04/20/23 Section 11.1 1

Section 11.1 The Fundamental Counting Principle

Objective

1. Use the Fundamental Counting Principle to determine the number of possible outcomes in a given situation.

04/20/23 Section 11.1 2

Fundamental Counting Principle

• DefinitionIf 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.

• Tree DiagramA representation of all possiblechoices. This tree diagram shows

that there are 2·3 = 6 differentOutfits from 2 pairs of jeans and three T-shirts.

04/20/23 Section 11.1 3

Example 1Applying the Fundamental Counting

Principle

The Greasy Spoon Restaurant offers 6 appetizers and 14 main courses. In how many ways can a person order a two-course meal?

Solution:

Choosing from one of 6 appetizers and one of 14 main courses, the total number of two-course meals is:

6 ∙14 = 84

04/20/23 Section 11.1 4

The Fundamental Counting Principle with More than Two Groups of Items

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

• The number of possible outfitsfrom 2 pair of jeans, 3 T-shirts and 2 pairs of sneakers are: 2·3·2 = 12

04/20/23 Section 11.1 5

Example 2 Applying the Fundamental Principle of Counting with

More Than Two Groups of Items

Options in Planning a Course ScheduleNext semester, you are planning to take three courses – math, English and humanities. There are 8 sections of math,5 of English, and 4 of humanities that you find suitable.Assuming no scheduling conflicts, how many different three-course schedules are possible?Solution:This situation involves making choices with three groups of

items. Math English Humanities {8 choices} {5 choices} {4 choices}

There are 8∙5∙4 = 160 different three-course schedules.

04/20/23 Section 11.1 6

Example 3 A Multiple-Choice Test

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

• Solution:

This situation involves making choices with ten questions:Question 1 Question 2 Question 3 ∙ ∙ ∙ Question 9 Question 10{4 choices} {4 choices} {4 choices} {4 choices} {4 choices}

The number of different ways you can answer the questions is: 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 ∙ 4 = 410 = 1,048,576

04/20/23 Section 11.1 7

Example 4 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: This situation involves making choices with ten groups of items. Here are the choices for each of the ten groups of items: Area Code Local Telephone Number 8 10 10 8 10 10 10 10 10 10

The total number of different telephone numbers is: 8 ∙ 10 ∙ 10 ∙ 8 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 = 6,400,000,000

04/20/23 Section 11.2 8

Section 11.2Permutations

Objectives1. Use the Fundamental Counting Principle

to count permutations.

2. Evaluate factorial expressions.

3. Use the permutations formula.

4. Find the number of permutations of duplicate items.

04/20/23 Section 11.2 9

Permutations

• Permutation is an ordered arrangement of items that occurs when:

– No item is used more than once.

– The order of arrangement makes a difference.

04/20/23 Section 11.2 10

Example 1Counting Permutations

• You are in charge of planning a concert tour with U2, ‘N Sync, Aerosmith and the Rolling Stones. You decide that the Rolling Stones will perform last. How many different ways can you put together the concert?

• Solution:

You can now choose any one of the three groups, U2, ‘N sync or Aerosmith, as the opening act. That leaves two groups left to choose from for the second group and only one group left for the third group.

Using the Fundamental Counting Principle there are:

3∙2∙1∙1 = 6 different ways to arrange the concert.

04/20/23 Section 11.2 11

Factorial Notation

• The product 7∙6∙5∙4∙3∙2∙1 is called factorial and is written 7!• Definition:

If n is a positive integer, the notation n! (read “n factorial”) is the product of all positive integers from n down through 1.

n! = n(n - 1)(n - 2)···(3)(2)(1)

0! (zero factorial), by definition, is 1.

0! = 1

04/20/23 Section 11.2 12

Example 2Using Factorial Notation

• Evaluate without using your calculator:

∕

∕

∕

∕ 600,893,72223242526

!21

!212223242526

!21

!26

336678!5

!5678

!5

!8

04/20/23 Section 11.2 13

A Formula for Permutations

• Permutations of n things taken r at a time.

