Thinking Mathematically

1
Thinking Mathematically

• Chapter 11
• Counting Methods and Probability Theory

2
Thinking Mathematically

• Section 1
• The Fundamental Counting Principle

3
The Fundamental Counting Principle

• If you can choose one item from a group of M
  items and a second item from a group of N items,
  then the total number of two-item choices is M ?
  N.
• You the numbers!

MULTIPLY
4
The Fundamental Counting Principle
At breakfast, you can have eggs, pancakes or
cereal. You get a free juice with your meal
either OJ or apple juice. How many different
breakfasts are possible?

5
Example Applying the Fundamental Counting
Principle

• The Greasy Spoon Restaurant offers 6 appetizers
  and 14 main courses. How many different meals can
  be created by selecting one appetizer and one
  main course?
• Using the fundamental counting principle, there
  are 14 ? 6 84 different ways a person can order
  a two-course meal.

6
Example Applying the Fundamental Counting
Principle

• This is the semester that you decide to take your
  required psychology and social science courses.
• Because you decide to register early, there are
  15 sections of psychology from which you can
  choose. Furthermore, there are 9 sections of
  social science that are available at times that
  do not conflict with those for psychology. In
  how many ways can you create two-course schedules
  that satisfy the psychology-social science
  requirement?

7
Solution

• The number of ways that you can satisfy the
  requirement is found by multiplying the number of
  choices for each course.
• You can choose your psychology course from 15
  sections and your social science course from 9
  sections. For both courses you have
• 15 ? 9, or 135 choices.

8
The Fundamental Counting Principle

• The number of ways a series of successive things
  can occur is found by multiplying the number of
  ways in which each thing can occur.

9
Example Options in Planning a Course Schedule

• Next semester you are planning to take three
  courses - math, English, and humanities. Based
  on time blocks and highly recommended professors,
  there are 8 sections of math, 5 of English, and 4
  of humanities that you find suitable. Assuming
  no scheduling conflicts, there are
• 8 ? 5 ? 4 160 different three course schedules.

10
Example

• Car manufacturers are now experimenting with
  lightweight three-wheeled cars, designed for a
  driver and one passenger, and considered ideal
  for city driving. Suppose you could order such a
  car with a choice of 9 possible colors, with or
  without air-conditioning, with or without a
  removable roof, and with or without an onboard
  computer. In how many ways can this car be
  ordered in terms of options?

11
Solution

• This situation involves making choices with four
  groups of items.
• color - air-conditioning - removable roof -
  computer
• 9 ? 2 ? 2 ? 2 72
• Thus the car can be ordered in 72 different ways.

12
Example A Multiple Choice Test

• You are taking a multiple-choice test that has
  ten questions. Each of the questions has four
  choices, with one correct choice per question.
  If you select one of these options per question
  and leave nothing blank, in how many ways can you
  answer the questions?

13
Solution

• We DONT blindly multiply the first two numbers
  we see. The answer is not 10 ? 4 40.
• We use the Fundamental Counting Principle to
  determine the number of ways you can answer the
  test. Multiply the number of choices, 4, for
  each of the ten questions
• 4 ? 4 ? 4 ? 4 ? 4 ? 4 ? 4 ? 4 ? 4 ? 4 1,048,576

14
Example Telephone Numbers in the United States

• Telephone numbers in the United States begin
  with three-digit area codes followed by
  seven-digit local telephone numbers. Area codes
  and local telephone numbers cannot begin with 0
  or 1. How many different telephone numbers are
  possible?

15
Solution

• We use the Fundamental Counting Principle to
  determine the number of different telephone
  numbers that are possible.
• 8 ? 10 ? 10 ? 8 ? 10 ? 10 ? 10 ? 10 ? 10 ? 10
  6,400,000,000

16
Thinking Mathematically

• Section 2
• Permutations

17
Permutations

• A permutation is an arrangement of objects.
• No item is used more than once.
• The order of arrangement makes a difference.

18
Example Counting Permutations

• Based on their long-standing contribution to
  rock music, you decide that the Rolling Stones
  should be the last group to perform at the
  four-group Offspring, Pink Floyd, Sublime,
  Rolling Stones concert. Given this decision, in
  how many ways can you put together the concert?

