4. What is the probability of picking an ace?

5. Probability =

6. What is the probability of picking an ace? 4 / 52 =.077 or 7.7 chances in 100

7. Every card has the same probability of being picked

8. What is the probability of getting a 10, J, Q, or K?

9. (.077) + (.077) + (.077) + (.077) =.308 16 / 52 =.308

10. What is the probability of getting a 2 and then after replacing the card getting a 3 ?

11. (.077) * (.077) =.0059

12. What is the probability that the two cards you draw will be a black jack?

13. 10 Card = (.077) + (.077) + (.077) + (.077) =.308 Ace after one card is removed = 4/51 =.078 (.308)*(.078) =.024

14. Practice What is the probability of rolling a “1” using a six sided dice? What is the probability of rolling either a “1” or a “2” with a six sided dice? What is the probability of rolling two “1’s” using two six sided dice?

15. Practice What is the probability of rolling a “1” using a six sided dice? 1 / 6 =.166 What is the probability of rolling either a “1” or a “2” with a six sided dice? What is the probability of rolling two “1’s” using two six sided dice?

16. Practice What is the probability of rolling a “1” using a six sided dice? 1 / 6 =.166 What is the probability of rolling either a “1” or a “2” with a six sided dice? (.166) + (.166) =.332 What is the probability of rolling two “1’s” using two six sided dice?

17. Practice What is the probability of rolling a “1” using a six sided dice? 1 / 6 =.166 What is the probability of rolling either a “1” or a “2” with a six sided dice? (.166) + (.166) =.332 What is the probability of rolling two “1’s” using two six sided dice? (.166)(.166) =.028

18. Cards What is the probability of drawing an ace? What is the probability of drawing another ace? What is the probability the next four cards you draw will each be an ace? What is the probability that an ace will be in the first four cards dealt?

19. Cards What is the probability of drawing an ace? 4/52 =.0769 What is the probability of drawing another ace? 4/52 =.0769; 3/51 =.0588;.0769*.0588 =.0045 What is the probability the next four cards you draw will each be an ace?.0769*.0588*.04*.02 =.000003 What is the probability that an ace will be in the first four cards dealt?.0769+.078+.08+.082 =.3169

20. Probability 1.00.00 Event will not occur Event must occur

21. Probability In this chapter we deal with discreet variables –i.e., a variable that has a limited number of values Previously we discussed the probability of continuous variables (Z –scores) –It does not make sense to seek the probability of a single score for a continuous variable Seek the probability of a range of scores

22. Key Terms Independent event –When the occurrence of one event has no effect on the occurrence of another event e.g., voting behavior, IQ, etc. Mutually exclusive –When the occurrence of one even precludes the occurrence of another event e.g., your year in the program, if you are in prosem

23. Key Terms Joint probability –The probability of the co-occurrence of two or more events The probability of rolling a one and a six p (1, 6) p (Blond, Blue)

24. Key Terms Conditional probabilities –The probability that one event will occur given that some other vent has occurred e.g., what is the probability a person will get into a PhD program given that they attended Villanova –p(Phd l Villa) e.g., what is the probability that a person will be a millionaire given that they attended college –p($$ l college)

25. Example Owns a video game Does not own a video game Total No Children103545 Children253055 Total3565100

26. What is the simple probability that a person will own a video game? Owns a video game Does not own a video game Total No Children103545 Children253055 Total3565100

27. What is the simple probability that a person will own a video game? 35 / 100 =.35 Owns a video game Does not own a video game Total No Children103545 Children253055 Total3565100

28. What is the conditional probability of a person owning a video game given that he or she has children? p (video l child) Owns a video game Does not own a video game Total No Children103545 Children253055 Total3565100

29. What is the conditional probability of a person owning a video game given that he or she has children? 25 / 55 =.45 Owns a video game Does not own a video game Total No Children103545 Children253055 Total3565100

30. What is the joint probability that a person will own a video game and have children? p(video, child) Owns a video game Does not own a video game Total No Children103545 Children253055 Total3565100

31. What is the joint probability that a person will own a video game and have children? 25 / 100 =.25 Owns a video game Does not own a video game Total No Children103545 Children253055 Total3565100

