2. Probabilities
What is the probability that among 23 people (this class) there will be a shared birthday?

4. # people required(i.e. n) when m=365
The number of people required so that the probability that some pair will have a birthday separated by k days or fewer will be higher than 50% is: k
# people required(i.e. n) when m=365 23 1 14 2 11 3 9 4 8 5 6 7

5. Patrick's Casino

7. What is the probability of picking an ace?

8. Probability =

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

10. Every card has the same probability of being picked

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

12. (.077) + (.077) + (.077) + (.077) = .308
16 / 52 = .308

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

14. (.077) * (.077) = .0059

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

16. 10 Card = (.077) + (.077) + (.077) + (.077) = .308
Ace after one card is removed = 4/51 = .078
(.308)*(.078) = .024
But you could also get an Ace (.077) and then a ten (.078*4 = .312) or .077*.312 = .024
So the prob of either of these occurring and getting blackjack is = .048

17. 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?

18. 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?

19. 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?

20. 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

21. Cards What is the probability of drawing an ace?
What is the probability your first 2 cards are aces?
What is the probability your first 4 cards are aces?
What is the probability that out of 4 cards, at least one will be an ace?

22. 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 =
What is the probability that an ace will be in the first four cards dealt?
= .3169

23. Probability
.00
1.00
Event must occur
Event will not occur

24. 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

25. Key Terms Independent event Mutually exclusive
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

26. 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)

27. 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)

28. Does not own a video game
Example
Owns a video game
Does not own a video game
Total
No Children 10 35 45
Children 25 30 55 65
100

29. 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 Children 10 35 45
Children 25 30 55 65
100

30. Does not own a video game
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 Children 10 35 45
Children 25 30 55 65
100

31. Does not own a video game
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 Children 10 35 45
Children 25 30 55 65
100

32. Does not own a video game
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 Children 10 35 45
Children 25 30 55 65
100

33. Does not own a video game
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 Children 10 35 45
Children 25 30 55 65
100

34. Does not own a video game
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 Children 10 35 45
Children 25 30 55 65
100

35. Does not own a video game
25 / 100 = * .55 = .19
Owns a video game
Does not own a video game
Total
No Children 10 35 45
Children 25 30 55 65
100

36. Does not own a video game
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 Children 10 35 45
Children 25 30 55 65
100

37. Practice Republican Democrat Total Male 52 27 79 Female 18 65 83 70 92
162

38. p (republican). p(female) p (republican, male)
p (republican) p(female) p (republican, male) p(female, republican) p (republican l male) p(male l republican)
Republican
Democrat
Total
Male 52 27 79
Female 18 65 83 70 92
162

39. p (republican) = 70 / 162 =. 43 p (republican, male) = 52 / 162 =
p (republican) = 70 / 162 = .43 p (republican, male) = 52 / 162 = .32 p (republican l male) = 52 / 79 = .66
Republican
Democrat
Total
Male 52 27 79
Female 18 65 83 70 92
162

40. p(female) = 83 / 162 =. 51 p(female, republican) = 18 / 162 =
p(female) = 83 / 162 = .51 p(female, republican) = 18 / 162 = .11 p(male l republican) = 52 / 70 = .74
Republican
Democrat
Total
Male 52 27 79
Female 18 65 83 70 92
162

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

42. 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

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?

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
A, B, C, D, E, F Note: 0! = 1

46. 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, C A, C, D A, D, E B, C, A

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

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

49. 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?

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

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

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

53. Practice There are three different prizes
1st $1,000
2nd $500
3rd $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?

54. Practice
336 ways of awarding the three different prizes

55. 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?

56. Combinations
56 different ways to award these prizes

57. 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?

58. 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!

59. 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?

60. Practice
e+230 Or

61. 26,904,727,073,180,495,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000

62. 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? 

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

67. 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

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

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. 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

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

74. 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

75. 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

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
p =

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

78. 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?

79. 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

80. 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

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
p = .0429

82. 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)=
(.4988)(.5022)=.00095
pX q(N-X)

83. 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

84. 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

85. 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

86. 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 =
Each combination has the same probability – but how many combinations are there?

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

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

90. 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