)!(

!

rn

nPrn

04/20/23 Section 11.2 14

Example 3Using the Formula for Permutations

• You and 19 of your friends have decided to form a business. The group needs to choose three officers– a CEO, an operating manager, and a treasure. In how many ways can those offices be filled?

• Solution:

Your group is choosing r = 3 officers from a group of n = 20 people. The order matters because each officer has different responsibilities:

∕

∕6840181920

!17

!17181920

!17

!20

)!320(

!20320

PPrn

04/20/23 Section 11.2 15

A Formula for Permutations of Duplicate Items

• Permutations of Duplicate Items

The number of permutations of n items, where p items are identical, q items are identical, r items are identical, and so on, is given by:

!...!!

!

rqp

n

04/20/23 Section 11.2 16

Example 4Using the Formula for Permutations of Duplicate

Items

• In how many distinct ways can the letters of the word MISSISSIPPI be arranged?

• Solution:

The word contains 11 letters (n = 11) where four Is are identical (p = 4), four Ss are identical (q = 4) and 2 Ps are identical (r = 2). The number of distinct permutations are:

650,43121234!4

!4567891011

!2!4!4

!11

!!!

!

rqp

n

04/20/23 Section 11.3 17

Section 11.3Combinations

Objectives

1. Distinguish between permutation and combination problems.

2. Solve problems involving combinations using the combinations formula.

04/20/23 Section 11.3 18

Combinations

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

Note: • Permutation problems involve situations in which order

matters. • Combination problems involve situations in which the

order of items makes no difference.

04/20/23 Section 11.3 19

Comparing Combinations and Permutations

Given the letters: A,B,C,D: We can compare how many

permutations and how many combinations are possible if

we chose 3 letters at a time:

04/20/23 Section 11.3 20

Example 1Distinguishing Between Permutations and

CombinationsDetermine which involve permutations and which involve

combinations.• Six student are running for student government

president, vice-president and treasure. The student with the greatest number of votes becomes the president, the second highest vote-getter becomes vice-president, and the student who gets the third largest number of votes will be treasurer. How many different outcomes are possible?

• Solution: Order matters since the number of the three highest votes determine the office. Permutations

04/20/23 Section 11.3 21

Example 1Distinguishing Between Permutations and

Combinations (continued)

• Six people are on the board of supervisors for your neighborhood part. 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?

• Solution:

The order in which the three people are selected does not matter since they are not filling different roles. Combinations

04/20/23 Section 11.3 22

A Formula for Combinations

The number of possible combinations if r items are taken from n items is:

!)!(

!

rrn

nCrn

04/20/23 Section 11.3 23

Example 2Using the Formula for Combinations

• How many three-person committees could be formed from 8 people?

• Solution: We are selecting 3 people ( r = 3) from 8 people ( n = 8)

∕

∕ 56

123!5

!5678

!3!5

!8

!3)!38(

!838

CCrn

04/20/23 Section 11.3 24

Example 3Using the Formula for Combinations and the

Fundamental Counting Principle

• The U.S Senate of the 104th Congress

consisted of 54 Republicans and 46

Democrats.

How many committees can be formed

if each committee must have 3

Republicans and 2 Democrats?

• Solution:

The order in which members are selected does not matter so this is a problem of combinations.

04/20/23 Section 11.3 25

Example 3 continued

First, select 3 Republicans out of 54 Republicans:

Secondly, select 2 Democrats out of 46 Democrats:

Use the Fundamental Counting Principle to find the number of committees that can be formed.

804,24123

525354

123!51

!51525354

!3!51

!54

!3)!354(

!54354

CCrn

035,112

444546

12!44

!444546

!2!44

!46

!2)!246(

!46246

CCrn

140,672,25035,1804,24246354

CC

04/20/23 Section 11.4 26

Section 11.4Fundamentals of Probability

Objectives

1. Compute theoretical probability.

2. Compute empirical probability.

04/20/23 Section 11.4 27

Probability

• Probabilities are assigned values from 0 to 1.

• The closer the probability of a given event is to 1, the more likely it is that the event will occur.

• The closer the probability of a given event is to 0, the less likely that the event will occur.