32. 25 / 100 =.25.35 *.55 =.19 Owns a video game Does not own a video game Total No Children103545 Children253055 Total3565100

33. The multiplication rule assumes that the two events are independent of each other – it does not work when there is a relationship! Owns a video game Does not own a video game Total No Children103545 Children253055 Total3565100

34. Practice RepublicanDemocratTotal Male522779 Female186583 Total7092162

35. p (republican)p(female) p (republican, male)p(female, republican) p (republican l male)p(male l republican) RepublicanDemocratTotal Male522779 Female186583 Total7092162

36. p (republican) = 70 / 162 =.43 p (republican, male) = 52 / 162 =.32 p (republican l male) = 52 / 79 =.66 RepublicanDemocratTotal Male522779 Female186583 Total7092162

37. p(female) = 83 / 162 =.51 p(female, republican) = 18 / 162 =.11 p(male l republican) = 52 / 70 =.74 RepublicanDemocratTotal Male522779 Female186583 Total7092162

38. Foot Race Three different people enter a “foot race” A, B, C How many different combinations are there for these people to finish?

39. Foot Race A, B, C A, C, B B, A, C B, C, A C, B, A C, A, B 6 different permutations of these three names taken three at a time

40. Foot Race Six different people enter a “foot race” A, B, C, D, E, F How many different permutations are there for these people to finish?

41. Permutation Ingredients: N = total number of events r = number of events selected

42. Permutation Ingredients: N = total number of events r = number of events selected A, B, C, D, E, F Note: 0! = 1

43. Foot Race Six different people enter a “foot race” A, B, C, D, E, F How many different permutations are there for these people to finish in the top three? A, B, CA, C, DA, D, EB, C, A

44. Permutation Ingredients: N = total number of events r = number of events selected

45. Permutation Ingredients: N = total number of events r = number of events selected

46. Foot Race Six different people enter a “foot race” If a person only needs to finish in the top three to qualify for the next race (i.e., we don’t care about the order) how many different outcomes are there?

47. Combinations Ingredients: N = total number of events r = number of events selected

48. Combinations Ingredients: N = total number of events r = number of events selected

49. Note: Use Permutation when ORDER matters Use Combination when ORDER does not matter

50. Practice There are three different prizes –1 st $1,00 –2 nd $500 –3 rd $100 There are eight finalist in a drawing who are going to be awarded these prizes. A person can only win one prize How many different ways are there to award these prizes?

51. Practice 336 ways of awarding the three different prizes

52. Practice There are three prizes (each is worth $200) There are eight finalist in a drawing who are going to be awarded these prizes. A person can only win one prize How many different ways are there to award these prizes?

53. Combinations 56 different ways to award these prizes

54. Practice A shirt comes in four sizes and six colors. One also has the choice of a dragon, alligator, or no emblem on the pocket. How many different kinds of shirts can you order?

55. Practice A shirt comes in four sizes and six colors. One also has the choice of a dragon, alligator, or no emblem on the pocket. How many different kinds of shirts can you order? 4*6*3 = 72 Don’t make it hard on yourself!

56. Practice In a California Governor race there were 135 candidates. The state created ballots that would list candidates in different orders. How many different types of ballots did the state need to create?

57. Practice 2.6904727073180495e+230 Or

58. 26,904,727,073,180,495,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,00 0,000,000,000,000,000,000,000,000,000,0 00,000,000,000,000,000,000,000,000,000, 000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,00 0,000,000,000,000,000,000,000,000,000

59. Bonus Points Suppose you’re on a game show and you’re given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice?

60. http://www.nytimes.com/2008/04/08/scienc e/08monty.html?_r=1http://www.nytimes.com/2008/04/08/scienc e/08monty.html?_r=1

62. You pick #1 Door 1Door 2Door 3Results GAME 1AUTOGOAT Switch and you lose. GAME 2GOATAUTOGOAT Switch and you win. GAME 3GOAT AUTO Switch and you win. GAME 4AUTOGOAT Stay and you win. GAME 5GOATAUTOGOAT Stay and you lose. GAME 6GOAT AUTO Stay and you lose.