19
Solution

• We use the Fundamental Counting Principle to
  find the number of ways you can put together the
  concert. Multiply the choices
• 3 ? 2 ? 1 ? 1 6
• Thus, there are six different ways to arrange the
  concert if the Rolling Stones are the final group
  to perform.

3 choices
2 choices
1 choice
1 choice
whichever of the two remaining
stones
offspring pink floyd sublime
only one remaining
20
Example Counting Permutations

• You need to arrange seven of your favorite books
  along a small shelf. How many different ways can
  you arrange the books, assuming that the order of
  the books makes a difference to you?

21
Solution

• You may choose any of the seven books for the
  first position on the shelf. This leaves six
  choices for second position. After the first two
  positions are filled, there are five books to
  choose from for the third position, four choices
  left for the fourth position, three choices left
  for the fifth position, then two choices for the
  sixth position, and only one choice left for the
  last position.
• 7 ? 6 ? 5 ? 4 ? 3 ? 2 ? 1 5040
• There are 5040 different possible permutations.

22
Factorial Notation

• If n is a positive integer, the notation n! is
  the product of all positive integers from n down
  through 1.
• n! n(n-1)(n-2)(3)(2)(1)
• note that 0!, by definition, is 1.
• 0!1

23
Permutations of n Things Taken r at a Time

• The number of permutations possible if r items
  are taken from n items

n!
nPr n(n 1) (n 2) (n
3) . . . (n r 1)
(n r)!
n! n(n 1) (n 2) (n 3) . . . (n r 1)
(n - r) (n - r - 1) . . . (2)(1)
(n r)!
(n - r) (n - r - 1) . . . (2)(1)
24
Permutations of n Things Taken r at a Time

• The number of permutations possible if r items
  are taken from n items
• nPr starting at n, write down r numbers going
  down by one

nPr n(n 1) (n 2) (n 3) . . . (n r 1)
1
2
3
4
r
25
Problem

• A math club has eight members, and it must
  choose 5 officers --- president, vice-president,
  secretary, treasurer and student government
  representative. Assuming that each office is to
  be held by one person and no person can hold more
  than one office, in how many ways can those five
  positions be filled?

We are arranging 5 out of 8 people into the five
distinct offices. Any of the eight can be
president. Once selected, any of the remaining
seven can be vice-president.
Clearly this is an arrangement, or permutation,
problem.
8P5 8!/(8-5)! 8!/3! 8 7 6 5 4 6720
26
Permutations with duplicates.

• In how many ways can you arrange the letters of
  the word minty?
• That's 5 letters that have to be arranged, so the
  answer is 5P5 5! 120
• But how many ways can you arrange the letters of
  the word messes?
• You would think 6!, but you'd be wrong!

27
messes
here are six permutations of messes
m
e
s
s
e
s
1
m
e
s
s
e
s
well, all 3! arrangements of the s's look the
same to me!!!! This is true for any arrangement
of the six letters in messes, so every six
permutations should count only once. The same
applies for the 2! arrangement of the e's
2
m
e
s
s
e
s
m
e
s
s
e
s
3
m
e
s
s
e
s
4
m
e
s
s
e
s
5
6
28
Permutations with duplicates.

• How many ways can you arrange the letters of the
  word messes?
• The problem is that there are three s's and 2
  e's. It doesn't matter in which order the s's
  are placed, because they all look the same!
• This is called permutations with duplicates.

29
Permutations with duplicates.

• Since there are 3! 6 ways to arrange the s's,
  there are 6 permutations that should count as
  one. Same with the e's. There are 2! 2
  permutations of them that should count as 1.
• So we divide 6! by 3! and also by 2!
• There are 6!/3!2! 720/12 60 ways to arrange
  the word messes.

30
Permutations with duplicates.

• In general if we want to arrange n items, of
  which m1, m2, .... are identical, the number of
  permutations is

31
Problem

• A signal can be formed by running different
  colored flags up a pole, one above the other.
  Find the number of different signals consisting
  of 6 flags that can be made if 3 of the flags
  are white, 2 are red, and 1 is blue

6!/3!2!1! 720/(6)(2)(1) 720/12 60
32
Thinking Mathematically

• Section 3
• Combinations

33
Combination definition

• A combination of items occurs when
• The item are selected from the same group.
• No item is used more than once.
• The order of the items makes no difference.

34
How to know when the problem is a permutation
problem or a combination problem

• Permutation
• arrangement, arrange
• order matters
• Combination
• selection, select
• order does not matter.