04/20/23 Section 11.4 28

Theoretical Probability

• Experiment is any occurrence for which the outcome is uncertain.

• Sample space is the set of all possible outcomes of an experiment , denoted by S.

• Event, denoted by E is any subset of a sample space.

• Sum of the theoretical probabilities of all possible outcomes is 1.

04/20/23 Section 11.4 29

Computing Theoretical Probability

• If an event E has n(E) equally likely outcomes and its sample space S has n(S) equally-likely outcomes, the theoretical probability of event E, denoted by P(E), is:

• P(E) = number of outcomes in event E = n(E) total number of possible outcomes n(S)

04/20/23 Section 11.4 30

Example 1Computing Theoretical Probability

• A die is rolled once. The sample space is S = {1,2,3,4,5,6} with n(S) = 6. Find the probability of rolling

a. a 3 b. an even number• Solution to a: There is only one way to roll a 3 so n(E) = 1.P(3) = number of outcomes that result in 3 = n(E) = 1 total number of possible outcomes n(S) 6

• Solution to b:Rolling an even number describes the event E = {2,4,6}.This event can occur in 3 ways: n(E) = 3.P(even number) = number of outcomes that result in even number = n(E) = 3 =1 total number of possible outcomes n(S) 6 2

04/20/23 Section 11.4 31

Example 2Probability and a Deck of 52 Cards

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

a. KingSolution: P(3) = number of outcomes that result in a king = n(E) = 4 =1 total number of possible outcomes n(S) 52 13

b. HeartSolution:P(heart) = number of outcomes that result in a heart = n(E) = 13=1 total number of possible outcomes n(S) 52 4

04/20/23 Section 11.4 32

Empirical Probability

• Applies to situations in which we observe how frequently an event occurs.

• Computing Empirical Probability

The empirical probability of event E is:

• P(E) = observed number of times E occurs = n(E) total number of observed occurrences n(S)

04/20/23 Section 11.4 33

Example 4Computing Empirical Probability

If one person is randomly selected from the populationdescribed above, find the probability that the person is female.Solution. The probability of selecting a female is the observed

number of females, 110.1(million), divided by the total number of U.S. adults, 212.5 (million).

P (selecting a female) = number females_ = 110.1 ≈ 0.52 total number of adults 212.5

04/20/23 Section 11.5 34

Section 11.5Probability with the Fundamental counting Principle, Permutations, and Combinations

Objectives

1. Compute probabilities with permutations.

2. Compute probabilities with combinations,

04/20/23 Section 11.5 35

Example 1Probability with Permutations.

• Five groups agree to randomly select the order of performance by picking cards out of a hat, one at a time. What is the probability of the Rolling Stones performing fourth and the Beatles last?

• Solution:P (Rolling Stones fourth, Beatles last) =Number of permutations with Rolling Stones fourth, Beatles last

Total number of possible permutations

• Use the Fundamental Counting Principle to find the total number of possible permutations.

• There are 5·4·3·2·1 =120 possible permutations.

04/20/23 Section 11.5 36

Example 1 continued

• Use the Fundamental Counting Principle to find the number of permutations with the Rolling Stones performing fourth and the Beatles performing last.

• There are 3·2·1·1·1 = 6 possible permutations.• Putting both values in the original equation.

P (Rolling Stones fourth, Beatles last) = 6 = 1 120 20

04/20/23 Section 11.5 37

Example 2Probability and Combinations: Winning the Lottery

• Florida’s lottery game, LOTTO, is set up so that each player chooses six different numbers from 1 to 53. With one LOTTO ticket, what is the probability of winning this prize?

• Solution: Because the order of the six numbers does not matter,

this situation involves combinations.

P (winning) = number of ways of winning total number of possible combinations

04/20/23 Section 11.5 38

Example 2 continued

P (winning) = number of ways of winning total number of possible combinations

Using the combinations formula, r = 6 and n = 53.

If a person buys only one ticket, then that person has selected only one combination, thus

P (winning) = number of ways of winning ___ = __1___ total number of possible combinations 22,957,480

≈ 0.0000000436

22,957,48047!6!

53!

