Thinking Mathematically Chapter 11: Counting Methods and Probability Theory

Thinking Thinking MathematicallyMathematically

Chapter 11:Chapter 11:Counting Methods andCounting Methods and

Probability TheoryProbability Theory


Section 1:Section 1:

The Fundamental Counting The Fundamental Counting PrinciplePrinciple

The Fundamental Counting PrincipleThe Fundamental Counting Principle

If you can choose one item from a group of If you can choose one item from a group of MM items and a second item from a group of items and a second item from a group of NN items, then the total number of two-item items, then the total number of two-item choices is choices is MM NN..

You the numbers!You the numbers!MULTIPLYMULTIPLY

The Fundamental Counting PrincipleThe Fundamental Counting Principle

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

eggseggs cerealcerealpancakespancakes

appleappleOJOJ

11 22

OJOJ appleapple

33 44

appleappleOJOJ

55 66

Example: Applying the Fundamental Example: Applying the Fundamental Counting PrincipleCounting Principle

• The Greasy Spoon Restaurant offers The Greasy Spoon Restaurant offers 6 6 appetizers and appetizers and 1414 main courses. How many main courses. How many different meals can be created by selecting different meals can be created by selecting one appetizer and one main course? one appetizer and one main course?

• Using the fundamental counting principle, Using the fundamental counting principle, there are there are 14 14 6 = 84 6 = 84 different ways a different ways a person can order a two-course meal.person can order a two-course meal.

Example: Applying the Fundamental Example: Applying the Fundamental Counting PrincipleCounting Principle

• This is the semester that you decide to take your This is the semester that you decide to take your required psychology and social science courses. required psychology and social science courses.

• Because you decide to register early, there are Because you decide to register early, there are 1515 sections of psychology from which you can sections of psychology from which you can choose. Furthermore, there are choose. Furthermore, there are 99 sections of social sections of social science that are available at times that do not science that are available at times that do not conflict with those for psychology. In how many conflict with those for psychology. In how many ways can you create two-course schedules that ways can you create two-course schedules that satisfy the psychology-social science requirement?satisfy the psychology-social science requirement?

SolutionSolution

The number of ways that you can satisfy the The number of ways that you can satisfy the requirement is found by multiplying the requirement is found by multiplying the number of choices for each course. number of choices for each course. You can choose your psychology course You can choose your psychology course from from 1515 sections and your social science sections and your social science course from course from 99 sections. For both courses sections. For both courses you have: you have: 15 15 9, 9, or or 135135 choices. choices.

The Fundamental Counting The Fundamental Counting PrinciplePrinciple

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

Example: Options in Planning a Example: Options in Planning a Course ScheduleCourse Schedule

Next semester you are planning to take three Next semester you are planning to take three courses - math, English, and humanities. Based courses - math, English, and humanities. Based on time blocks and highly recommended on time blocks and highly recommended professors, there are professors, there are 88 sections of math, sections of math, 55 of of English, and English, and 44 of humanities that you find of humanities that you find suitable. Assuming no scheduling conflicts, there suitable. Assuming no scheduling conflicts, there are:are:

8 8 5 5 4 = 160 4 = 160 different three course schedules. different three course schedules.

ExampleExample

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

SolutionSolution

This situation involves making choices with This situation involves making choices with four groups of items.four groups of items.

color - air-conditioning - removable roof - computercolor - air-conditioning - removable roof - computer

9 9 2 2 2 2 2 = 72 2 = 72

Thus the car can be ordered in Thus the car can be ordered in 7272 different different ways.ways.

Example: A Multiple Choice TestExample: A Multiple Choice Test

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

SolutionSolution

We DON’T blindly multiply the first two numbers We DON’T blindly multiply the first two numbers we see. The answer is we see. The answer is notnot 10 10 4 = 40.4 = 40.

We use the Fundamental Counting Principle to We use the Fundamental Counting Principle to determine the number of ways you can answer the determine the number of ways you can answer the test. Multiply the number of choices, 4, for each of test. Multiply the number of choices, 4, for each of the ten questionsthe ten questions

4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 =1,048,576=1,048,576

Example: Telephone Numbers in Example: Telephone Numbers in the United Statesthe United States

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

SolutionSolution

We use the Fundamental Counting Principle We use the Fundamental Counting Principle to determine the number of different to determine the number of different telephone numbers that are possible.telephone numbers that are possible.