35
Example Distinguishing between Permutations and
Combinations

• For each of the following problems, explain if
  the problem is one involving permutations or
  combinations.
• Six students are running for student government
  president, vice-president, and treasurer. The
  student with the greatest number of votes becomes
  the president, the second biggest vote-getter
  becomes vice-president, and the student who gets
  the third largest number of votes will be student
  government treasurer. How many different
  outcomes are possible for these three positions?

36
Solution

• Students are choosing three student government
  officers from six candidates. The order in which
  the officers are chosen makes a difference
  because each of the offices (president,
  vice-president, treasurer) is different. Order
  matters. This is a problem involving
  permutations.

37
Example Distinguishing between Permutations and
Combinations

• Six people are on the volunteer board of
  supervisors for your neighborhood park. A
  three-person committee is needed to study the
  possibility of expanding the park. How many
  different committees could be formed from the six
  people on the board of supervisors?

38
Solution

• A three-person committee is to be formed from the
  six-person board of supervisors. The order in
  which the three people are selected does not
  matter because they are not filling different
  roles on the committee. Because order makes no
  difference, this is a problem involving
  combinations.

39
Example Distinguishing between Permutations and
Combinations

• Baskin-Robbins offers 31 different flavors of ice
  cream. One of their items is a bowl consisting
  of three scoops of ice cream, each a different
  flavor. How many such bowls are possible?

40
Solution

• A three-scoop bowl of three different flavors is
  to be formed from Baskin-Robbins 31 flavors.
  The order in which the three scoops of ice cream
  are put into the bowl is irrelevant. A bowl with
  chocolate, vanilla, and strawberry is exactly the
  same as a bowl with vanilla, strawberry, and
  chocolate. Different orderings do not change
  things, and so this problem is combinations.

41
Combinations of n Things Taken r at a Time
n!
n
nCr
r!(n r)!
r
Note that the sum of the two numbers on the
bottom (denominator) should add up to the number
on the top (numerator).
42
Computing Combinations

• Suppose we need to compute 9C3
• r 3, n r 6
• The denominator is the factorial of smaller of
  the two 3!

43
Computing Combinations

• Suppose we need to compute 9C3
• r 3, n r 6
• In the numerator write (the product of) all the
  numbers from 9 down to n - r 1 6 1 7
• There should be the same number of terms in the
  numerator and denominator 9 ? 8 ? 7

44
Computing Combinations

• If called upon, there's a fairly easy way to
  compute combinations.
• Given nCr , decide which is bigger r or n r.
• Take the smaller of the two and write out the
  factorial (of the number you picked) as a
  product.
• Draw a line over the expression you just wrote.

45
Computing Combinations

• If called upon, there's a fairly easy way to
  compute combinations.
• Now, put n directly above the line and directly
  above the leftmost number below.
• Eliminate common factors in the numerator and
  denominator.
• Do the remaining multiplications.
• You're done!

46
Computing Combinations

• Suppose we need to compute 9C3 .
• n r 6, and the smaller of 3 and 6 is 3.

3
4
9
? 8 ? 7
3 ? 4 ? 7
84
3 ? 2 ? 1
1
1
47
Problem

• A three-person committee is to be formed from the
  eight-person board of supervisors.

We saw that this is a combination problem
2
8 ? 7 56
48
A deck of cards
suits
ranks
aces
face cards
49
Problem

• How many poker hands (five cards) are possible
  using a standard deck of 52 cards?
• We need to pick any five cards from the deck.
  Therefore we are selecting 5 out of 52 cards.

50
Problem

• A poker hand of a full house consists of three
  cards of the same rank (e.g. three 8's or three
  kings or three aces.) and two cards of another
  rank. How many full houses are possible?
• First we need to select 1 out of the 13 possible
  ranks. 13
• Then we need to select 3 out of the 4 cards of
  that rank. 4C3 4
• By the fundamental counting principle, there are
  13 ? 4 52 ways to do this part of the problem.

51
Problem

• Similarly, we then need to select 1 out of the 12
  remaining ranks. 12
• Then select 2 out of the 4 cards of that rank.
  4C2 4!/2!2! 24/4 6
• Once again by the fundamental counting principle,
  there are 12 ? 6 72 ways to solve this part of
  the problem.