6)!6!(53

53!CC

653rn

04/20/23 Section 11.5 39

Example 3Probability 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 3 men.

• Solution:

Order of selection does not matter, so this is a problem involving combinations.

P(3 men) = number of ways of selecting 3 men total number of possible combinations

04/20/23 Section 11.5 40

Example 3 continuedConsider the denominator: We are selecting r = 3 people

from a total group of n = 12.

Consider the numerator. We are interested in selecting 3 (r = 3)men from 5 (n = 5) men.

Therefore: P(3 men ) = 10 = 1

220 22

102!3!

5!

3)!3!(5

5!C

35

2209!3!

12!

3)!3!(12

12!C

312

04/20/23 Section 11.6 41

11.6Events Involving Not and Or; Odds

Objectives

1. Find the probability that an event will not occur.

2. Find the probability of one event or a second event occurring.

3. Understand and use odds.

04/20/23 Section 11.6 42

Probability of an Event Not Occurring

• Complement of E: If we know P(E), the probability of an event E, we can determine the probability that the event will not occur, denoted by P(not E).

• The probability that an event E will not occur is equal to 1 minus the probability that it will occur.

P(not E) = 1 – P(E)

• The probability that an event E will occur is equal to 1 minus the probability that it will not occur.

P(E) = 1 – P(not E)

Using set notation, if E’ is the complement of E, then P(E’) = 1 – P(E) and P(E) = 1 – P(E’)

04/20/23 Section 11.6 43

Example 1The Probability of an Event Not Occurring

• If you are dealt one card from a standard 52-card deck, find the probability that you are not dealt a queen.

• Solution: Because P(not E) = 1 – P(E) then

P(not a queen) = 1 – P(queen).

• There are 4 queens in a standard deck. • The probability of being dealt a queen is 4 =1 52 13

• P(not a queen) = 1 – P(queen) = 1 – 1 = 12 13 13

04/20/23 Section 11.6 44

Or Probabilities with Mutually Exclusive Events

• Mutually Exclusive Events: Events A and B are mutually exclusive if it is impossible for them to occur simultaneously.

• Or Probabilities with Mutually Exclusive Events:– If A and B are mutually exclusive events, then

P(A or B) = P(A) + P(B)

– Using set notation,

P(A U B) = P(A) + P(B)

04/20/23 Section 11.6 45

Example 2The Probability of Either of Two Mutually

Exclusive Events Occurring

• If one card is randomly selected from a deck of cards, what is the probability of selecting a king or a queen?

• Solution:

P ( king or queen) = P(king) + P(queen)

= 4 + 4 = 8 = 2 52 52 52 13

04/20/23 Section 11.6 46

Or Probabilities with Events That Are Not Mutually Exclusive

• If A and B are not mutually exclusive events then the probability that A or B will occur is determined by adding their individual probabilities and then subtracting the probability that A and B occur simultaneously.

– P(A or B) = P(A) + P(B) – P(A and B)

– Using set notation,

P(A U B) = P(A) + P(B) – P(A ∩B)

04/20/23 Section 11.6 47

Example 3An OR Probability With Events That Are Not

Mutually Exclusive

• In a group of 25 baboons, 18 enjoy grooming their neighbors, 16 enjoy screeching wildly, while 10 enjoy doing both. If one baboon is selected at random, find the probability that it enjoy grooming its neighbors or screeching wildly.

• Solution: Since 10 of the baboons enjoy both grooming their neighbors and screeching wildly, these events are not mutually exclusive.

04/20/23 Section 11.6 48

Example 3 continued

04/20/23 Section 11.6 49

Probability to Odds• If we know the probability of an event, we can determine

the odds in favor, or the odds against, the event.Probability to Odds:

1. The odds in favor of E are found by taking the probability that E will occur and dividing by the probability that E will not occur.

Odds in favor of E = P(E) . P(not E)

• The odds against E are found by taking the probability that E will not occur and dividing by the probability that E will occur.

Odds against E = P(not E) P(E)

The odds against E can also be found by reversing the ratio representing the odds in favor of E.

