1. Statistics for Business and Economics
Chapter 3
Probability

2. Contents Events, Sample Spaces, and Probability
Unions and Intersections
Complementary Events
The Additive Rule and Mutually Exclusive Events
Conditional Probability
The Multiplicative Rule and Independent Events
Bayes’s Rule
As a result of this class, you will be able to ...

3. Learning Objectives Develop probability as a measure of uncertainty
Introduce basic rules for finding probabilities
Use probability as a measure of reliability for an inference
Provide an advanced rule for finding probabilities
As a result of this class, you will be able to ...

4. DEFINITION OF PROBABILITY
PROBABILITY IS THE STUDY OF RANDOM OR NONDETERMINISTIC EXPERIMENTS. IT MEASURES THE NATURE OF UNCERTAINTY.

5. SOME KEY PROBABILISTIC TERMINOLOGIES
RANDOM EXPERIMENT
AN EXPERIMENT IN WHICH ALL OUTCOMES (RESULTS) ARE KNOWN BUT SPECIFIC OBSERVATIONS CANNOT BE KNOWN IN ADVANCE.
EXAMPLES:
TOSS A COIN
ROLL A DIE

6. RANDOM VARIABLE ILLUSTRATION
THE OUTCOME OF AN EXPERIMENT IS CALLED A RANDOM VARIABLE.
IT CAN ALSO BE DEFINED AS A QUANTITY THAT CAN TAKE ON DIFFERENT VALUES.
ILLUSTRATION

7. Probability As A Long – run Relative Frequency
What’s the probability of getting a head on the toss of a single fair coin? Use a scale from 0 (no way) to 1 (sure thing).
So toss a coin twice. Do it! Did you get one head & one tail? What’s it all mean?

8. Many Repetitions!* Total Heads Number of Tosses Number of Tosses 1.00
0.75
0.50
0.25
0.00 25 50 75
100
125
Number of Tosses

9. Events, Sample Spaces, and Probability
3.1
Events, Sample Spaces, and Probability
:1, 1, 3

10. Experiments & Sample Spaces
Process of observation that leads to a single outcome that cannot be predicted with certainty
Sample point
Most basic outcome of an experiment
Sample space (S)
Collection of all sample points
Sample Space Depends on Experimenter!

11. Visualizing Sample Space
1. Listing
S = {Head, Tail}
2. Venn Diagram H T S

12. Sample Space Examples Experiment Sample Space
Toss a Coin, Note Face {Head, Tail}
Toss 2 Coins, Note Faces {HH, HT, TH, TT}
Select 1 Card, Note Kind {2♥, 2♠, ..., A♦} (52)
Select 1 Card, Note Color {Red, Black}
Play a Football Game {Win, Lose, Tie}
Inspect a Part, Note Quality {Defective, Good}
Observe Gender {Male, Female}

13. Events Specific collection of sample points
Subset of a sample space (S)
Simple Event
Contains only one sample point
Compound Event
Contains two or more sample points

14. S Venn Diagram Experiment: Toss 2 Coins. Note Faces.
Sample Space S = {HH, HT, TH, TT}
Compound Event: At least one Tail
Other compound events could be formed:
Tail on the second toss {HT, TT}
At least 1 Head {HH, HT, TH} TH HT
Outcome HH TT S

15. Event Examples Experiment: Toss 2 Coins. Note Faces.
Sample Space: HH, HT, TH, TT
Event Outcomes in Event
Typically, the last event (Heads on Both) is called a simple event.
1 Head & 1 Tail HT, TH
Head on 1st Coin HH, HT
At Least 1 Head HH, HT, TH
Heads on Both HH

16. Probabilities

17. What is Probability?
1. Numerical measure of the likelihood that event will occur
P(Event)
P(A)
Prob(A)
2. Lies between 0 & 1
3. Sum of sample points is 1 1
Certain .5
Impossible

18. Probability Rules for Sample Points
Let pi = P(A) represent the probability of sample point i, that is, event A
1. All sample point or event probabilities must lie between 0 and 1 (i.e., 0 ≤ pi ≤ 1 or 0 ≤ P(A) ≤ 1 ).
2. The probabilities of all sample points within a sample space must sum to 1 (i.e.,  pi = 1).

