Probability and Counting Fundamentals Ginger Holmes Rowell, Middle TN State University MSP Workshop June 2006

ProbabilityProbabilityand and

Counting FundamentalsCounting Fundamentals

Ginger Holmes Rowell, Ginger Holmes Rowell, Middle TN State UniversityMiddle TN State University

MSP WorkshopMSP WorkshopJune 2006June 2006

OverviewOverview

• Probability IntroductionProbability Introduction

• Fundamentals of CountingFundamentals of Counting• Permutations: Permutations: ordered arrangementsordered arrangements• Combinations: Combinations: unordered unordered

arrangementsarrangements

• Selected ActivitiesSelected Activities

Probability ReviewProbability Review

• DefinitionsDefinitions

• Classical ProbabilityClassical Probability

• Relative Frequency ProbabilityRelative Frequency Probability

• Probability Fundamentals andProbability Fundamentals and Probability Rules Probability Rules

What is Probability?What is Probability?

• ProbabilityProbability the study of chance associated the study of chance associated

with the occurrence of eventswith the occurrence of events

• Types of ProbabilityTypes of Probability• Classical (Theoretical)Classical (Theoretical)• Relative Frequency (Experimental)Relative Frequency (Experimental)

Classical ProbabilityClassical Probability

Rolling dice and tossing a coin are Rolling dice and tossing a coin are activities associated with a activities associated with a classical approach to probability. classical approach to probability. In these cases, you can list all the In these cases, you can list all the possible outcomes of an possible outcomes of an experiment and determine the experiment and determine the actual probabilities of each actual probabilities of each outcome. outcome.

Listing Listing All Possible OutcomesAll Possible Outcomes of a Probabilistic Experiment of a Probabilistic Experiment

• There are various ways to list all There are various ways to list all possible outcomes of an possible outcomes of an experiment experiment • EnumerationEnumeration• Tree diagramsTree diagrams• Additional methods – counting Additional methods – counting

fundamentalsfundamentals

Three Children ExampleThree Children Example

• A couple wants to have exactly 3 A couple wants to have exactly 3 children. Assume that each children. Assume that each child is either a boy or a girl and child is either a boy or a girl and that each is a single birth. that each is a single birth.

• List all possible orderings for List all possible orderings for the 3 children.the 3 children.

EnumerationEnumeration11stst Child Child 22ndnd Child Child 33rdrd Child Child

EnumerationEnumeration11stst Child Child 22ndnd Child Child 33rdrd Child Child

BB BB BB

GG BB BB

BB GG BB

BB BB GG

GG GG BB

GG BB GG

BB GG GG

GG GG GG

Tree DiagramsTree Diagrams

1st Child 2nd Child 3rd Child BBBBBB

BB

G

B

GB

G

BBGBBGBGBBGBBGGBGGGBBGBBGBGGBGGGBGGBGGGGGG

GB

G

B

GB

G

DefinitionsDefinitions

• Sample SpaceSample Space - the list of all - the list of all possible outcomes from a possible outcomes from a probabilistic experiment. probabilistic experiment. • 3-Children Example:3-Children Example:

S = {BBB, BBG, BGB, BGG, S = {BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG} GBB, GBG, GGB, GGG}

• Each individual item in the list is called Each individual item in the list is called a a Simple EventSimple Event or or Single Event.Single Event.

Probability NotationProbability Notation

P(P(eventevent) = Probability of the ) = Probability of the eventevent occurringoccurring

Example: P(Boy) = P(B) = ½Example: P(Boy) = P(B) = ½

Probability of Single Events Probability of Single Events with Equally Likely Outcomeswith Equally Likely Outcomes

• If each outcome in the sample space If each outcome in the sample space is is equally likelyequally likely, then , then the probability of the probability of any one outcome is 1 divided by the any one outcome is 1 divided by the total number of outcomestotal number of outcomes..

outcomes ofnumber total

1event) simple(

outcomes,likely equally For

P

Three Children Example Three Children Example ContinuedContinued

• A couple wants 3 children. A couple wants 3 children. Assume the chance of a boy or Assume the chance of a boy or girl is girl is equally likelyequally likely at each birth. at each birth.