04/20/23 Section 11.6 50

Example 4From Probability to Odds

• You roll a single, six-sided die.Find the odds in favor of rolling a 2Solution: Let E represent the event of rolling a 2• First, find the probability of E occurring and the

probability of E not occurring.S = {1,2,3,4,5,6}E = {2}P(E) = 1 and P(not E) = 1 – P(E) = 1 – 1 = 5

6 6 6 1

• Odds in favor of E( rolling a 2 ) = P(E) . = 6 = 1 .6 =1 P(not E) 5 6 5 5 6

04/20/23 Section 11.6 51

Example 5From Odds to Probability

Odds to Probability If the odds in favor of event E are a to b, then the probability of

the event is given by:P(E) = a . a + b

• The odds in favor of a particular horse winning a race are 2 to 5. What is the probability this horse will win the race?

• Solution. Because odds in favor, a to b, means a probability of a , then the odds in favor, 2 to 5, a+bmeans a probability of:

2 = 2 2 + 5 7

04/20/23 Section 11.7 52

Section 11.7Events Involving And; Conditional Probability

Objectives

1. Find the probability of one event and a second event occurring.

2. Compute conditional probabilities.

04/20/23 Section 11.7 53

And Probabilities with Independent Events

• Independent Events: Two events are independent events if the occurrence of either of them has no effect on the probability of the other.

• And Probabilities with Independent Events

If A and B are independent events, then

P(A and B) = P(A) ∙P(B)

04/20/23 Section 11.7 54

Example 1Independent Events on a Roulette Wheel

• A U.S. roulette wheel has 38 numbered slots (1 through 36, 0, and 00. 18 are black, 18 are red, and 2 are green. The ball can land on any slot with equal probability. What is the probability of red occurring on 2 consecutive plays.

• Solution: the probability of red occurring on a play is 18 or 9 .

32 19Thus,

P( red and red) = P( red)∙P( red) = 9 . 9 = 81 ≈ 0.224

19 19 361

04/20/23 Section 11.7 55

Example 2Independent Events In a Family

• If two or more events are independent, we can find the probability of them all occurring by multiplying their probabilities. The probability of a baby girl is ½, so the probability of nine girls in a row is ½ used as a factor nine times.

• Solution:

P( nine girls in a row)

= ½∙ ½∙ ½∙ ½∙ ½∙ ½∙ ½∙ ½∙ ½

= (½)² = 1 . 512

04/20/23 Section 11.7 56

Example 3The Probability of an Event Happening At

Least Once• P( event happening at least once) = 1 – P( event does not happen)The probability that South Florida will not be hit by a

hurricane in ten years was computed to be 0.047.What is the probability that South Florida will be hit by a

hurricane at least once in the next 10 years?

Solution: P( at least one hurricane in 10 years) = 1 – P( no hurricane for ten years)= 1 – 0.047 = 0.953

04/20/23 Section 11.7 57

And Probabilities with Dependent Events

• Dependent Events: Two events are dependent events if the occurrence of one of them has an effect on the probability of the other.

• 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).

04/20/23 Section 11.7 58

Example 4And Probabilities with Dependent Events

• There are 20 chocolates in a box,

5 of which are chocolate-covered cherries.

What is the probability of choosing

2 chocolate-covered cherries?

• Solution:

P( chocolate-covered cherry and chocolate-covered cherry) =

P( chocolate-covered cherry) ∙ P( chocolate-covered cherry) = (given that one was selected )

1 ∙ 4 = 1 4 19 19

04/20/23 Section 11.7 59

Example 5And Probability With Three Dependent

Events

• If two or more events are dependent, we can find the probability of them all occurring by multiplying their probabilities.

• Three people are randomly selected, one person at a time, from 5 freshmen, 2 sophomores, and 4 juniors. Find the probability that the first two people selected are freshmen and the third is a junior.

• Solution:

04/20/23 Section 11.7 60

Conditional Probability

• The probability of event B, assuming that the event A has already occurred, is called the conditional probability of B, given A. This probability is denoted by P(B|A).

P(B|A) = n(B A) = number of outcomes common to B and A n(A)number of outcomes in A

• Conditional probability is the probability that event B occurs if the sample space is restricted to the outcomes associated with event A.

