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 = 6˛ = 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}

• How many different ways can someone roll at 10? ( 3 ways ) You can roll   (4,6) , (5,5), or   (6,4) .
• What number has the most amount of combinations available?  ( 7 ) There are 6 ways to roll a 7.
• What number(s) has the least amount of combinations available?  ( 2  and 12 )
There is only one way to roll 2.  (1,1)
There is only one way to roll 12. (6,6)
•

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 E1 = rolling a 10, then E1 = { (4,6),(5,5),(6,4) } .

Let's say E2 = rolling a 5 on both dice,  then E2 = { (5,5), } .

Let's say E3 = rolling a 12,  then E3 = { (6,6), }

Determine the following events.

 The event occurs when either E1 or E2 occurs.  That is when rolling  (4,6),(5,5),or (6,4) . The event occurs when both E1 and E2 occurs.  That is when rolling (5,5). The event never occurs and is the empty set.   Ř   When   the intersection is empty, the sets are mutually exclusive  (or disjoint). The event E1' 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.

• Experimental probability is derived from the actual data collected from an experiment.
• Probability  is the predicted outcome of the experiment.

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 E1 = rolling a 10, then E1 = { (4,6),(5,5),(6,4) } .

Let's say E2 = rolling a 5 on both dice,  then E2 = { (5,5), } .

Let's say E3 = rolling a 7   (6 ways to roll a 7)  E3 = { (6,1),(5,2),(4,3),(3,4),(2,5),(1,6) }.

Let's say E4 = rolling a 11   (2 ways to roll a 11)     E4 = { (5,6),(6,5) }.

 Pr(E1) = Pr(E2) = Pr(E3) = Pr(E4) = 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
1. What is the probability of Liz winning?
The sum of the probabilities add up to 1, (page 253)
Pr(Liz wins) = 1 -  Pr(Sam wins) - Pr(Sue wins) - Pr(Tom wins)
Pr(Liz wins) = 1 - 0.18 - 0.23 - 0.31 = 0.28
2. What is the probability that a boy wins?
Pr(boy wins) = Pr(Sam wins) +  Pr(Tom wins)  0.18 0.31 = 0.49
3. What is the probability that Tom loses?
That means someone else wins.
Pr(Tom loses) = 1 - Pr(Tom wins) = 1 -  0.31 = 0.69
4. What are the odds that Sue loses?
Pr(Sue loses) = 1 - Pr(Sue wins) = 10.23 = 0.77 ; a = 77, a + b = 100,  77 + b =100,  b = 23 The odds are 77 to 23
5. What are the odds that a girl wins?
Pr(Sam wins) = 0.18 ; a = 51, a + b = 100,  51 + b =100,  b = 49 The odds are 51 to 49
6. What are the odds that Sam  wins?
Pr(Sam wins) = ; a = 9, a + b = 50,  9 + b =50,   b = 41 The odds are 9 to 41

Inclusion-Exclusion 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 E1 = rolling a 10, then E1 = { (4,6),(5,5),(6,4) } .
Let's say E2 = rolling the same number on both dice,  then E2 = {(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(E1) = Pr(E2) = 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, inclusivewith replacement

• The sample space S = {1,2,3,4,5,6,7,8,9,10,11,12,13}
• With replacement means if the number 5 is chosen, it is put back into the sample space before another number is chosen.
• Inclusive means including the first and last numbers.
1. What is the probability that all the numbers are even?
The set of even numbers = {2,4,6,8,10,12}. There are 6 even numbers.
Since the probability of getting any even number is the same and we pick 5 numbers with replacement. 2. What is the probability that all the numbers are odd?
The set of odd numbers = {1,3,5,7,9,11,13}. There are 7 odd numbers.
Since the probability of getting any odd number is the same and we pick 5 numbers with replacement. 3. What is the probability that at least one of the numbers is odd?
It is easier to figure out the probability of no odd numbers, that is all even numbers.
Pr(no odd)  = 1 - Pr( all even) = 1 - 0.02 = 0.98   or Question #5 Five numbers are chosen at random from the whole numbers between 1 and 13, inclusivewithout replacement

• The sample space S = {1,2,3,4,5,6,7,8,9,10,11,12,13}
• Without replacement means if the number 5 is chosen, it can't be chosen again.
• Inclusive means including the first and last numbers.
1. What is the probability that all the numbers are even?
The set of even numbers = {2,4,6,8,10,12}. There are 6 even numbers.  But, once we pick the first even number, there are only 5 even numbers left to choose from. This is a combination. There 6 ways to choose 5 numbers out of 6.
There 1,287 ways to choose 5 numbers out of 13.
2. What is the probability that all the numbers are odd?
The set of odd numbers = {1,3,5,7,9,11,13}. There are 7 odd numbers. There 21 ways to choose 5 numbers out of 7.
There 1,287 ways to choose 5 numbers out of 13.
3. What is the probability that at least one of the numbers is odd?
It is easier to figure out the probability of no odd numbers, that is all even numbers.
Pr(no odd)  = 1 - Pr( all even) = 1 - 0.005 = 0.995 