• What is the What is the probabilityprobability that they that they will have will have exactly 3 girlsexactly 3 girls? ?

• What is the What is the probabilityprobability of ofhaving having exactly 3 boysexactly 3 boys??

Probability of Combinations of Probability of Combinations of Single EventsSingle Events

• An An EventEvent can be a combination can be a combination of of Single EventsSingle Events..

• The probability of such an event The probability of such an event is the sum of the individual is the sum of the individual probabilities.probabilities.

Three Children Example Three Children Example ContinuedContinued

P(exactly 2 girls) = __P(exactly 2 girls) = __

P(exactly 2 boys) = __P(exactly 2 boys) = __

P(at least 2 boys) = __P(at least 2 boys) = __

P(at most 2 boys) = __P(at most 2 boys) = __

P(at least 1 girl) = __P(at least 1 girl) = __

P(at most 1 girl) = __P(at most 1 girl) = __

• Sample Sample space =space =

Types of ProbabilityTypes of Probability

• Classical (Theoretical)Classical (Theoretical)

• Relative Frequency Relative Frequency (Experimental, Empirical)(Experimental, Empirical)

Relative Frequency ProbabilityRelative Frequency Probability

• Uses actual experience to Uses actual experience to determine the likelihood of an determine the likelihood of an outcome.outcome.

• What isWhat isthe chancethe chanceof makingof makinga B or better?a B or better?

GradeGrade FrequencyFrequency

AA 2020

BB 3030

CC 4040

Below CBelow C 1010

Relative Frequency Probability Relative Frequency Probability is Great Fun for Teachingis Great Fun for Teaching

• Rolling DiceRolling Dice

• Flipping CoinsFlipping Coins

• Drawing from Bags without Looking Drawing from Bags without Looking (i.e. Sampling)(i.e. Sampling)

• Sampling with M&M's Sampling with M&M's ((http://mms.com/cai/mms/faq.html#whhttp://mms.com/cai/mms/faq.html#what_percentat_percent))

Empirical ProbabilityEmpirical Probability

• Given a frequency distribution, Given a frequency distribution, the probability of an event, E, the probability of an event, E, being in a given group isbeing in a given group is

n

xP

ondistributi in the sfrequencie total

group theoffrequency E)(

Two-way Tables and Two-way Tables and ProbabilityProbability

• FindFind

P(M) P(M)

P(A)P(A)

P(A and M)P(A and M)

Made Made AA

Made Made

< A< ATotalTotal

MaleMale 3030 4545

FemaleFemale 6060 6565

TotalTotal

Teaching IdeaTeaching Idea

• Question: How Can You Win at Question: How Can You Win at Wheel of Fortune?Wheel of Fortune?

• Answer: Use Relative Frequency Answer: Use Relative Frequency Probability (see handout)Probability (see handout)

Source. Krulik and Rudnick. “Teaching Middle School Mathematics Activities, Materials and Problems.” p. 161. Allyn & Bacon, Boston. 2000.

Probability FundamentalsProbability Fundamentals

• What is What is wrongwrong with the statements? with the statements?• The probability of rain today is -10%.The probability of rain today is -10%.• The probability of rain today is 120%.The probability of rain today is 120%.• The probability of rain or no rain today is The probability of rain or no rain today is

90%.90%.

1) (

1)(

0)(

spacesampleP

eventP

eventP

Probability RulesProbability Rules

Let A and B be eventsLet A and B be events

Complement Rule:Complement Rule:

P(A) + P(not A) = 1P(A) + P(not A) = 1

Set NotationSet Notation

Union: A or B Union: A or B (inclusive “or”)(inclusive “or”)

BA

BA

Intersection: A and BIntersection: A and B

Probability RulesProbability Rules

Union P(AUB) = P(A or B)Union P(AUB) = P(A or B)