04/20/23 Section 11.7 61

Example 6Finding Conditional Probability

• A letter is randomly selected from the letters of the English alphabet. Find the probability of selecting a vowel, given that the outcome is a letter that precedes h.

• Solution:

We are looking for P( vowel | letter precedes h).

This is the probability of a vowel if the sample space is

restricted to the set of letters that precede h. Thus, the

sample space is given by: S = {a,b,c,d,e,f,g}.

There are 7 possible outcomes in the sample space. We

can select a vowel from this set in one of two ways: a or e.

Therefore, P( vowel | letter precedes h) = 2 7

04/20/23 Section 11.7 62

Example 7Conditional Probabilities with Real-World Data

• Assuming that these numbers are representative of all U.S. women age 40 to 50, find the probability that a women in this age range has a positive mammogram, given that she does not have breast cancer.

• Solution: The data states that there are 6944 women with positive mammograms and

6,944 + 92,256, or 99,200 women without breast cancer. Thus,

P( positive mammogram | no breast cancer) = 6,944 = 0.07 99,200

Mammography Screening on 100,000 U.S. Women, Ages 40 to 50

Breast Cancer No Breast Cancer Total

Positive Mammogram 720 6,944 7,664

Negative Mammogram 80 92,256 92,336

Total 800 99,200 100,000

04/20/23 Section 11.8 63

Section 11.8Expected Value

Objectives1. Compute expected value.

2. Use expected value to solve applied problems.

3. Use expected value to determine the average payoff or loss in a game of chance.

04/20/23 Section 11.8 64

Expected Value

• Expected value is a mathematical way to use probabilities to determine what to expect in various situations over the long run.

• Expected value is used to– determine premiums on insurance policies.– weigh the risks versus the benefits in alternatives in

business ventures. – indicate to a player of any game of chance what will

happen if the game is played repeatedly.

• Standard way to compute expected value is to multiply each possible outcome by its probability and then add these products.

04/20/23 Section 11.8 65

Example 1Computing Expected Value

• Find the expected value for the number of girls for a family with three children.

• Solution:A family with three children can have 0,1,2, or 3 girls. There are eight ways these outcomes can occur.

The expected value, E, is computed by multiplying each outcome by its probability and then adding these products.

04/20/23 Section 11.8 66

Example 1 continued

• The expected value is 1.5. This means that if we record the number of girls in many different three-child families, the average number of girls for all these families will be 1.5 or half the children.

1.523

812

83630

81

383

283

181

0E

04/20/23 Section 11.8 67

Example 2Determining an Insurance Premium

• An automobile insurance company hasdetermined the probabilities for various claim amounts (to the nearest $2000) fordrivers ages 16 through 21 as shown in the table. Calculate the expected value and describe what this means in practical terms.E = $0(0.70) + $2000(0.15) +

$4000(0.08) + $6000(0.05) + $8000(0.01)+$10,000(0.01) = $0+$300+ $320+$300+ $80+$100=$1100. • This means that in the long run, the average cost of a claim

is $1100 which is the very least the insurance company should charge to break even.

Amount of Claim

Probability

$0 0.70

$2000 0.15

$4000 0.08

$6000 0.05

$8000 0.01

$10,000 0.01

04/20/23 Section 11.8 68

Example 3Expect Value and Games of Chance

• To find the expected value of a game, multiply the gain or loss for each possible outcome by its probability. Then add the products.

• Find the expected value of betting $1 on the number 20 in roulette.

– If the ball lands on that number , you are awarded $35 and get to keep the $1 that you paid to play.

– If the ball lands on any of the other 37 slots, you are awarded nothing and the $1 that you bet is collected.

04/20/23 Section 11.8 69

Example 3 continued

38

3738

1

• The expected value is approximately -$0.05. This means that in the long run, a player can expect to lose about 5¢ for each game played.

Playing one Number with a 35 to 1 payoff in roulette

Outcome Gain or Loss Probability

Ball lands on 20 $35

Ball doesn’t land on 20 -$1

$0.0538

$2

38

$37$35

38

37$1)(

38

135E