Transcript Chapter 11 Powerpoint - Trimble County Schools

Thinking
Mathematically
Chapter 11:
Counting Methods and
Probability Theory
Thinking
Mathematically
Section 1:
The Fundamental Counting
Principle
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 MULTIPLY the numbers!
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?
eggs
OJ
1
cereal
pancakes
apple
OJ
2
3
apple
4
OJ
5
apple
6
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.
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?
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.
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.
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.
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?
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.
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?
Solution
We DON’T 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
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?
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
Thinking
Mathematically
Section 2:
Permutations
Permutations
• A permutation is an arrangement of
objects.
– No item is used more than once.
– The order of arrangement makes a difference.
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?
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
3 choices
offspring
pink floyd
sublime
2 choices
whichever
of the two
remaining
1 choice
only one
remaining
1 choice
stones
Thus, there are six different ways to arrange the
concert if the Rolling Stones are the final group to
perform.
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?
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.
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
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 – r)!
= n(n – 1) (n – 2) (n – 3) . . . (n – r + 1)
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)
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
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
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!
messes
here are six permutations of messes
me s s e s 1
me s s e s 2
me s s e s
3
me s s e s
4
me s s e s
5
6
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
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.
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.
Permutations with duplicates.
• In general if we want to arrange n items, of which
m1, m2, .... are identical, the number of
permutations is
n!
m1!m2!m3!
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
Thinking
Mathematically
Section 3:
Combinations
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.
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.
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?
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.
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?
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.
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?
Solution
• A three-scoop bowl of three different flavors is to
be formed from Baskin-Robbin’s 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.
Combinations of n Things Taken r at a
Time
n
r
= nCr =
n!
r!(n – r)!
Note that the sum of the two numbers on the bottom
(denominator) should add up to the number on the
top (numerator).
Computing Combinations
• Suppose we need to compute 9C3
9!
9!

99C 33 
3!(9  3)! 3!6!
3!6!
• r = 3, n – r = 6
• The denominator is the factorial of smaller of
the two: 3!
Computing Combinations
• Suppose we need to compute 9C3
9!
9!

99C 33 
3!(9  3)! 3!6!
3!6!
• 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
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.
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!
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21
1 1
= 3  4  7 = 84
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!
8

7

6

= 8  7 = 56
8 C3 
3!5! 3  2  1
A deck of cards
s
u
i
t
s
ranks
aces
face cards
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.
52  51  50  49  48  52  51  5  49  48  52  51  5  49  4  2,598,960
C

52 5
5 4 3 21
4 31
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.
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.
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!!!
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.
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.
4  4  4  4  4 =1024
So there are 10  1024 = 10,240 ways you can do
this.
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
Thinking
Mathematically
Section 4:
Fundamentals of Probability
Computing Theoretical
Probability
number of outcomes in event E
P( E ) 
number of outcomes in sample space S
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.
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
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
(number of ways a number less than 5 can occur)
(total number of possible outcomes)
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.
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
Empirical Probability
observed number of times E occurs
P( E ) 
total number of observed occurrences
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.
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
Thinking
Mathematically
Section 5:
Probability with the Fundamental
Counting Principle, Permutations, and
Combinations
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 band’s 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?
Solution
We begin by applying the definition of
probability to this situation.
P(Rolling Stones fourth, Beatles last) =
(permutations with Rolling Stones fourth, Beatles last)
(total number of possible permutations)
We can use the Fundamental Counting Principle to
find the total number of possible permutations.
5  4  3  2  1 = 120
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.
Solution cont.
Now we can return to our probability fraction.
P(Rolling Stones fourth, Beatles last) =
(permutations with Rolling Stones fourth, Beatles last)
(total number of possible permutations)
= 6/120 = 1/20
The probability of the Rolling Stones performing
fourth and the Beatles last is 1/20.
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:
a. three men.
b. one man and two women.
Solution
We begin with the probability of selecting three
men.
number of ways of selecting 3 men
P(3 men)=
total number of possible combinations
12C3
= 12!/((12-3)!3!) = 220
5C3 = 5!/((5-3)!(3!)) = 10
P(3 men) = 10/220 = 1/22
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) =
number of ways of selecting 1 man and 2 women
total number of possible combinations
= 105/220 = 21/44
Thinking
Mathematically
Section 6:
Events Involving Not and Or; Odds
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)
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,
then
P(A or B) = P(A) + P(B).
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)
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)?
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.
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.
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
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.
P( E )
P( E )