)()()()( BAPBPAPBAP

• Venn DiagramsVenn Diagrams

• Kyle Siegrist’s Venn Diagram Kyle Siegrist’s Venn Diagram AppletApplet

http://www.math.uah.edu/stat/applhttp://www.math.uah.edu/stat/applets/index.xmlets/index.xml

Teaching IdeaTeaching Idea

Two-way Tables and Two-way Tables and ProbabilityProbability

• FindFind

P(M)P(M)

P(A)P(A)

P(A and M)P(A and M)

P(A if M)P(A if M)

Made Made AA

Made Made

< A< A

TotalTotal

MaleMale 3030 4545 7575

FemaleFemale 6060 6565 125125

TotalTotal 9090 110110 200200

Conditional ProbabilityConditional Probability

P(A|B) = the conditional P(A|B) = the conditional probability of event A happening probability of event A happening given that event B has happenedgiven that event B has happened

“ “probability of A given B”probability of A given B”

)(

)()|(

BP

BAPBAP

IndependenceIndependence

• Events A and B are Events A and B are “Independent” if and only if“Independent” if and only if

)()|( APBAP

• Using the data in the two-way Using the data in the two-way table, is making an “A” table, is making an “A” independent from being male? independent from being male?

OverviewOverview

• Probability ReviewProbability Review

• Fundamentals of CountingFundamentals of Counting• Permutations: Permutations: ordered ordered

arrangementsarrangements• Combinations: Combinations: unordered unordered



Counting TechniquesCounting Techniques

• Fundamentals of CountingFundamentals of Counting

• PermutationsPermutations: : ordered arrangementsordered arrangements

• CombinationsCombinations: : unordered arrangementsunordered arrangements

Fundamentals of CountingFundamentals of Counting

• QQ: Jill has 9 shirts and 4 pairs of : Jill has 9 shirts and 4 pairs of pants. How many different pants. How many different outfits does she have?outfits does she have?

• AA: 9 x 4 = 36 : 9 x 4 = 36

36 different outfits36 different outfits


• Multiplication PrincipleMultiplication Principle:: If there are If there are aa ways of choosing one ways of choosing one

thing, and thing, and bb ways of choosing a ways of choosing a second thing after the first is chosen, second thing after the first is chosen, then the total number of choice then the total number of choice patterns is: patterns is:

aa x x bb


• QQ: 3 freshman, 4 sophomores, 5 : 3 freshman, 4 sophomores, 5 juniors, and 2 seniorsjuniors, and 2 seniors are running for are running for SGA representative. One individualSGA representative. One individual will be selected from each class. How will be selected from each class. How many different representative many different representative orderings are possible?orderings are possible?

• AA: 3 · 4 · 5 · 2 = : 3 · 4 · 5 · 2 = 120 different 120 different representative orderingsrepresentative orderings

Generalized Multiplication Generalized Multiplication PrinciplePrinciple

• If there are If there are aa ways of choosing one ways of choosing one thing, thing, bb ways of choosing a second ways of choosing a second thing after the first is chosen, and thing after the first is chosen, and cc ways of choosing a third thing after ways of choosing a third thing after the first two have been chosen…and the first two have been chosen…and zz ways of choosing the last item after ways of choosing the last item after the earlier choices, then the total the earlier choices, then the total number of choice patterns is number of choice patterns is

aa x x bb x x cc x … x x … x zz

ExampleExample

• Q: When I lived in Madison Co., Q: When I lived in Madison Co., AL, the license plates had 2 fixed AL, the license plates had 2 fixed numbers, 2 variable letters and 3 numbers, 2 variable letters and 3 variable numbers. How many variable numbers. How many different license plates were different license plates were possible?possible?

• A: 26 x 26 x 10 x 10 x 10 = A: 26 x 26 x 10 x 10 x 10 = 676,000 different plates676,000 different plates


Q: How many more license plate Q: How many more license plate numbers will Madison County numbers will Madison County gain by changing to 3 letters and gain by changing to 3 letters and 2 numbers?2 numbers?

