Finite Mathematics Lesson 8
Chapters 6.1  6.4
Probability theory is essential to all the casinos in our valley. How does a casino owner pick the payoffs for their games? It would be important to know the likelihood of getting a flush in poker, a blackjack in 21, or a 7 in craps. Let me assure you. Casinos wouldn't stay in business very long if the odds of each game isn't thoroughly researched. This is a fun chapter.
Course Notes
Section 6.1 Introduction
Read the section. No assignments for this section.
Section 6.2 Experiments , Outcomes and Events
Read the section. Example 3 is a good problem to examine. I will going back to it throughout the chapter.
An event E is a subset of the sample space. We say that the event occurs when the outcome of the experiment is an element of set E. 
An experiment (craps) consists of throwing two dice, one red and one green, and observing the uppermost face. Notice, there are 6 sides
of each die. The possible different rolls would 6
× 6 = 6² = 36
The reason for using 2 different color dice is to emphasize
that (1,2)
and (2,1)
are 2 different throws.
(1,1)  (1,2)  (1,3)  (1,4)  (1,5)  (1,6)  
(2,1)  (2,2)  (2,3)  (2,4)  (2,5)  (2,6)  
(3,1)  (3,2)  (3,3)  (3,4)  (3,5)  (3,6)  
(4,1)  (4,2)  (4,3)  (4,4)  (4,5)  (4,6)  
(5,1)  (5,2)  (5,3)  (5,4)  (5,5)  (5,6)  
(6,1)  (6,2)  (6,3)  (6,4)  (6,5)  (6,6) 
Looking at the sample space above...Let a new sample space be the sum of the 2 dice. S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11,12}
The event E is a subset of a sample space. We say an event occurs when the outcome of the experiment is an element of E.
Let's say E_{1} = rolling a 10, then E_{1} = { (4,6),(5,5),(6,4) } .
Let's say E_{2} = rolling a 5 on both dice, then E_{2} = { (5,5), } .
Let's say E_{3} = rolling a 12, then E_{3} = { (6,6), }
Determine the following events.

occurs when either E_{1} or E_{2} occurs. That is when rolling (4,6),(5,5),or (6,4) .  

occurs when both E_{1} and E_{2} occurs. That is when rolling (5,5).  

never occurs and is the empty set. Ø When the intersection is empty, the sets are mutually exclusive (or disjoint).  

E_{1}' 
occurs when any number other than 10 is thrown. 
Section 6.3 Assignments of Probabilities
Of course , read the section first. Note the difference between experimental probability and probability.
These are not necessarily the same. We predict that tossing a fair coin has 50% likelihood of landing on heads. But I could throw a coin 10 times in a row and have it land on tails each time.
Let's look at the craps example again.
(1,1)  (1,2)  (1,3)  (1,4)  (1,5)  (1,6)  
(2,1)  (2,2)  (2,3)  (2,4)  (2,5)  (2,6)  
(3,1)  (3,2)  (3,3)  (3,4)  (3,5)  (3,6)  
(4,1)  (4,2)  (4,3)  (4,4)  (4,5)  (4,6)  
(5,1)  (5,2)  (5,3)  (5,4)  (5,5)  (5,6)  
(6,1)  (6,2)  (6,3)  (6,4)  (6,5)  (6,6) 
We denote the probability of an event E by Pr(E). Each of these pairs is equally likely to occur. (fair dice) So the probability of throw any one these pairs is 1/36.
Let's say E_{1} = rolling a 10, then E_{1} = { (4,6),(5,5),(6,4) } .
Let's say E_{2} = rolling a 5 on both dice, then E_{2} = { (5,5), } .
Let's say E_{3 }= rolling a 7 (6 ways to roll a 7) E_{3} = { (6,1),(5,2),(4,3),(3,4),(2,5),(1,6) }.
Let's say E_{4 }= rolling a 11 (2 ways to roll a 11) E_{4} = { (5,6),(6,5) }.
Pr(E_{1}) = 

Pr(E_{2}) = 

Pr(E_{3}) = 

Pr(E_{4}) = 

The Addition Principle (page 256) allows us to add the probabilities given an Event. The sum of the probabilities always add up to 1 (100%). In the example above, we have 36 different outcomes each with a probability of 1/36. Therefore, 36×(1/36) = 1 
Odds
We here bookmakers talk frequently about odds. The Tyson fight pays 1 to 8, the horse pays 5 to 2, or the Redskins are a 200 to 1 underdog to win next years superbowl. Use this formula to convert odds to probability.
If the odds in favor of an event are a to b, the probability of the event is  
If the odds of winning a race is 3 to 2 the probability is  
If the odds of winning a race is 3 to 7 the probability is 
Example problem
#16 page 261
Four people are running for class president, Liz, Sam, Sue , and Tom.
The probabilities of Sam, Sue , and Tom winning are .18, .23, and .31, respectively.
Organize data...
Pr(Sam wins) = 0.18  Pr(Sue wins) = 0.23  Pr(Tom wins) = 0.31 
; a = 77, a + b = 100, 77 + b =100, b = 23  
The odds are 77 to 23 
; a = 51, a + b = 100, 51 + b =100, b = 49  
The odds are 51 to 49 
; a = 9, a + b = 50, 9 + b =50, b = 41  
The odds are 9 to 41 
InclusionExclusion Principle Revisited
We have to make sure that we don't add the same probability more than once if an event belongs to more than one set. We will go back to the craps example to illustrate this point.
Let's say E_{1} = rolling a 10, then E_{1} = { (4,6),(5,5),(6,4) } .
Let's say E_{2} = rolling the same number on both dice, then E_{2}
= {(1,1),(2,2), (3,3),(4,4)(5,5),(6,6) } .
Find the probability of rolling either a 10 or the same number on both dice.
Pr(E_{1}) = 

Pr(E_{2}) = 
6 possible combinations out of 36 samples  
only 1 possible throw (5,5)  
subtract the throw in common to both events  
Section 6.4 Calculating Probabilities of Events
Experiments with Equally Likely Outcomes
If a sample space has N equally likely outcomes, the the
probability of each outcome is 1÷N.
Given an Event E in that sample space consisting of M outcomes:
Complement Rule
Let E be any event, E
' its complement. Complement means not in E. Then,
Pr(E) = 1  Pr(E ' )
If Pr(E ') = 0.6 , then Pr(E) = 1  0.6 = 0.4
Example Compare problem #4 and #5 on page 268
Question #4 Five numbers are chosen at random from the whole numbers between 1 and 13, inclusive, with replacement
Question #5 Five numbers are chosen at random from the whole numbers between 1 and 13, inclusive, without replacement
When you are ready, click here for the assignment for this section.
Please notify me of any errors on this page joe@joemath.com