52
Problem

• Finally, in order to select a full house, we
  merely have to choose any of the 52 possible
  threes of a kind and choose any of the remaining
  72 pairs.
• Therefore there are 52 ? 72 3744 full houses!!!

53
Another Problem

• In poker, a straight is
• 5 cards in sequence, for example
• 3, 4, 5, 6 and a 7
• 8, 9, 10, jack, queen
• ace, 2, 3, 4, 5 (ace as low card)
• 10, jack, queen, king, ace. (ace as high card)
• and not all of the same suit (that's a straight
  flush!)
• How many possible straights can you be dealt?
• Think about it. I won't give the answer now.

54
Another Problem

• One part of the solution is straightforward.
  In how many ways can you select five cards in
  sequence?
• the low card of the sequence can not be greater
  than 10. Getting a 10 as low card means that
  the high card is an ace. You can't go higher
  than ace!
• there are 4 possible suits for each card. So
  there are 10 ? 1024 10,240 ways you can do
  this.

4 ? 4 ? 4 ? 4 ? 4 1024
55
Another Problem

• But some of those include the ones where all the
  cards are of the same suit.
• Remember, there are 10 possible low cards
• There are 4 possible suits for that low card
• Once we have chosen one of those 40 possible
  cards, there is exactly one possibility for the
  next four cards.
• 10,240 40 10,200
• That's the answer

56
Thinking Mathematically

• Section 4
• Fundamentals of Probability

57
Computing Theoretical Probability

58
Remember

• A probability can never be greater than 1 or
  less than 0.

A probability can never be greater than 1 or
less than 0.
A probability can never be greater than 1 or
less than 0.
59
Example Computing Theoretical Probability

• A die is rolled once. Find the probability of
  getting a number less than 5.
• Sample space all possible outcomes 1, 2, 3, 4,
  5, 6
• Event the roll yields a number less than 5 1,
  2, 3, 4

60
Solution

• The event of getting a number less than 5 can
  occur in 4 ways 1, 2, 3, 4.
• P(less than 5)
• 4/6 2/3

61
Example Probability and a Deck of 52 Cards

• You are dealt one card from a standard 52-card
  deck. Find the probability of being dealt a
  King.

62
Solution

• Because there are 52 cards, the total number of
  possible ways of being dealt a single card is 52.
  We use 52, the total number of possible outcomes,
  as the number in the denominator. Because there
  are 4 Kings in the deck, the event of being dealt
  a King can occur 4 ways.
• P(King) 4/52 1/13

63
Empirical Probability
64
Example Computing Empirical Probability

• There are approximately 3 million Arab Americans
  in America. The circle graph shows that the
  majority of Arab Americans are Christians. If an
  Arab American is selected at random, find the
  empirical probability of selecting a Catholic.

65
Solution

• The probability of selecting a Catholic is the
  observed number of Arab Americans who are
  Catholic, 1.26 (million), divided by the total
  number of Arab Americans, 3 (million).
• P(selecting a Catholic from the Arab American
  Population) 1.26/3 0.42

66
Thinking Mathematically

• Section 5
• Probability with the Fundamental Counting
  Principle, Permutations, and Combinations

67
Example Probability and Permutations

• Five groups in a tour, Offspring, Pink Floyd,
  Sublime, the Rolling Stones, and the Beatles,
  agree to determine the order of performance based
  on a random selection. Each bands name is
  written on one of five cards. The cards are
  placed in a hat and then five cards are drawn
  out, one at a time. The order in which the cards
  are drawn determines the order in which the bands
  perform. What is the probability of the Rolling
  Stones performing fourth and the Beatles last?

68
Solution

• We begin by applying the definition of
  probability to this situation.
• P(Rolling Stones fourth, Beatles last)
• We can use the Fundamental Counting Principle to
  find the total number of possible permutations.
• 5 ? 4 ? 3 ? 2 ? 1 120

69
Solution cont.

• We can also use the Fundamental Counting
  Principle to find the number of permutations with
  the Rolling Stones performing fourth and the
  Beatles performing last. You can choose any one
  of the three groups as the opening act. This
  leaves two choices for the second group to
  perform, and only one choice for the third group
  to perform. Then we have one choice for fourth
  and last.
• 3 ? 2 ? 1 ? 1 ? 1 6
• There are six lineups with Rolling Stones fourth
  and Beatles last.