8 8 10 10 10 10 8 8 10 10 10 10 10 10 10 10 10 10 10 =6,400,000,000 10 =6,400,000,000



PermutationsPermutations

PermutationsPermutations

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

Example: Counting Example: Counting PermutationsPermutations

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

SolutionSolution

We use the Fundamental Counting Principle to We use the Fundamental Counting Principle to find the number of ways you can put together the find the number of ways you can put together the concert. Multiply the choices:concert. Multiply the choices:

3 3 2 2 1 1 1 = 6 1 = 6

Thus, there are six different ways to arrange the Thus, there are six different ways to arrange the

concert if the Rolling Stones are the final group to concert if the Rolling Stones are the final group to perform.perform.

stonesoffspring pink floydsublime

whichever of the two remaining

only one remaining

3 choices 2 choices 1 choice 1 choice

Example: Counting Example: Counting PermutationsPermutations

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

SolutionSolution

You may choose any of the seven books for the You may choose any of the seven books for the first position on the shelf. This leaves six choices first position on the shelf. This leaves six choices for second position. After the first two positions for second position. After the first two positions are filled, there are five books to choose from for are filled, there are five books to choose from for the third position, four choices left for the fourth the third position, four choices left for the fourth position, three choices left for the fifth position, position, three choices left for the fifth position, then two choices for the sixth position, and only then two choices for the sixth position, and only one choice left for the last position.one choice left for the last position.

7 7 6 6 5 5 4 4 3 3 2 2 1 = 5040 1 = 5040There are 5040 different possible permutations.There are 5040 different possible permutations.

Factorial NotationFactorial Notation

If If nn is a positive integer, the notation is a positive integer, the notation nn! is ! is the product of all positive integers from n the product of all positive integers from n down through 1.down through 1.

nn! = ! = nn((nn-1)(-1)(nn-2)…(3)(2)(1)-2)…(3)(2)(1)

note that 0!, by definition, is 1note that 0!, by definition, is 1..0!=10!=1

Permutations of Permutations of nn Things Taken Things Taken rr at a at a TimeTime

The number of permutations possible if The number of permutations possible if rr items are taken from items are taken from nn items: items:

nnPPrr = = = = nn((nn – 1) ( – 1) (nn – 2) ( – 2) (nn – 3) . . . ( – 3) . . . (nn – – rr + 1) + 1)

nn!!

((nn – – rr)!)!

nn! = ! = nn((nn – 1) ( – 1) (nn – 2) ( – 2) (nn – 3) . . . ( – 3) . . . (nn – – rr + 1) ( + 1) (nn - - rr) () (nn - - rr -- 1) . . . (2)(1) 1) . . . (2)(1)

((nn – – rr)! )! = = ((nn - - rr) () (nn - - rr -- 1) . . . (2)(1) 1) . . . (2)(1)

Permutations of Permutations of nn Things Taken Things Taken rr at a at a TimeTime

The number of permutations possible if The number of permutations possible if rr items are taken from items are taken from nn items: items:

nnPPrr: starting at : starting at nn, write down , write down rr numbers numbers

going down by one:going down by one:

nnPPrr = = nn((nn – 1) ( – 1) (nn – 2) ( – 2) (nn – 3) . . . ( – 3) . . . (nn – – rr + 1) + 1)

11 22 33 44 rr

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

We are arranging 5 out of 8 people into the five distinct We are arranging 5 out of 8 people into the five distinct offices. Any of the eight can be president. Once selected, offices. Any of the eight can be president. Once selected, any of the remaining seven can be vice-president.any of the remaining seven can be vice-president.

Clearly this is an arrangement, or permutation, problem.Clearly this is an arrangement, or permutation, problem.

88PP55 = 8!/(8-5)! = 8!/3! = 8 = 8!/(8-5)! = 8!/3! = 8 · 7 · 7 · 6 · 5 · 4 = 6720· 6 · 5 · 4 = 6720

Permutations with duplicates.Permutations with duplicates.

• In how many ways can you arrange the In how many ways can you arrange the letters of the word letters of the word mintyminty??

• That's 5 letters that have to be arranged, so That's 5 letters that have to be arranged, so the answer is the answer is 55PP55 = 5! = 120 = 5! = 120

• But how many ways can you arrange the But how many ways can you arrange the letters of the word letters of the word messesmesses??

• You would think 6!, but you'd be wrong!You would think 6!, but you'd be wrong!

messesmesses

mm ee ss ss ee ssmm ee ss ss ee ss

mm ee ss ss ee ss

mm ee ss ss ee ss

mm ee ss ss ee ss

mm ee ss ss ee ss

11

22

33

44

55

66

here are six permutations of here are six permutations of messesmesses

well, all 3! arrangements of the well, all 3! arrangements of the ss's 's look the same to me!!!!look the same to me!!!!

This is true for any arrangement This is true for any arrangement of the six letters in of the six letters in messesmesses, , so so every six permutations should every six permutations should count only once.count only once.

The same applies for the 2! The same applies for the 2! arrangement of the arrangement of the ee's's


• How many ways can you arrange the letters How many ways can you arrange the letters of the word of the word messesmesses??

