Sample Spaces
• Collection of all possible outcomes
– e.g.: All six faces of a dice:
– e.g.: All 52 cards in a deck:
1
Events
• Simple event
– Outcome from a sample space with one
characteristic
– e.g.: A red card from a deck of cards
• Joint event
– Involves two outcomes simultaneously
– e.g.: An ace that is also red from a deck of
cards
2
Visualizing Events
• Contingency tables
Ace
• Tree diagrams
Not Ace
Total
Black
Red
2
2
24
24
26
26
Total
4
48
52
Ace
Full
Deck
of Cards
Red
Cards
Black
Cards
Not an Ace
Ace
Not an Ace
3
Simple and Joint Events
The Event of a Triangle
The event of a triangle AND blue in color
There are 5 triangles in this collection of 18 objects
Two triangles that are blue
4
Special Events
Null Event
• Impossible event
e.g.: Club & diamond on one card
draw
• Complement of event
– For event A, all events not in A
– Denoted as A’
– e.g.: A: queen of diamonds
A’: all cards in a deck that are
not queen of diamonds

5
Special Events
• Mutually exclusive events
– Two events cannot occur together
– e.g. -- A: queen of diamonds; B: queen of clubs
• Events A and B are mutually exclusive
• Collectively exhaustive events
– One of the events must occur
– The set of events covers the whole sample space
– e.g. -- A: all the aces; B: all the black cards; C: all
the diamonds; D: all the hearts
• Events A, B, C and D are collectively exhaustive
• Events B, C and D are also collectively
exhaustive
6
Probability
• Sample of 1,000 households in
terms of purchase behaviour for
big-screen TV sets.
Planned to purchase
Yes
No
Total
Yes
200
100
300
Actually purchased
No
50
650
700
Total
250
750
1000
7
Decision Tree
8
• Sample of 300 households whether
the TV set purchased was an
HDTV and whether they also
purchased a DVD player.
Purchased HDTV
HDTV
Not HDTV
Total
Yes
38
70
108
Purchased DVD
No
42
150
192
Total
80
220
300
Draw the decision tree for purchased a DVD
player and an HDTV.
9
Joint Probability Using
Contingency Table
Event
B1
Event
B2
Total
A1
P(A1 and B1) P(A1 and B2) P(A1)
A2
P(A2 and B1) P(A2 and B2) P(A2)
Total
Joint Probability
P(B1)
P(B2)
1
Marginal (Simple) Probability
10
Computing Compound
Probability
• Probability of a compound event, A or B:
P( A or B)  P( A  B)
number of outcomes from either A or B or both

total number of outcomes in sample space
11
Compound Probability
(Addition Rule)
P(A1 or B1 ) = P(A1) + P(B1) - P(A1 and B1)
Event
Event
B1
B2
Total
A1
P(A1 and B1) P(A1 and B2) P(A1)
A2
P(A2 and B1) P(A2 and B2) P(A2)
Total
P(B1)
P(B2)
1
For Mutually Exclusive Events: P(A or B) = P(A) + P(B)
12
Computing Conditional
Probability
• The probability of event A given that event B has
occurred:
P( A and B)
P( A | B) 
P( B)
13
Conditional Probability and
Statistical Independence
• Conditional probability:
P( A and B)
P( A | B) 
P( B)
• Multiplication rule:
P( A and B)  P( A | B) P( B)
 P( B | A) P( A)
14
Conditional Probability and
Statistical Independence
• Events A and B are independent if
P( A | B)  P ( A)
or P ( B | A)  P ( B )
or P ( A and B )  P ( A) P ( B )
• Events A and B are independent when the probability
of one event, A, is not affected by another event, B
15
Bayes’s Theorem
P  Bi | A  

Same
Event
P  A | Bi  P  Bi 
P  A | B1  P  B1       P  A | Bk  P  Bk 
P  Bi and A 
P  A
Adding up
the parts
of A in all
the B’s
16
Bayes’s Theorem
Using Contingency Table
Fifty percent of borrowers repaid their loans. Out of those
who repaid, 40% had a college degree. Ten percent of
those who defaulted had a college degree. What is the
probability that a randomly selected borrower who has a
college degree will repay the loan?
P  R   .50
P  C | R   .4
P  C | R   .10
PR | C  ?
17
Bayes’s Theorem
Using Contingency Table
Repay
Repay
Total
College
.2
.05
.25
College
.3
.45
.75
Total
.5
.5
1.0
PR | C 
P C | R  P  R 
P C | R  P  R   P C | R  P  R 
.4 .5 

.2


 .8
.4 .5  .1.5 .25
18