70
Solution cont.

• Now we can return to our probability fraction.
• P(Rolling Stones fourth, Beatles last)
• 6/120 1/20
• The probability of the Rolling Stones performing
  fourth and the Beatles last is 1/20.

71
Example Probability and Combinations

• A club consists of five men and seven women.
  Three members are selected at random to attend a
  conference. Find the probability that the
  selected group consists of
• three men.
• one man and two women.

72
Solution

• We begin with the probability of selecting three
  men.
• P(3 men)
• 12C3 12!/((12-3)!3!) 220
• 5C3 5!/((5-3)!(3!)) 10
• P(3 men) 10/220 1/22

73
Solution part b cont.

• There are 5 men. We can select 1 man in 5C1
  ways.
• There are 7 women. We can select 2 women in 7C2
  ways.
• By the Fundamental Counting Principle, the number
  of ways of selecting 1 man and 2 women is
• 5C1 ? 7C2 5 ? 21 105
• Now we can fill in the numbers in our probability
  fraction. P(1 man and 2 women)
• 105/220 21/44

74
Thinking Mathematically

• Section 6
• Events Involving Not and Or Odds

75
The Probability of an Event Not Occurring

• The probability that an event E will not occur is
  equal to one minus the probability that it will
  occur.
• P(not E) 1 - P(E)

76
Mutually Exclusive Events

• If it is impossible for events A and B to occur
  simultaneously, the events are said to be
  mutually exclusive.

If A and B are mutually exclusive events,
thenP(A or B) P(A) P(B).
77
Or Probabilities with Events That Are Not
Mutually Exclusive

• If A and B are not mutually exclusive events,
  then
• P(A or B) P(A) P(B) - P(A and B)

78
Examples

• In picking a card from a standard deck,
• 1. What is the probability of picking a red
  King?
• 2. What is the probability of not picking a red
  King?
• 3. What is the probability of picking a Heart
  or a face card (a face card is either a Jack,
  Queen or King)?

79
Example

• In picking a card from a standard deck,
• 1. What is the probability of picking a red
  King?
• A red King is either a King of Hearts or a King
  of Diamonds. Since a card can't be both a Heart
  and a Diamond, the events are mutually exclusive
• the probability is 1/52 1/52 1/26.

80
Example

• In picking a card from a standard deck,
• 2. What is the probability of not picking a red
  King?
• Since P(not A) 1 P(A) and we saw that P(A)
  1/26,
• the probability is 1 1/26 25/26.

81
Example

• In picking a card from a standard deck,
• 3. What is the probability of picking a Heart
  or a face card (a face card is either a Jack,
  Queen or King)?
• Since some Hearts are face cards, the events are
  not mutually exclusive.
• There are 13 Hearts, so P(Heart) 13/52.
• There are 12 face cards, so P(Face Card)
  12/52.
• There are 3 cards that are both a Heart and a
  Face Card. (Namely the Jack, Queen and King of
  Hearts).
• P(Heart or Face Card) 13/52 12/52 3/52
  22/52

82
Odds

• For some given event, we sometimes express the
  chances for an outcome by odds.
• The odds in favor of something happening is the
  ratio of the probability that it will happen to
  the probability that it won't.
• Odds in favor of E
• When the Odds in favor of E are a to b, we write
  it as ab.

83
Odds

• Just as we can talk about the odds in favor of
  something happening, we can also talk about the
  odds against something happening
• Odds against E
• When the Odds against E are b to a, we write it
  as ba.

84
Odds

• Suppose you roll a pair of dice.
• There are 6 ? 6 36 possible outcomes.
• There are three ways of rolling an 11 or higher
  a 5 and a 6, a 6 and a 5, and two 6's.
• The probability of rolling an 11 or higher is
  , or .
• The probability of not rolling an 11 or higher is
  1 , or .

85
Odds

• Suppose you roll a pair of dice.
• The odds in favor of rolling an 11 or better
  are
• The odds against rolling an 11 or better are

86
Finding Probabilities from Odds

• If the odds in favor of an event E are a to b,
  then the probability of the event is given by

87
Finding Probabilities from Odds

• If the odds against an event E are a to b, then
  the probability of the event is given by

88
Finding Probabilities from Odds

• Example
• Suppose Bluebell is listed as 71 in the third
  race at the Meadowlands.
• The odds listed on a horse are odds against that
  horse winning, that is, losing.
• The probability of him losing is
• 7 / (71) 7/8.
• The probability of him winning is 1/8.