64. Practice The probability of winning “Blingoo” is.30 What is the probability that you will win 20 of the next 30 games of Blingoo ? Note: previous probability methods do not work for this question

65. Binomial Distribution Used with situations in which each of a number of independent trials results in one of two mutually exclusive outcomes

66. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

67. Game of Chance The probability of winning “Blingoo” is.30 What is the probability that you will win 20 of the next 30 games of Blingoo ? Note: previous probability methods do not work for this question

68. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

69. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

70. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

71. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

72. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

73. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events p =.000029

74. What does this mean? p =.000029 This is the probability that you would win EXACTLY 20 out of 30 games of Blingoo

75. Game of Chance Playing perfect black jack – the probability of winning a hand is.498 What is the probability that you will win 8 of the next 10 games of blackjack?

76. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

77. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

78. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events p =.0429

79. Excel

80. Binomial Distribution What is this doing? Its combining together what you have learned so far! One way to fit our 8 wins would be (joint probability): W, W, W, W, W, W, W, W, L, L = (.498)(.498)(.498)(.498)(.498)(.498)(.498)(.498)(.502)(.502)= (.498 8 )(.502 2 )=.00095 p X q (N-X)

81. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

82. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

83. Binomial Distribution Other ways to fit our question W, L, L, W, W, W, W, W L, W, W, W, W, L, W, W W, W, W, L, W, W, W, L L, L, W, W, W, W, W, W W, L, W, L, W, W, W, W W, W, L, W, W, W, L, W

84. Binomial Distribution Other ways to fit our question W, L, L, W, W, W, W, W =.00095 L, W, W, W, W, L, W, W =.00095 W, W, W, L, W, W, W, L =.00095 L, L, W, W, W, W, W, W =.00095 W, L, W, L, W, W, W, W =.00095 W, W, L, W, W, W, L, W =.00095 Each combination has the same probability – but how many combinations are there?

85. Combinations Ingredients: N = total number of events r = number of events selected 45 different combinations

86. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

87. Binomial Distribution Any combination would work. 00095+ 00095+ 00095+ 00095+ 00095+ 00095+ 00095+ 00095+...... 00095 Or 45 *. 00095 =.04

88. Binomial Distribution Ingredients: N = total number of events p = the probability of a success on any one trial q = (1 – p) = the probability of a failure on any one trial X = number of successful events

89. Practice You bought a ticket for a fire department lottery and your brother has bought two tickets. You just read that 1000 tickets were sold. –a) What is the probability you will win the grand prize? –b) What is the probability that your brother will win? –c) What is the probably that you or your bother will win?

90. 5.2 A) 1/1000 =.001 B)2/1000 =.002 C).001 +.002 =.003

91. Practice Assume the same situation at before except only a total of 10 tickets were sold and there are two prizes. –a) Given that you didn’t win the first prize, what is the probability you will win the second prize? –b) What is the probability that your borther will win the first prize and you will win the second prize? –c) What is the probability that you will win the first prize and your brother will win the second prize? –d) What is the probability that the two of you will win the first and second prizes?

92. 5.3 A) 1/9 =.111 B) 2/10 * 1/9 = (.20)*(.111) =.022 C) 1/10 * 2/9 = (.10)*(.22) =.022 D).022 +.022 =.044

93. Practice In some homes a mother’s behavior seems to be independent of her baby's, and vice versa. If the mother looks at her child a total of 2 hours each day, and the baby looks at the mother a total of 3 hours each day, and if they really do behave independently, what is the probability that they will look at each other at the same time?

94. 5.8 2/24 =.083 3/24 =.125.083*.125 =.01

95. Practice Abe ice-cream shot has six different flavors of ice cream, and you can order any combination of any number of them (but only one scoop of each flavor). How many different ice-cream cone combinations could they truthfully advertise (note, we don’t care about the order of the scoops and an empty cone doesn’t count).

96. 5.29 6 + 15 + 20 +15 + 6 + 1 = 63

97. Extra Brownie Points! Lottery To Win: choose the 5 winnings numbers –from 1 to 49 AND Choose the "Powerball" number –from 1 to 42 What is the probability you will win? –Use combinations to answer this question