selected. d) Find the probability that a purple ball followed by a green ball is selected.

40. Slide 12-40 Copyright © 2005 Pearson Education, Inc. Solutions a) 3 2 = 6 ways b) c) d) P(purple then green = 1/6 Note: The probability of each simple event is B P B G B G P G P PB PG BP BG GP GB

41. Copyright © 2005 Pearson Education, Inc. 12.6 Or and And Problems

42. Slide 12-42 Copyright © 2005 Pearson Education, Inc. Or Problems P(A or B) = P(A) + P(B)  P(A and B) Example: Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 is written on a separate piece of paper. The 10 pieces of paper are then placed in a bowl and one is randomly selected. Find the probability that the piece of paper selected contains an even number or a number greater than 5.

43. Slide 12-43 Copyright © 2005 Pearson Education, Inc. Solution P(A or B) = P(A) + P(B)  P(A and B) Thus, the probability of selecting an even number or a number greater than 5 is 7/10.

44. Slide 12-44 Copyright © 2005 Pearson Education, Inc. Example Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 is written on a separate piece of paper. The 10 pieces of paper are then placed in a bowl and one is randomly selected. Find the probability that the piece of paper selected contains a number less than 3 or a number greater than 7.

45. Slide 12-45 Copyright © 2005 Pearson Education, Inc. Solution There are no numbers that are both less than 3 and greater than 7. Therefore,

46. Slide 12-46 Copyright © 2005 Pearson Education, Inc. Mutually Exclusive Two events A and B are mutually exclusive if it is impossible for both events to occur simultaneously.

47. Slide 12-47 Copyright © 2005 Pearson Education, Inc. Example One card is selected from a standard deck of playing cards. Determine the probability of the following events.  a) selecting a 3 or a jack  b) selecting a jack or a heart  c) selecting a picture card or a red card  d) selecting a red card or a black card

48. Slide 12-48 Copyright © 2005 Pearson Education, Inc. Solutions a) 3 or a jack b) jack or a heart

49. Slide 12-49 Copyright © 2005 Pearson Education, Inc. Solutions continued c) picture card or red card d) red card or black card

50. Slide 12-50 Copyright © 2005 Pearson Education, Inc. And Problems P(A and B) = P(A) P(B) Example: Two cards are to be selected with replacement from a deck of cards. Find the probability that two red cards will be selected.

51. Slide 12-51 Copyright © 2005 Pearson Education, Inc. Example Two cards are to be selected without replacement from a deck of cards. Find the probability that two red cards will be selected.

52. Slide 12-52 Copyright © 2005 Pearson Education, Inc. Independent Events Event A and Event B are independent events if the occurrence of either event in no way affects the probability of the occurrence of the other event. Experiments done with replacement will result in independent events, and those done without replacement will result in dependent events.

53. Slide 12-53 Copyright © 2005 Pearson Education, Inc. Example A package of 30 tulip bulbs contains 14 bulbs for red flowers, 10 for yellow flowers, and 6 for pink flowers. Three bulbs are randomly selected and planted. Find the probability of each of the following.  All three bulbs will produce pink flowers.  The first bulb selected will produce a red flower, the second will produce a yellow flower and the third will produce a red flower.  None of the bulbs will produce a yellow flower.  At least one will produce yellow flowers.

54. Slide 12-54 Copyright © 2005 Pearson Education, Inc. Solution 30 tulip bulbs, 14 bulbs for red flowers, 10 for yellow flowers, and 6 for pink flowers. All three bulbs will produce pink flowers.

55. Slide 12-55 Copyright © 2005 Pearson Education, Inc. Solution 30 tulip bulbs, 14 bulbs for red flowers, 0010 for yellow flowers, and 6 for pink flowers. The first bulb selected will produce a red flower, the second will produce a yellow flower and the third will produce a red flower.

56. Slide 12-56 Copyright © 2005 Pearson Education, Inc. Solution 30 tulip bulbs, 14 bulbs for red flowers, 0010 for yellow flowers, and 6 for pink flowers. None of the bulbs will produce a yellow flower.

57. Slide 12-57 Copyright © 2005 Pearson Education, Inc. Solution 30 tulip bulbs, 14 bulbs for red flowers, 0010 for yellow flowers, and 6 for pink flowers. At least one will produce yellow flowers. P(at least one yellow) = 1  P(no yellow)

58. Copyright © 2005 Pearson Education, Inc. 12.7 Conditional Probability

59. Slide 12-59 Copyright © 2005 Pearson Education, Inc. Conditional Probability In general, the probability of event E 2 occurring, given that an event E 1 has happened is called conditional probability and is written P(E 2 |E 1 ).