89
Finding Probabilities from Odds

• Example
• Suppose Bluebell is listed as 71 in the third
  race at the Meadowlands. (ab against)
• The odds listed on a horse are odds against that
  horse winning, that is, losing.
• The probability of him losing is
• 7 / (71) 7/8.
• The probability of him winning is 1/8.

90
Thinking Mathematically

• Section 7
• Events Involving And Conditional Probability

91
Independent Events

• Two events are independent events if the
  occurrence of either of them has no effect on the
  probability of the other.
• For example, if you roll a pair of dice two
  times, then the two events are independent. What
  gets rolled on the second throw is not affected
  by what happened on the first throw.

92
And Probabilities with Independent Events

• If A and B are independent events, then
• P(A and B) P(A) ? P(B)
• The example of choosing from four pairs of socks
  and then choosing from three pairs of shoes (
  12 possible combinations) is an example of two
  independent events.

93
Dependent Events

• Two events are dependent events if the occurrence
  of one of them does have an effect on the
  probability of the other.
• Selecting two Kings from a deck of cards by
  selecting one card, putting it aside, and then
  selecting a second card, is an example of two
  dependent events.
• The probability of picking a King on the second
  selection changes because the deck now contains
  only 51, not 52, cards.

94
And Probabilities with Dependent Events

• If A and B are dependent events, then
• P(A and B) P(A) ? P(B given that A has
  occurred)
• written as P(A) ? P(BA)

95
Conditional Probability

• The conditional probability of B, given A,
  written P(BA), is the probability that event B
  will occur computed on the assumption that event
  A has occurred.
• Notice that when the two events are independent,
  P(BA) P(B).

96
Conditional Probability

• Example
• Suppose you are picking two cards from a deck of
  cards. What is the probability you will pick a
  King and then another face card?
• The probability of an King is .
• Once the King is selected, there are 11 face
  cards left in a deck holding 51 cards.
• P(A) . P(BA)
• The probability in question is ?

97
Applying Conditional Probability to Real-World
Data

• P(BA)
• observed number of times B and A occur together
• observed number of times A occurs

98
Review
99
Thinking Mathematically

• Section 8
• Expected Value

100
Expected Value

• Expected value is a mathematical way to use
  probabilities to determine what to expect in
  various situations over the long run.
• For example, we can use expected value to find
  the outcomes of the roll of a fair dice.
• The outcomes are 1, 2, 3, 4, 5, and 6, each with
  a probability of . The expected value, E, is
  computed by multiplying each outcome by its
  probability and then adding these products.
• E 1? 2? 3? 4? 5? 6?
• (123456)/6 3.5

101
Expected Value

• E 1? 2? 3? 4? 5? 6?
• (1 2 3 4 5 6)/6 3.5
• Of course, you can't roll a 3½ . But the
  average value of a roll of a die over a long
  period of time will be around 3½.

102
Example Expected Value and Roulette

• A roulette wheel has 38 different "numbers."

• One way to bet in roulette is to place 1 on a
  single number.
• If the ball lands on that number, you are awarded
  35 and get to keep the 1 that you paid to play
  the game.
• If the ball lands on any one of the other 37
  slots, you are awarded nothing and the 1 you bet
  is collected.

103
Example Expected Value and Roulette

• 38 different numbers.
• If the ball lands on your number, you win awarded
  35 and you keep the 1 you paid to play the
  game.
• If the ball lands on any of the other 37 slots,
  you are awarded nothing and you lose the 1 you
  bet.

• Find the expected value for playing roulette if
  you bet 1 on number 11 every time. Describe what
  this means.

104
Solution

• E 35( ) (-1)( )
• - - -0.05
• This means that in the long run, a player can
  expect to lose about 5 cents for each game
  played.

105
Expected Value

• A real estate agent is selling a house. She gets
  a 4-month listing. There are 3 possibilities
• she sells the house (30 chance) earns 25,000
• another agent sellsthe house (20 chance)
  earns 10,000
• house not sold (50 chance) loses 5,000
• What is the expected profit (or loss)?
• If the expected profit is at least 6000 she
  would consider it a good deal.

106
Expected Value
7,500
2,000
-2,500
7,000
The realtor can expect to make 7,000.Make the
deal!!!!