• The problem is that there are three The problem is that there are three ss's and 2 's and 2 ee's. It doesn't matter in which order the 's. It doesn't matter in which order the ss's 's are placed, because they all look the same!are placed, because they all look the same!

• This is called permutations with duplicates.This is called permutations with duplicates.


• Since there are 3! = 6 ways to arrange the Since there are 3! = 6 ways to arrange the ss's, there are 6 permutations that should 's, there are 6 permutations that should count as one. Same with the count as one. Same with the ee's. There are 's. There are 2! = 2 permutations of them that should 2! = 2 permutations of them that should count as 1.count as 1.

• So we divide 6! by 3! and also by 2!So we divide 6! by 3! and also by 2!• There are 6!/3!2! = 720/12 = 60 ways to There are 6!/3!2! = 720/12 = 60 ways to

arrange the word arrange the word messesmesses. .

1 2 3

nm m m

!! ! !


• In general if we want to arrange In general if we want to arrange nn items, of which items, of which mm11, m, m22,, .... .... are identical, the number of are identical, the number of

permutations is permutations is

1 2 3

nm m m

!! ! !

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

6!/3!2!1! = 720/(6)(2)(1) = 720/12 = 606!/3!2!1! = 720/(6)(2)(1) = 720/12 = 60



CombinationsCombinations

Combination: definitionCombination: definition

A A combinationcombination of items occurs when: of items occurs when: • The item are selected from the same The item are selected from the same

group.group.

• No item is used more than once.No item is used more than once.

• The order of the items makes no The order of the items makes no difference.difference.

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

combination problemcombination problem• Permutation: Permutation:

– arrangement, arrange arrangement, arrange – order mattersorder matters

• Combination Combination – selection, selectselection, select– order does not matter.order does not matter.

Example: Distinguishing between Example: Distinguishing between Permutations and CombinationsPermutations and Combinations

• For each of the following problems, explain if the For each of the following problems, explain if the problem is one involving permutations or problem is one involving permutations or combinations.combinations.

• Six students are running for student government Six students are running for student government president, vice-president, and treasurer. The student president, vice-president, and treasurer. The student with the greatest number of votes becomes the with the greatest number of votes becomes the president, the second biggest vote-getter becomes president, the second biggest vote-getter becomes vice-president, and the student who gets the third vice-president, and the student who gets the third largest number of votes will be student government largest number of votes will be student government treasurer. How many different outcomes are treasurer. How many different outcomes are possible for these three positions?possible for these three positions?

SolutionSolution

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


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

SolutionSolution

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


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

SolutionSolution

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

Combinations of Combinations of nn Things Taken Things Taken rr at a at a TimeTime

nn!!rr!(!(nn – – rr)!)!

nnrr

= = nnCCrr = =

Note that the sum of the two numbers on the bottom Note that the sum of the two numbers on the bottom (denominator) should add up to the number on the (denominator) should add up to the number on the top (numerator).top (numerator).

C9 39! 9!

3!(9 3)! 3!6!

Computing CombinationsComputing Combinations

• Suppose we need to compute Suppose we need to compute 99CC33

• rr = 3, = 3, n – rn – r = 6 = 6• The denominator is the factorial of smaller of The denominator is the factorial of smaller of

the two: the two: 3!3!

C9 39! 9!

3!(9 3)! 3!6!

C9 39! 9!

3!(9 3)! 3!6!


• Suppose we need to compute Suppose we need to compute 99CC33

• r r = 3, = 3, n – rn – r = 6 = 6• In the numerator write (the product of) all the In the numerator write (the product of) all the

numbers from 9 down to numbers from 9 down to n - rn - r + 1 = 6 + 1 = 7 + 1 = 6 + 1 = 7::• There should be the same number of terms in There should be the same number of terms in

the numerator and denominator: the numerator and denominator: 9 9 8 8 7 7

C9 39! 9!

3!(9 3)! 3!6!


• If called upon, there's a fairly easy way to If called upon, there's a fairly easy way to compute combinations.compute combinations.

– Given Given nnCCr r , decide which is bigger: , decide which is bigger: rr or or n – rn – r..

– Take the smaller of the two and write out the Take the smaller of the two and write out the factorial (of the number you picked) as a factorial (of the number you picked) as a product.product.

– Draw a line over the expression you just wrote.Draw a line over the expression you just wrote.


• If called upon, there's a fairly easy way to If called upon, there's a fairly easy way to compute combinations.compute combinations.– Now, put Now, put nn directly above the line and directly directly above the line and directly

above the leftmost number below.above the leftmost number below.– Eliminate common factors in the numerator and Eliminate common factors in the numerator and

denominator.denominator.– Do the remaining multiplications.Do the remaining multiplications.– You're done!You're done!