19. Example
A COIN IS WEIGHTED SO THAT HEADS IS TWICE AS LIKELY TO APPEAR AS TAILS. FIND P(T) AND P(H).
2. THREE STUDENTS A, B AND C ARE IN A SWIMMING RACE. A AND B HAVE THE SAME PROBABILITY OF WINNING AND EACH IS TWICE AS LIKELY TO WIN AS C. FIND THE PROBABILITY THAT B OR C WINS.
Solution

20. Equally Likely Probability
P(Event) = X / T
X = Number of outcomes in the event
T = Total number of sample points in Sample Space
Each of T sample points is equally likely
— P(sample point) = 1/T
© T/Maker Co.

21. Probability of An Event
The probability of an event A, denoted by P(A), is obtained by adding the probabilities of the individual outcomes in the event.
When all the possible outcomes are equally likely,

22. Example
A PAIR OF FAIR DICE IS TOSSED. FIND THE PROBABILITY THAT THE MAXIMUM OF THE TWO NUMBERS IS GREATER THAN 4.
ONE CARD IS SELECTED AT RANDOM FROM 50 CARDS NUMBERED 1 TO 50. FIND THE PROBABILITY THAT THE NUMBER ON THE CARD IS (I) DIVISIBLE BY 5, (II) PRIME, (III) ENDS IN THE DIGIT 2.
Solution

23. Steps for Calculating Probability
1. Define the experiment; describe the process used to make an observation and the type of observation that will be recorded
2. List the sample points
3. Assign probabilities to the sample points
4. Determine the collection of sample points contained in the event of interest
5. Sum the sample points probabilities to get the event probability

24. Combinations Rule
A sample of n elements is to be drawn from a set of N elements. The number of different samples possible
is denoted by
and is equal to
where the factorial symbol (!) means that
For example,
0! is defined to be 1.

25. Example
Compute

26. Unions and Intersections
3.2
Unions and Intersections
:1, 1, 3

27. Unions & Intersections
Outcomes in either events A or B or both
‘OR’ statement
Denoted by  symbol (i.e., A  B)
2. Intersection
Outcomes in both events A and B
‘AND’ statement
Denoted by  symbol (i.e., A  B)

28. ILLUSTRATING THE INTERSECTION AND UNION OF EVENTS THROUGH VENN DIAGRAMS

29. Example Solution Toss a fair die once. Define the following events:
E: event outcomes;
O: odd primes;
F: event primes;
M: multiple of 3;
Find
E or O;
O or F;
O and M;
F and E;
Solution

30. Event Union: Venn Diagram
Experiment: Draw 1 Card. Note Kind, Color & Suit.
Ace
Black
Event Black: 2, 2,..., A
Sample Space: 2, 2, 2, ..., A S
Event Ace: A, A, A, A
Event Ace  Black: A, ..., A, 2, ..., K

31. Event Union: Two–Way Table
Experiment: Draw 1 Card. Note Kind, Color & Suit.
Color
Simple Event Ace:
A, A, A, A
Sample Space (S): 2, 2, 2, ..., A
Type
Red
Black
Total
Ace
Ace &
Ace &
Ace
Red
Black
Non-Ace
Non &
Non &
Non-
Red
Black
Ace
Total
Red
Black S
Event Ace  Black: A,..., A, 2, ..., K
Simple Event Black: 2, ..., A

32. Event Intersection: Venn Diagram
Experiment: Draw 1 Card. Note Kind, Color & Suit.
Ace
Black
Event Black: 2,...,A
Sample Space: 2, 2, 2, ..., A S
Event Ace: A, A, A, A
Event Ace  Black: A, A

33. Event Intersection: Two–Way Table
Experiment: Draw 1 Card. Note Kind, Color & Suit.
Color
Simple Event Ace:
A, A, A, A
Sample Space (S): 2, 2, 2, ..., A
Type
Red
Black
Total
Ace
Ace &
Ace &
Ace
Red
Black
Non-Ace
Non &
Non &
Non-
Red
Black
Ace
Event Ace  Black: A, A
Total
Red
Black S
Simple Event Black: 2, ..., A

34. Compound Event Probability
1. Numerical measure of likelihood that compound event will occur
2. Can often use two–way table
Two variables only

35. Event Probability Using Two–Way Table
Total 1 2 A
P(A  B )
P(A  B )
P(A ) 1 1 1 1 2 1 A
P(A  B )
P(A  B )
P(A ) 2 2 1 2 2 2
Total
P(B )
P(B ) 1 1 2
Joint Probability
Marginal (Simple) Probability

36. Two–Way Table: Example
Experiment: Draw 1 Card. Note Kind & Color.
Color
Type
Red
Black
Total
Ace
2/52
2/52
4/52
Non-Ace
24/52
24/52
48/52
P(Ace)
Total
26/52
26/52
52/52
P(Red)
P(Ace  Red)

37. Thinking Challenge What’s the Probability? P(A) = P(D) = P(C  B) =
P(A  D) =
P(B  D) =
What’s the Probability?
Event C D
Total A 4 2 6 B 1 3 5 10
Let students solve first. Allow about 20 minutes for this.