– Odds in favor of E =
P ( not E ) 1- P ( E )
– When the Odds in favor of E are a to b, we write
it as a:b.
Odds
• Just as we can talk about the odds in favor
of something happening, we can also talk
about the odds against something
happening:
P ( not E ) 1  P ( E )
– Odds against E =

P( E )
P( E )
– When the Odds against E are b to a, we write
it as b:a.
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 3 ,
36
1
or .
12
– The probability of not rolling an 11 or higher is
1 – 1 , or 11.
12
12
Odds
• Suppose you roll a pair of dice.
– The odds in favor of rolling an 11 or better are
1
12
11
12
1

or 1 : 11
11
– The odds against rolling an 11 or better are
11
12
1
12
11

or 11 : 1
1
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
a
P( E ) 
ab
Finding Probabilities from
Odds
• If the odds against an event E are a to b,
then the probability of the event is given
by
b
P( E ) 
ab
Finding Probabilities from
Odds
• Example:
– Suppose Bluebell is listed as 7:1 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 / (7+1) = 7/8.
– The probability of him winning is 1/8.
Finding Probabilities from
Odds
• Example:
– Suppose Bluebell is listed as 7:1 in the third
race at the Meadowlands. (a:b against)
– The odds listed on a horse are odds against that
horse winning, that is, losing.
– The probability of him losing is
a
7 / (7+1) = 7/8. a  b
b
– The probability of him winning is 1/8. a  b
ab
Thinking
Mathematically
Section 7:
Events Involving And;
Conditional Probability
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.
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.
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.
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(B|A)
Conditional Probability
• The conditional probability of B, given A,
written P(B|A), 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(B|A) = P(B).
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?
1 .
– The probability of an King is 4 =
52 13
– Once the King is selected, there are 11 face cards
left in a deck holding 51 cards.
– P(A) = 1 . P(B|A) = 11
51
13
– The probability in question is 1  11
13 51
Applying Conditional
Probability to Real-World Data
P(B|A) =
observed number of times B and A occur together
observed number of times A occurs
Review
P(not E)
P(A or B)
P(A and B)
Odds in favor a:b
Odds against a:b
1 – P(E)
P(A) + P(B)
mutually
– P(A and B) exclusive:
P(A) + P(B)
P(A)  P(B|A)
independent:
P(A)  P(B)
P(E) / P(not E)
probability is
a/(a+b)
P(not E) / P(E)
probability is
b/(a+b)
Thinking
Mathematically
Section 8:
Expected Value
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 16. The expected value, E, is computed
by multiplying each outcome by its probability and
then adding these products.
• E = 1 16 + 2 16 + 3 16 + 4 16 + 5 16 + 6 16
= (1+2+3+4+5+6)/6 = 216 = 3.5
Expected Value
E = 1 16 + 2 16 + 3 16 + 4 16 + 5 16 + 6 16
= (1 + 2 + 3 + 4 + 5 + 6)/6 = 216 = 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.
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.
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.
Solution
Outcome
Gain/Loss
$35
 $1
11
Not 11
Probability
1
38
37
38
E = $35( 1 ) + (-$1)( 37)
38
=
35
$
38
37
$ =
38
38
2
-$ ≈
38
-$0.05
This means that in the long run, a player can
expect to lose about 5 cents for each game
played.
Expected Value
• A real estate agent is selling a house. She gets a 4month listing. There are 3 possibilities:
– she sells the house:
(30% chance) earns $25,000
– another agent sells
the 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.
Expected Value
Outcome
Probability
she sells
0.3
Profit or
loss
+$25,000
product
other sells
0.2
+$10,000
+$2,000
doesn't sell 0.5
-$5,000
-$2,500
+$7,500
+$7,000
The realtor can expect to make $7,000.
Make the deal!!!!