• Suppose we need to compute Suppose we need to compute 99CC3 3 ..

– n – rn – r = 6= 6, and the, and the smaller ofsmaller of 3 3 andand 6 6 isis 3 3. .

3 2 1

9 8 73

1

4

1

= 3 4 7 = 84

ProblemProblem

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

We saw that this is a combination problemWe saw that this is a combination problem

C8 38! 8 7 6

3!5! 3 2 1 = 8 7 = 56

2

A deck of cardsA deck of cards

face cardsface cards

ranksranks

ssuuiittss

acesaces

ProblemProblem

• How many poker hands (five cards) are How many poker hands (five cards) are possible using a standard deck of 52 cards?possible using a standard deck of 52 cards?

• We need to pick any five cards from the We need to pick any five cards from the deck. Therefore we are selecting 5 out of deck. Therefore we are selecting 5 out of 52 cards.52 cards.

C52 552 51 50 49 48 52 51 5 49 48 52 51 5 49 4 2,598,960

5 4 3 2 1 4 3 1

ProblemProblem• A poker hand of aA poker hand of a full housefull house consists of three consists of three

cards of the same rank (e.g. three 8's or three kings cards of the same rank (e.g. three 8's or three kings or three aces.) and two cards of another rank. or three aces.) and two cards of another rank. How many full houses are possible?How many full houses are possible?

• First we need to select 1 out of the 13 possible First we need to select 1 out of the 13 possible ranks. ranks. 1313

• Then we need to select 3 out of the 4 cards of that Then we need to select 3 out of the 4 cards of that rank. rank. 44CC33 = 4 = 4

• By the fundamental counting principle, there are By the fundamental counting principle, there are 13 13 4 = 52 4 = 52 ways to do this part of the problem. ways to do this part of the problem.

ProblemProblem

• Similarly, we then need to select 1 out of Similarly, we then need to select 1 out of the 12 remaining ranks. the 12 remaining ranks. 1212

• Then select 2 out of the 4 cards of that rank. Then select 2 out of the 4 cards of that rank.

44CC22 = 4!/2!2! = 24/4 = 6 = 4!/2!2! = 24/4 = 6

• Once again by the fundamental counting Once again by the fundamental counting principle, there are principle, there are 12 12 6 = 72 6 = 72 ways to ways to solve this part of the problem.solve this part of the problem.

ProblemProblem

• Finally, in order to select a full house, we Finally, in order to select a full house, we merely have to choose any of the 52 merely have to choose any of the 52 possible threes of a kind and choose any of possible threes of a kind and choose any of the remaining 72 pairs.the remaining 72 pairs.

• Therefore there are Therefore there are 52 52 72 = 3744 72 = 3744 full full houses!!!houses!!!

Another ProblemAnother Problem• In poker, a In poker, a straightstraight is is

– 5 cards 5 cards in sequencein sequence, for example , for example • 3, 4, 5, 6 and a 7 3, 4, 5, 6 and a 7

• 8, 9, 10, jack, queen 8, 9, 10, jack, queen

• ace, 2, 3, 4, 5 (ace as low card)ace, 2, 3, 4, 5 (ace as low card)

• 10, jack, queen, king, ace. (ace as high card)10, jack, queen, king, ace. (ace as high card)

– and not all of the same suit (that's a straight flush!)and not all of the same suit (that's a straight flush!)

• How many possible straights can you be How many possible straights can you be dealt?dealt?

• Think about it. I won't give the answer now.Think about it. I won't give the answer now.

Another ProblemAnother Problem• One part of the solution is straightforward. One part of the solution is straightforward.

In how many ways can you select five cards in In how many ways can you select five cards in sequence?sequence?– the low card of the sequence can not be greater the low card of the sequence can not be greater

than 10. Getting a 10 as low card means that the than 10. Getting a 10 as low card means that the high card is an ace. You can't go higher than ace!high card is an ace. You can't go higher than ace!

– there are 4 possible suits for each card. there are 4 possible suits for each card.

So there are So there are 10 10 1024 = 10,240 1024 = 10,240 ways you can do ways you can do this. this.

4 4 4 4 4 4 4 4 4 =1024 4 =1024

Another ProblemAnother Problem• But some of those include the ones where all But some of those include the ones where all

the cards are of the same suit.the cards are of the same suit.• Remember, there are Remember, there are 1010 possible low cards possible low cards• There are There are 44 possible suits for that low card possible suits for that low card• Once we have chosen one of those Once we have chosen one of those 4040 possible possible

cards, there is exactly one possibility for the cards, there is exactly one possibility for the next four cards.next four cards.