38. Solution* The Probabilities Are: P(A) = 6/10 P(D) = 5/10
P(C  B) = 1/10
P(A  D) = 9/10
P(B  D) = 3/10
Event C D
Total A 4 2 6 B 1 3 5 10

39. 3.3
Complementary Events
:1, 1, 3

40. S Complementary Events AC A Complement of Event A
The event that A does not occur
All events not in A
Denote complement of A by AC S AC A

41. S Rule of Complements AC A
The sum of the probabilities of complementary events equals 1:
P(A) + P(AC) = 1 S AC A

42. Complement of Event: Example
Experiment: Draw 1 Card. Note Color.
Black
Sample Space: 2, 2, 2, ..., A S
Event Black: 2, 2, ..., A
Complement of Event Black, BlackC: 2, 2, ..., A, A

43. Complement of An Event Example
What is the probability of getting at least 1 head in the experiment of tossing a fair coin 3 times?
The Sample Space Is
HHH HHT HTH HTT THH THT TTH TTT
Solution

44. The Additive Rule and Mutually Exclusive Events
3.4
The Additive Rule and Mutually Exclusive Events
:1, 1, 3

45. Mutually Exclusive Events
Events do not occur simultaneously
A  B does not contain any sample points 

46. Mutually Exclusive Events Example
Experiment: Draw 1 Card. Note Kind & Suit. 
Outcomes in Event Heart: 2, 3, 4 , ..., A
Sample Space: 2, 2, 2, ..., A
Mutually Exclusive
What is the intersection of mutually exclusive events?
The null set.  S
Event Spade: 2, 3, 4, ..., A
Events  and are Mutually Exclusive

47. Addition And General Addition Rules
Used to get compound probabilities for union of events
P(A OR B) = P(A  B) = P(A) + P(B) – P(A  B)
For mutually exclusive events: P(A OR B) = P(A  B) = P(A) + P(B)

48. Venn Diagram Illustration of The General Addition Rule

49. General Addition Rule: Example
A class contains 10 men and 20 women of which half the men and half the women have brown eyes. Find the probability that a person chosen at random is a man or has brown eyes.
Solution

50. General Addition Rule Example
Experiment: Draw 1 Card. Note Kind & Color.
Color
Type
Red
Black
Total
Ace 2 4
Non-Ace 24 48 26 52
Try other examples using this table.
P(Ace  Black) =
P(Ace) +
P(Black) –
P(Ace 
Black)
= – =

51. Addition Rule: Example
A class contains 5 freshmen, 4 sophomores, 8 juniors and 3 seniors. A student is chosen at random to represent the class. Find the probability that the student chosen is a sophomore or a junior.
Solution

52. Thinking Challenge Using the additive rule, what is the probability?
P(A  D) =
P(B  C) =
Using the additive rule, what is the probability?
Event C D
Total A 4 2 6 B 1 3 5 10
Let students solve first. Allow about 10 minutes for this.

53. Solution* Using the additive rule, the probabilities are: 1. 6 5 2 9
P(A  D) = P(A) + P(D) – P(A  D)
= – = 2.
P(B  C) = P(B) + P(C) – P(B  C)
= – =

54. Conditional Probability
3.5
Conditional Probability
:1, 1, 3

55. Conditional Probability
1. Event probability given that another event occurred
2. Revise original sample space to account for new information
Eliminates certain outcomes
3. P(A | B) = P(A and B) = P(A  B) P(B) P(B)

56. Venn Diagram Illustration

57. Example
In a certain college, 25% of the students failed mathematics, 15% of the students failed chemistry, and 10% of the students failed both mathematics and chemistry. A student is selected at random.
(a) If he failed chemistry, what is the probability that he failed mathematics?
(b) If he failed mathematics, what is the probability that he failed chemistry?
Solution

58. Example
You draw a card at random from a standard deck of 52 cards. Find each of the following probabilities:
The card is a heart, given that it is red.
The card is red, given that it is a heart.
The card is an ace, given that it is red.
The card is a queen, given that it is a face card.
Solution

59. Conditional Probability Using Two–Way Table
Experiment: Draw 1 Card. Note Kind & Color.
Color
Type
Red
Black
Total
Ace 2 4
Non-Ace 24 48 26 52
Revised Sample Space
Try other examples using this table.