A: 26 x 26 x 26 x 10 x 10 = A: 26 x 26 x 26 x 10 x 10 = 1,757,600 1,757,600

• 1,757,600 – 676,000 = 1,757,600 – 676,000 = 1,081,600 1,081,600 more license plate numbersmore license plate numbers

Permutations: Permutations: Ordered ArrangementsOrdered Arrangements

• Q: Given 6 people and 6 chairs in a line, Q: Given 6 people and 6 chairs in a line, how many seating arrangements how many seating arrangements (orderings) are possible?(orderings) are possible?

• A: 6 · 5 · 4 · 3 · 2 · 1 = 6! A: 6 · 5 · 4 · 3 · 2 · 1 = 6! = = 720 orderings720 orderings

Permutations: Permutations: Ordered ArrangementsOrdered Arrangements

• Q: Given 6 people and 4 chairs in a line, Q: Given 6 people and 4 chairs in a line, how many different orderings are how many different orderings are possible?possible?

• A: 6 · 5 · 4 · 3 = A: 6 · 5 · 4 · 3 = 360 different orderings360 different orderings

Permutations:Permutations:Ordered ArrangementsOrdered Arrangements

• Permutation of Permutation of nn objects taken objects taken rr at a time: at a time:

rr-permutation, P(-permutation, P(nn,,rr), ), nnPPrr

Q: Given 6 people and 5 chairs in a line, Q: Given 6 people and 5 chairs in a line, how many different orderings are how many different orderings are possible?possible?

A: 6 · 5 · 4 · 3 · 2 = A: 6 · 5 · 4 · 3 · 2 = 720 different orderings720 different orderings


nnPPr r = = nn((nn-1)···(-1)···(nn-(-(rr-1))-1)) = = nn((nn-1)···(-1)···(nn--rr+1)+1) = = nn((nn-1)···(-1)···(nn--rr+1) +1) ((nn--rr)!)! ((nn--rr)!)! = = nn((nn-1)···(-1)···(nn--rr+1)(+1)(nn--rr)···(3)(2)(1))···(3)(2)(1) ((nn--rr)!)! = = nn! ! ((nn--rr)!)!

)!(

!

rn

nPrn


• Q: How many different batting orders are Q: How many different batting orders are possible for a baseball team consisting possible for a baseball team consisting of 9 players?of 9 players?

• A: 9 · 8 · 7 ··· 3 · 2 · 1 = 9!A: 9 · 8 · 7 ··· 3 · 2 · 1 = 9! = = 362,880 batting orders362,880 batting orders

• Note: this is equivalent to Note: this is equivalent to 99PP99.. 99PP99 = = 9! 9! = = 9! 9! = 9! = 9! (9-9)! 0!(9-9)! 0!


• Q: How many different batting Q: How many different batting orders are possible for the leading orders are possible for the leading fourfour batters? batters?

• A: 9 · 8 · 7 · 6 = A: 9 · 8 · 7 · 6 = 3,024 orders 3,024 orders

99PP44 = = 9! 9! = = 9! 9! = = 9! 9!

(9-4)! 5! 5!(9-4)! 5! 5!

Permutations:Permutations:Indistinguishable ObjectsIndistinguishable Objects

• Q: How many different letter Q: How many different letter arrangements can be formed using the arrangements can be formed using the letters letters T E N N E S S E ET E N N E S S E E ? ?

• A: There are 9! Permutations of the A: There are 9! Permutations of the letters letters T E N N E S S E ET E N N E S S E E if the letters if the letters are are distinguishabledistinguishable..

• However, 4 E’s are iHowever, 4 E’s are indistinguishablendistinguishable. . There are 4! ways to order the E’s.There are 4! ways to order the E’s.

Permutations:Permutations:Indistinguishable Objects, Cont.Indistinguishable Objects, Cont.• 2 S’s and 2 N’s are 2 S’s and 2 N’s are indistinguishable. indistinguishable.