• 10,240 – 40 = 10,20010,240 – 40 = 10,200• That's the answerThat's the answer



Fundamentals of ProbabilityFundamentals of Probability

Computing Theoretical Computing Theoretical ProbabilityProbability

number of outcomes in event EP E

number of outcomes in sample space S( )

RememberRemember

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



Example: Computing Theoretical Example: Computing Theoretical ProbabilityProbability

A die is rolled once. Find the probability of A die is rolled once. Find the probability of getting a number less than getting a number less than 55..

Sample space: all possible outcomes: Sample space: all possible outcomes: 1, 2, 3, 1, 2, 3, 4, 5, 64, 5, 6

Event: the roll yields a number less than 5: Event: the roll yields a number less than 5: 1, 1, 2, 3, 42, 3, 4

SolutionSolution

The event of getting a number less than 5 can occur The event of getting a number less than 5 can occur in 4 ways: in 4 ways: 1, 2, 3, 4.1, 2, 3, 4.

PP(less than 5) =(less than 5) =

= 4/6 = 2/3= 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 Example: Probability and a Deck of 52 CardsCards

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

SolutionSolution

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

PP(King) = 4/52 = 1/13(King) = 4/52 = 1/13

Empirical ProbabilityEmpirical Probability

observed number of times E occursP E

total number of observed occurrences( )

Example: Computing Empirical Example: Computing Empirical ProbabilityProbability

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

SolutionSolution

The probability of selecting a Catholic is the The probability of selecting a Catholic is the observed number of Arab Americans who are observed number of Arab Americans who are Catholic, 1.26 (million), divided by the total Catholic, 1.26 (million), divided by the total number of Arab Americans, 3 (million).number of Arab Americans, 3 (million).

PP(selecting a Catholic from the Arab (selecting a Catholic from the Arab American Population) = 1.26/3 = 0.42American Population) = 1.26/3 = 0.42


Section 5:Section 5:Probability with the Fundamental Probability with the Fundamental

Counting Principle, PermutationsCounting Principle, Permutations, , and and CombinationsCombinations

Example: Probability and Example: Probability and PermutationsPermutations

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

SolutionSolution

We begin by applying the definition of We begin by applying the definition of probability to this situation.probability to this situation.

PP(Rolling Stones fourth, Beatles last) =(Rolling Stones fourth, Beatles last) =

We can use the Fundamental Counting Principle to We can use the Fundamental Counting Principle to find the total number of possible permutations.find the total number of possible permutations.

5 5 4 4 3 3 2 2 1 = 120 1 = 120

(permutations with Rolling Stones fourth, Beatles last)(total number of possible permutations)

Solution cont.Solution cont.We can also use the Fundamental Counting Principle to We can also use the Fundamental Counting Principle to

find the number of permutations with the Rolling find the number of permutations with the Rolling Stones performing fourth and the Beatles performing Stones performing fourth and the Beatles performing last. You can choose any one of the three groups as last. You can choose any one of the three groups as the opening act. This leaves two choices for the the opening act. This leaves two choices for the second group to perform, and only one choice for the second group to perform, and only one choice for the third group to perform. Then we have one choice for third group to perform. Then we have one choice for fourth and last.fourth and last.

3 3 2 2 1 1 1 1 1 = 6 1 = 6There are six lineups with Rolling Stones fourth and There are six lineups with Rolling Stones fourth and

Beatles last.Beatles last.

Solution cont.Solution cont.

Now we can return to our probability fraction.Now we can return to our probability fraction.PP(Rolling Stones fourth, Beatles last) =(Rolling Stones fourth, Beatles last) =

= 6/120 = 1/20= 6/120 = 1/20The probability of the Rolling Stones performing The probability of the Rolling Stones performing

fourth and the Beatles last is 1/20.fourth and the Beatles last is 1/20.

(permutations with Rolling Stones fourth, Beatles last)(total number of possible permutations)

Example: Probability and Example: Probability and CombinationsCombinations

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

a.a. three men.three men.

b.b. one man and two women.one man and two women.

SolutionSolution

We begin with the probability of selecting three We begin with the probability of selecting three men. men.

PP(3 men)=(3 men)=

1212CC33 = 12!/((12-3)!3!) = 220 = 12!/((12-3)!3!) = 220

55CC33 = 5!/((5-3)!(3!)) = 10 = 5!/((5-3)!(3!)) = 10

PP(3 men) = 10/220 = 1/22(3 men) = 10/220 = 1/22

number of ways of selecting 3 mentotal number of possible combinations

Solution part b cont.Solution part b cont.There are 5 men. We can select 1 man in There are 5 men. We can select 1 man in 55CC11 waysways..

There are 7 women. We can select 2 women in There are 7 women. We can select 2 women in 77CC22 ways.ways.