60. Slide 12-60 Copyright © 2005 Pearson Education, Inc. Example Given a family of two children, and assuming that boys and girls are equally likely, find the probability that the family has  a) two girls.  b) two girls if you know that at least one of the children is a girl.  c) two girls given that the older child is a girl.

61. Slide 12-61 Copyright © 2005 Pearson Education, Inc. Solutions a)two girls There are four possible outcomes BB, BG, GB, and GG. b) two girls if you know that at least one of the children is a girl

62. Slide 12-62 Copyright © 2005 Pearson Education, Inc. Solutions continued two girls given that the older child is a girl S = {BB, BG, GB, GG} Older is a Girl BG, GG Two girls given older is a girl GG

63. Slide 12-63 Copyright © 2005 Pearson Education, Inc. Conditional Probability For any two events, E 1, and E 2, S = {BB, BG, GB, GG} Older is a Girl BG, GG Two girls given older is a girl GG

64. Slide 12-64 Copyright © 2005 Pearson Education, Inc. Example Use the results of the taste test given at a local mall. If one person from the sample is selected at random, find the probability the person selected 257127130Total 1327260Women 1255570Men TotalPrefers Wintergreen Prefers Peppermint

65. Slide 12-65 Copyright © 2005 Pearson Education, Inc. Example continued a) prefers peppermint b) is a woman

66. Slide 12-66 Copyright © 2005 Pearson Education, Inc. Example continued c) prefers peppermint, given that a woman is selected d) is a man, given that the person prefers wintergreen

67. Copyright © 2005 Pearson Education, Inc. 12.8 The Counting Principle and Permutations

68. Slide 12-68 Copyright © 2005 Pearson Education, Inc. Counting Principle If a first experiment can be performed in M distinct ways and a second experiment can be performed in N distinct ways, then the two experiments in that specific order can be performed in M N ways.

69. Slide 12-69 Copyright © 2005 Pearson Education, Inc. Example A password to logon to a computer system is to consist of 3 letters followed by 3 digits. Determine how many different passwords are possible if  a) repetition of letters and digits is permitted  b) repetition of letters and digits is not permitted  c) the first letter must be a vowel (a, e, I, o, u) and the first digit cannot be 0, and repetition of letters and digits is not permitted.

70. Slide 12-70 Copyright © 2005 Pearson Education, Inc. Solutions repetition of letters and digits is permitted.  There are 26 letters and 10 digits. We have 6 positions to fill.

71. Slide 12-71 Copyright © 2005 Pearson Education, Inc. Solution repetition of letters and digits is not permitted.

72. Slide 12-72 Copyright © 2005 Pearson Education, Inc. Solution the first letter must be a vowel (a, e, i, o, u) and the first digit cannot be 0, and repetition of letters and digits is not permitted.

73. Slide 12-73 Copyright © 2005 Pearson Education, Inc. Permutations A permutation is any ordered arrangement of a given set of objects. (Order is important) Number of Permutations The number of permutations of n distinct items is n factorial, symbolized n!, where n! = n(n  1)(n  2) … (3)(2)(1)

74. Slide 12-74 Copyright © 2005 Pearson Education, Inc. Example How many different ways can 6 stuffed animals be arranged in a line on a shelf? 6! = 6 5 4 3 2 1 = 720 The 6 stuffed animals can be arranged in 720 different ways.

75. Slide 12-75 Copyright © 2005 Pearson Education, Inc. Example Consider the six numbers 1, 2, 3, 4, 5 and 6. In how many distinct ways can three numbers be selected and arranged if repetition is not allowed? 6 5 4 = 120 Thus, there are 120 different possible ordered arrangements, or permutations.

76. Slide 12-76 Copyright © 2005 Pearson Education, Inc. Permutation Formula The number of permutations possible when r objects are selected from n objects is found by the permutation formula

77. Slide 12-77 Copyright © 2005 Pearson Education, Inc. Example The swimming coach has 8 swimmers who can compete in the 100 m butterfly, he can only enter 3 swimmers in the event. In how many ways could he select the 3 swimmers?