There are 2! orderings of each.There are 2! orderings of each.• Once all letters are ordered, there is Once all letters are ordered, there is

only one place for the T.only one place for the T.• If the E’s, N’s, & S’s are If the E’s, N’s, & S’s are

indistinguishableindistinguishable among themselves, among themselves, then there arethen there are

9! 9! = 3,780 different orderings of= 3,780 different orderings of (4!·2!·2!) (4!·2!·2!) T E N N E S S E ET E N N E S S E E

Permutations:Permutations:Indistinguishable ObjectsIndistinguishable Objects

Subsets of Indistinguishable ObjectsSubsets of Indistinguishable Objects

Given Given nn objects of which objects of which aa are alike, are alike, bb are alike, …, and are alike, …, and zz are alike are alike

There are There are nn!! permutations. permutations. a!·b!···z!a!·b!···z!

Combinations:Combinations:Unordered ArrangementsUnordered Arrangements

• Combinations:Combinations: number of different number of different groups of size groups of size rr that can be chosen that can be chosen from a set of from a set of nn objects (order is objects (order is irrelevant)irrelevant)

• Q: From a group of 6 people, select Q: From a group of 6 people, select 4. How many different possibilities 4. How many different possibilities are there?are there?

• A: There are A: There are 66PP44=360 different =360 different orderingsorderings of 4 people out of 6. of 4 people out of 6.

6·5·4·3 = 360 = 6·5·4·3 = 360 = 66PP44 = = n!

(n -r)!

Unordered Example continuedUnordered Example continued

• However the However the orderorder of the chosen 4 people is of the chosen 4 people is irrelevant. There are 24 different orderings irrelevant. There are 24 different orderings of 4 objects.of 4 objects.

4 · 3 · 2 · 1 = 24 = 4! =4 · 3 · 2 · 1 = 24 = 4! =rr!!

• Divide the Divide the total numbertotal number of orderings by the of orderings by the number of number of orderings of the 4 chosen orderings of the 4 chosen people.people. 360360 = = 15 different groups of 4 people15 different groups of 4 people..

2424

Combinations: Combinations: Unordered ArrangementsUnordered Arrangements

The number of ways to choose The number of ways to choose rr objects from a group of objects from a group of n n objectsobjects..

C(n,r)C(n,r) or or nnCCrr, read as “, read as “nn choose choose rr””

)!(!

!

rnr

n

r

nCrn


• Q: From a group of 20 people, a Q: From a group of 20 people, a committee of 3 is to be chosen. committee of 3 is to be chosen. How many different committees How many different committees are possible?are possible?

• A: A:

committeesdifferent 1140)!320(!3

!20

3

20320

C


• Q: From a group of 5 men & 7 women, Q: From a group of 5 men & 7 women, how many different committees of 2 how many different committees of 2 men & 3 women can be found?men & 3 women can be found?

• A: There are A: There are 55CC22 groups of men & groups of men & 77CC33

groups of women. Using the groups of women. Using the multiplication principle multiplication principle

women3 &men 2 of committees possible 350

35035*10)!37(!3

!7*

)!25(!2

!5

3

7*

2

5

ReviewReview


• Fundamentals of CountingFundamentals of Counting• Permutations: Permutations: ordered arrangementsordered arrangements• Combinations: Combinations: unordered unordered



Practice ProblemPractice Problem

• You have 30 students in your class, You have 30 students in your class, which will be arranged in 5 rows of which will be arranged in 5 rows of 6 people. Assume that any student 6 people. Assume that any student can sit in any seat. can sit in any seat. • How many different seating charts How many different seating charts

could you have for the first row?could you have for the first row?• How many different seating charts How many different seating charts

could you have for the whole class?could you have for the whole class?

It’s Your TurnIt’s Your Turn

• Make up three counting problems Make up three counting problems which would interest your students, which would interest your students, include one permutation and one include one permutation and one combination and one of your combination and one of your choice. choice.

• Calculate the answer for these Calculate the answer for these problems.problems.

OverviewOverview


• Fundamentals of CountingFundamentals of Counting• Permutations: Permutations: ordered arrangementsordered arrangements• Combinations: Combinations: unordered arrangementsunordered arrangements


HomeworkHomework

Documents

Probability and Counting Fundamentals Ginger Holmes Rowell, Middle TN State University MSP Workshop June 2006