By the Fundamental Counting Principle, the number of By the Fundamental Counting Principle, the number of ways of selecting 1 man and 2 women isways of selecting 1 man and 2 women is

55CC11 77CC22 = 5 = 5 21 = 105 21 = 105

Now we can fill in the numbers in our probability fraction. Now we can fill in the numbers in our probability fraction. PP(1 man and 2 women) = (1 man and 2 women) =

= 105/220 = 21/44= 105/220 = 21/44

number of ways of selecting 1 man and 2 womentotal number of possible combinations



Events Involving Not and Or; OddsEvents Involving Not and Or; Odds

The Probability of an Event Not The Probability of an Event Not OccurringOccurring

The probability that an event The probability that an event EE will not occur will not occur is equal to one minus the probability that it is equal to one minus the probability that it will occur.will occur.

PP(not (not EE) = 1 - ) = 1 - PP((EE))

Mutually Exclusive EventsMutually Exclusive Events

If it is impossible for events If it is impossible for events AA and and BB to occur to occur simultaneously, the events are said to be simultaneously, the events are said to be mutually exclusivemutually exclusive..

If If AA and and BB are mutually exclusive events, are mutually exclusive events, thenthen

PP((AA or or BB) = ) = PP((AA) + ) + PP((BB).).

Or: Probabilities with Events That Or: Probabilities with Events That Are Not Mutually ExclusiveAre Not Mutually Exclusive

If If AA and and BB are not mutually exclusive events, are not mutually exclusive events, thenthen

PP((AA or or BB) = ) = PP((AA) + ) + PP((BB) - ) - PP((AA and and BB))

ExamplesExamplesIn picking a card from a standard deck,In picking a card from a standard deck,

1. What is the probability of picking a 1. What is the probability of picking a red Kingred King??

2. What is the probability of 2. What is the probability of notnot picking a picking a redred KingKing??

3. What is the probability of picking a 3. What is the probability of picking a HeartHeart or a or a faceface cardcard (a face card is either a (a face card is either a JackJack, , QueenQueen or or KingKing)?)?

ExampleExampleIn picking a card from a standard deck,In picking a card from a standard deck,

1. What is the probability of picking a 1. What is the probability of picking a red Kingred King??

A A red Kingred King is either a is either a KingKing ofof HeartsHearts or a or a KingKing ofof DiamondsDiamonds. . Since a card can't be both a Since a card can't be both a HeartHeart and a and a DiamondDiamond, the events , the events are are mutually exclusive:mutually exclusive:

the probability is 1/52 + 1/52 = 1/26.the probability is 1/52 + 1/52 = 1/26.


2. What is the probability of 2. What is the probability of notnot picking a picking a redred KingKing??

Since P(Since P(notnot AA) = 1 – P() = 1 – P(AA) and we saw that P() and we saw that P(AA)) = 1/26,= 1/26,

the probability is 1 – 1/26 = 25/26. the probability is 1 – 1/26 = 25/26.


3. What is the probability of picking a 3. What is the probability of picking a HeartHeart or a or a faceface cardcard (a (a face card is either a face card is either a JackJack, , QueenQueen or or KingKing)?)?

Since some Since some HeartsHearts are face cards, the events are are face cards, the events are notnot mutually mutually exclusive.exclusive.

There are 13 There are 13 HeartsHearts, so P(, so P(HeartHeart) = 13/52. ) = 13/52.

There are 12 face cards, so P(There are 12 face cards, so P(FaceFace CardCard) = 12/52. ) = 12/52.

There are 3 cards that are both a There are 3 cards that are both a HeartHeart and a and a FaceFace CardCard. . (Namely the (Namely the JackJack, , QueenQueen and and King King ofof Hearts Hearts).).

P(P(HeartHeart oror FaceFace CardCard) = 13/52 + 12/52 – 3/52 = 22/52) = 13/52 + 12/52 – 3/52 = 22/52

OddsOdds• For some given event, we sometimes express For some given event, we sometimes express

the chances for an outcome by the chances for an outcome by oddsodds..– The odds in favor of something happening is the The odds in favor of something happening is the

ratio of the probability that it will happen to the ratio of the probability that it will happen to the probability that it won't.probability that it won't.

– Odds in favor of EOdds in favor of E = =

– When the When the OddsOdds in favor ofin favor of EE are are aa to to bb, we write it , we write it as as a:ba:b..

P E P EP not E P E

( ) ( )( ) 1- ( )

OddsOdds

• Just as we can talk about the odds in favor Just as we can talk about the odds in favor of something happening, we can also talk of something happening, we can also talk about the odds about the odds againstagainst something something happening:happening:

– Odds against EOdds against E = =

– When the When the OddsOdds against Eagainst E are are bb to to aa, we write , we write it as it as b:ab:a..