78. Slide 12-78 Copyright © 2005 Pearson Education, Inc. Permutations of Duplicate Objects The number of distinct permutations of n objects where n 1 of the objects are identical, n 2 of the objects are identical, …, n r of the objects are identical is found by the formula

79. Slide 12-79 Copyright © 2005 Pearson Education, Inc. Example In how many different ways can the letters of the word “CINCINNATI” be arranged? Of the 10 letters, 2 are C’s, 3 are N’s, and 3 are I’s.

80. Copyright © 2005 Pearson Education, Inc. 12.9 Combinations

81. Slide 12-81 Copyright © 2005 Pearson Education, Inc. Combination A combination is a distinct group (or set) of objects without regard to their arrangement. The number of combinations possible when r objects are selected from n objects is found by

82. Slide 12-82 Copyright © 2005 Pearson Education, Inc. Example A student must select 4 of 7 essay questions to be answered on a test. In how many ways can this selection be made? There are 35 different ways that 4 of 7 questions can be selected.

83. Slide 12-83 Copyright © 2005 Pearson Education, Inc. Example Toastline Bakery is testing 5 new wheat breads, 4 bran breads and 3 oat breads. If it plans to market 2 of the wheat breads, 2 of the bran breads and one of the oat breads, how many different combinations are possible?

84. Copyright © 2005 Pearson Education, Inc. 12.10 Solving Probability Problems by Using Combinations

85. Slide 12-85 Copyright © 2005 Pearson Education, Inc. Example A club consists of 5 men and 6 women. Four members are to be selected at random to form a committee. What is the probability that the committee will consist of two women?

86. Slide 12-86 Copyright © 2005 Pearson Education, Inc. Example The Honey Bear is testing 10 new flavors of ice cream. They are testing 5 vanilla based, 3 chocolate based and 2 strawberry based ice creams. If we assume that each of the 10 flavors has the same chance of being selected and that 4 new flavors will be produced, find the probability that  a) no chocolate flavors are selected.  b) at least 1 chocolate is selected.  c) 2 vanilla and 2 chocolate are selected.

87. Slide 12-87 Copyright © 2005 Pearson Education, Inc. Solution 5 vanilla, 3 chocolate, 2 strawberry selecting 4 flavors a) b)

88. Slide 12-88 Copyright © 2005 Pearson Education, Inc. Solution continued

89. Copyright © 2005 Pearson Education, Inc. 12.11 Binomial Probability Formula

90. Slide 12-90 Copyright © 2005 Pearson Education, Inc. To Use the Binomial Probability Formula There are n repeated independent trials. Each trial has two possible outcomes, success and failure. For each trial, the probability of success (and failure) remains the same.

91. Slide 12-91 Copyright © 2005 Pearson Education, Inc. Binomial Probability Formula The probability of obtaining exactly x successes, P(x), in n independent trials is given by where p is the probability of success on a single trial and q is the probability of failure on a single trial.

92. Slide 12-92 Copyright © 2005 Pearson Education, Inc. Example A basket contains 5 pens: one of each color red, blue, black, green and purple. Five pens are going to be selected with replacement from the basket. Find the probability that  no blue pens are selected  exactly 1 blue pen is selected  exactly 2 blue pens are selected  exactly 3 blue pens are selected

93. Slide 12-93 Copyright © 2005 Pearson Education, Inc. Solution no blue pens exactly 1 blue pen

94. Slide 12-94 Copyright © 2005 Pearson Education, Inc. Solution continued exactly 2 blue pens exactly 3 blue pens

95. Slide 12-95 Copyright © 2005 Pearson Education, Inc. Example A manufacturer of lunch boxes knows that 0.7% of the lunch boxes are defective.  Write the binomial probability formula that would be use to determine the probability that exactly x our of n lunch boxes produced are defective.  Write the binomial probability formula that would be used to find the probability that exactly 4 lunch boxes out of 60 produced will be defective.

96. Slide 12-96 Copyright © 2005 Pearson Education, Inc. Solution

97. Slide 12-97 Copyright © 2005 Pearson Education, Inc. Example The probability that an egg selected at random has a weak shell and will crack is 0.2. Find the probability that  none of 8 eggs selected at random has a weak shell.  at least one of 8 eggs selected has a weak shell.

98. Slide 12-98 Copyright © 2005 Pearson Education, Inc. Solution p = 0.2q = 1  0.2 = 0.8 We want 0 successes in 8 trials. at least 1egg = 1  0.167772 = 0.832228