60. Thinking Challenge
Using the table then the formula, what’s the probability?
P(A|D) =
P(C|B) =
Event C D
Total A 4 2 6 B 1 3 5 10
Let students solve first. Allow about 20 minutes for this.

61. Solution*
Using the formula, the probabilities are:

62. The Multiplicative Rule and Independent Events
3.6
The Multiplicative Rule and Independent Events
:1, 1, 3

63. Multiplicative Rule
1. Used to get compound probabilities for intersection of events 2. P(A and B) = P(A  B) = P(A)  P(B|A) = P(B)  P(A|B) 3. For Independent Events: P(A and B) = P(A  B) = P(A)  P(B)

64. Example (Multiplicative rule)
(1) A man is dealt 5 cards one after the other from an ordinary deck of 52 cards. What is the probability that they are all spades?
(2) The students in a class are selected at random, one after the other, for an examination. Find the probability that the boys and girls in the class alternate if
(i) the class consists of 4 boys and 3 girls.
(ii) the class consists of 3 boys and 3 girls.
Solution

65. Multiplicative Rule: Example
Experiment: Draw 1 Card. Note Kind & Color.
Color
Type
Red
Black
Total
Ace 2 2 4
Try other examples using this table.
Non-Ace 24 24 48
Total 26 26 52
P(Ace  Black) = P(Ace)∙P(Black | Ace)

66. Example(Independence)
The baseball World Series ends when a team wins 4 games. Suppose that sports analysts consider one team a bit stronger, with a 55% chance to win any individual game. Find the probability that the underdog will win the series by winning the first 4 games.
Solution

67. Example(Independence)
A slot machine has three wheels that spin independently. Each has 10 equally likely symbols: 4 bars, 3 lemons, 2 cherries, and a bell. If you play, what is the probability that
1. you get three lemons?
2. you get no fruit symbols?
3. you get no bells?
4. you get at least one bar?
Solution

68. Statistical Independence
1. Event occurrence does not affect probability of another event
Toss 1 coin twice
2. Causality not implied
3. Tests for independence
P(A | B) = P(A)
P(B | A) = P(B)
P(A  B) = P(A)  P(B)

69. Thinking Challenge
Using the multiplicative rule, what’s the probability?
Event C D
Total A 4 2 6 B 1 3 5 10
P(C  B) =
P(B  D) =
P(A  B) =
Let students solve first. Allow about 10 minutes for this.

70. Solution*
Using the multiplicative rule, the probabilities are:

71. Tree Diagram
Experiment: Select 2 pens from 20 pens: 14 blue & 6 red. Don’t replace.
Dependent! R
P(R  R)=(6/20)(5/19) =3/38
5/19 R
6/20
14/19 B
P(R  B)=(6/20)(14/19) =21/95 R
6/19
P(B  R)=(14/20)(6/19) =21/95
14/20 B
13/19 B
P(B  B)=(14/20)(13/19) =91/190

72. 3.7
Bayes’s Rule
:1, 1, 3

73. Bayes’s Rule
Given k mutually exclusive and exhaustive events B1, B1, Bk , such that P(B1) + P(B2) + … + P(Bk) = 1, and an observed event A, then

74. Bayes Rule Via Tree Diagram (Example)
In a certain college, 4% of the men and 1% of the women are taller than 6 feet. Furthermore, 60% of the students are women. Now if a student is selected at random and is taller than 6 feet, what is the probability that the student is a woman?
Solution

75. Bayes Rule Via Tree Diagram (Example)
Suppose a test for TB has the following properties: If the person tested has TB, 99% of time the test will return a positive result; if a person tested does not have TB, 95% of the time the test will return a negative result. Among the test group, 2% actually have TB. What proportion of those tested positive actually have TB?
Solution

76. Bayes’ Rule Example
A company manufactures MP3 players at two factories. Factory I produces 60% of the MP3 players and Factory II produces 40%. Two percent of the MP3 players produced at Factory I are defective, while 1% of Factory II’s are defective. An MP3 player is selected at random and found to be defective. What is the probability it came from Factory I?

77. Bayes’ Rule: Example Defective 0.02 Factory I 0 .6 0.98 Good Defective
0.01
0 .4
Factory II
0.99
Good

78. Key Ideas
Probability Rules for k Sample Points, S1, S2, S3, , Sk 1. 0 ≤ P(Si) ≤ 1 2.

79. Key Ideas
Random Sample All possible such samples have equal probability of being selected.

80. Key Ideas
Combinations Rule Counting number of samples of n elements selected from N elements

81. Key Ideas
Bayes’s Rule