P not E P EP E P E( ) 1 ( )

( ) ( )

OddsOdds

• Suppose you roll a pair of dice. Suppose you roll a pair of dice. – There are There are 6 6 6 = 36 6 = 36 possible outcomes. possible outcomes.– There are There are threethree ways of rolling an 11 or higher: ways of rolling an 11 or higher:

a 5 and a 6, a 6 and a 5, and two 6's.a 5 and a 6, a 6 and a 5, and two 6's.– The probability of rolling an 11 or higher is The probability of rolling an 11 or higher is , ,

or .or .– The probability of not rolling an 11 or higher is The probability of not rolling an 11 or higher is

1 1 –– , or . , or .

336

1112

112

112

OddsOdds

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

– The The odds against rolling an 11 or betterodds against rolling an 11 or better are are

or 1

121112

1 1 :1111

or 11121

12

11 11 :11

Finding Probabilities from Finding Probabilities from OddsOdds

• If the odds in favor of an event If the odds in favor of an event EE are are aa to to bb, , then the probability of the event is given bythen the probability of the event is given by

aP Ea b

( )


• If the odds against an event If the odds against an event EE are are aa to to bb, , then the probability of the event is given then the probability of the event is given byby

bP Ea b

( )


• ExampleExample: : – Suppose Bluebell is listed as Suppose Bluebell is listed as 7:17:1 in the third in the third

race at the Meadowlands.race at the Meadowlands.– The odds listed on a horse are odds The odds listed on a horse are odds againstagainst

that horse winning, that is, that horse winning, that is, losinglosing..– The probability of him losing is The probability of him losing is

7 / (7+1) = 7/87 / (7+1) = 7/8..– The probability of him winning is The probability of him winning is 1/81/8..


• ExampleExample: : – Suppose Bluebell is listed as Suppose Bluebell is listed as 7:17:1 in the third in the third

race at the Meadowlands. (race at the Meadowlands. (aa::bb againstagainst))– The odds listed on a horse are odds The odds listed on a horse are odds againstagainst that that

horse winning, that is, horse winning, that is, losinglosing..– The probability of him losing is The probability of him losing is

7 / (7+1) = 7/87 / (7+1) = 7/8. . – The probability of him winning is The probability of him winning is 1/81/8. .

b

a bb

a b

a

a ba

a b



Events Involving And; Events Involving And; Conditional ProbabilityConditional Probability

Independent EventsIndependent Events

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

• For example, if you roll a pair of dice two For example, if you roll a pair of dice two times, then the two events are independent. times, then the two events are independent. What gets rolled on the second throw is not What gets rolled on the second throw is not affected by what happened on the first affected by what happened on the first throw.throw.

AndAnd Probabilities with Probabilities with Independent EventsIndependent Events

• If If AA and and BB are independent events, then are independent events, then

PP((A A andand B B)) = P = P((AA) ) P P((BB))

• The example of choosing from four pairs of The example of choosing from four pairs of socks and then choosing from three pairs of socks and then choosing from three pairs of shoes (= 12 possible combinations) is an shoes (= 12 possible combinations) is an example of two independent events.example of two independent events.

Dependent EventsDependent Events

• Two events are Two events are dependent eventsdependent events if the occurrence if the occurrence of one of them of one of them does does have an effect on the have an effect on the probability of the other.probability of the other.

• Selecting two Kings from a deck of cards by Selecting two Kings from a deck of cards by selecting one card, putting it aside, and then selecting one card, putting it aside, and then selecting a second card, is an example of two selecting a second card, is an example of two dependent events.dependent events.

• The probability of picking a King on the second The probability of picking a King on the second selection changes because the deck now contains selection changes because the deck now contains only 51, not 52, cards.only 51, not 52, cards.

AndAnd Probabilities with Probabilities with Dependent EventsDependent Events

• If If AA and and B B are dependent events, thenare dependent events, then

• PP((A and BA and B)) = = P P((AA)) PP((B B given thatgiven that A A has occurred)has occurred)

• written aswritten as PP((AA)) PP((BB||AA))

Conditional ProbabilityConditional Probability

• The The conditional probabilityconditional probability of of BB, given , given AA, , written written PP((B|AB|A)),, is the probability that event is the probability that event BB will occur computed on the assumption will occur computed on the assumption that event that event AA has occurred. has occurred.

• Notice that when the two events are Notice that when the two events are independent, independent, PP((B|AB|A)) == PP((BB))..

Conditional ProbabilityConditional Probability• Example:Example:

– Suppose you are picking two cards from a deck of cards. Suppose you are picking two cards from a deck of cards. What is the probability you will pick a What is the probability you will pick a KingKing and then and then another another face cardface card??

– The probability of an The probability of an KingKing is = . is = .

– Once the Once the KingKing is selected, there are 11 face cards left in a is selected, there are 11 face cards left in a deck holding 51 cards.deck holding 51 cards.

– PP((AA)) = . = . PP((B|AB|A)) = =

– The probability in question is The probability in question is

452

113

113

1151

1151

113

Applying Conditional Applying Conditional Probability to Real-World DataProbability to Real-World Data

PP((B|AB|A)) = =

observed number of timesobserved number of times B B andand A A occur togetheroccur togetherobserved number of timesobserved number of times A A occurs occurs

ReviewReview

PP(not (not EE) ) 1 – 1 – PP((EE))

PP((AA or or BB)) PP((AA) + ) + PP((BB) ) – – PP((AA and and BB))

mutually mutually exclusiveexclusive: : PP((AA) + ) + PP((BB))

PP((AA and and BB)) PP((AA) ) PP((B|AB|A)) independentindependent::PP((AA) ) PP((BB))

Odds in favor - Odds in favor - aa::bb

PP((EE) / ) / PP(not (not EE)) probability is probability is a/a/((a+ba+b))

Odds against - Odds against - aa::bb

PP(not (not EE) / P() / P(EE)) probability is probability is b/b/((a+ba+b))



Expected ValueExpected Value

Expected ValueExpected Value

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

• For example, we can use expected value to find the For example, we can use expected value to find the outcomes of the roll of a fair dice.outcomes of the roll of a fair dice.

• The outcomes are 1, 2, 3, 4, 5, and 6, each with a The outcomes are 1, 2, 3, 4, 5, and 6, each with a probability of . The expected value, probability of . The expected value, EE, is computed , is computed by multiplying each outcome by its probability and by multiplying each outcome by its probability and then adding these products.then adding these products.

• EE = 1= 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + 6 + 6 = (1+2+3+4+5+6)/6 = = 3.5= (1+2+3+4+5+6)/6 = = 3.5

16

16

16

16

16

16

216

16

ExpectedExpected ValueValue

EE = 1 = 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + 6 + 6= (1 + 2 + 3 + 4 + 5 + 6)/6 = = 3.5= (1 + 2 + 3 + 4 + 5 + 6)/6 = = 3.5

Of course, you can't roll a 3½ . But the Of course, you can't roll a 3½ . But the average value of a roll of a die over a long average value of a roll of a die over a long period of time will be around 3½. period of time will be around 3½.

216

16

16

16

16

16

16

ExampleExample ExpectedExpected ValueValue andand RouletteRoulette

A roulette wheel has 38 different A roulette wheel has 38 different "numbers.""numbers."

• One way to bet in roulette is to place $1 One way to bet in roulette is to place $1 on a single number. on a single number.

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

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

ExampleExample ExpectedExpected ValueValue andand RouletteRoulette

• 38 different numbers.38 different numbers.• If the ball lands on your number, If the ball lands on your number,

you win awarded $35 and you keep you win awarded $35 and you keep the $1 you paid to play the game. the $1 you paid to play the game.

• If the ball lands on any of the other If the ball lands on any of the other 37 slots, you are awarded nothing 37 slots, you are awarded nothing and you lose the $1 you bet.and you lose the $1 you bet.

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

SolutionSolution

EE = $35( ) + (-$1)( ) = $35( ) + (-$1)( )

= $ - $ = -$ = $ - $ = -$ ≈≈ -$0.05 -$0.05This means that in the long run, a player can This means that in the long run, a player can expect to lose about 5 cents for each game expect to lose about 5 cents for each game played. played.

35$37381$

138

OutcomeOutcome Gain/LossGain/Loss ProbabilityProbability

1111

Not 11Not 11

38

3738

1

38

2

38

37

38

35


• A real estate agent is selling a house. She gets a 4-A real estate agent is selling a house. She gets a 4-month listing. There are 3 possibilities:month listing. There are 3 possibilities:– she sells the house: she sells the house: (30% chance) earns $25,000(30% chance) earns $25,000– another agent sellsanother agent sells

the house:the house: (20% chance) earns $10,000(20% chance) earns $10,000– house not sold:house not sold: (50% chance) loses $5,000(50% chance) loses $5,000

• What is the expected profit (or loss)?What is the expected profit (or loss)?

• If the expected profit is at least $6000 she would If the expected profit is at least $6000 she would consider it a good dealconsider it a good deal..


OutcomeOutcome ProbabilityProbability Profit or Profit or lossloss

productproduct

she sellsshe sells 0.30.3 +$25,000+$25,000

other sellsother sells 0.20.2 +$10,000+$10,000

doesn't selldoesn't sell 0.50.5 -$5,000-$5,000

+$7,500

-$2,500

+$2,000

+$7,000

The realtor can expect to make $7,000.The realtor can expect to make $7,000.

Make the deal!!!!Make the deal!!!!

Documents

Thinking Mathematically Chapter 11: Counting Methods and Probability Theory