3.13 Probability workbook 2018 - Mathematics with Ms Walker · 2018-02-04 · Theoretical...

3.13Probability

Name:

2

3.13Probability

Credits:4

Assessment:External

FormativeAssessmentdate:

Thisisanexternalexamination;youwillcompleteaformativeassessmentattheendofthisunit,amockexaminationinTerm3andtheactualpaperinTerm4NZQAexaminations.

Achievement AchievementwithMerit AchievementwithExcellence

• Applyprobabilityconceptsinsolvingproblems

• Applyprobabilityconceptsusingrelationalthinking,insolvingproblems.

• Applyprobabilityconcepts,usingextendedabstractthinking,insolvingproblems.

3

Introduction

Thisstandardwillrequireyoutoapplyingprobabilityconceptsinsolvingproblems.Thiswillinvolvetheuseof:

•trueprobabilityversusmodelestimatesversusexperimentalestimates•randomness•independence•mutuallyexclusiveevents•conditionalprobabilities•probabilitydistributiontablesandgraphs•twowaytables•probabilitytrees•Venndiagrams

WhatisProbability?

4

Thelanguageofprobability

Whatdothesetermsmean?

P(x>3)

P(x<4)

P(x=6)

P(x³2)

P(x£1)

P(3<x<7)

P(1£x£5)

Doesitmatterwhichwaythearrowsgointhelasttwo?

Page6-8

5

ProbabilityRevision

Probabilitiescanbeexpressedasaprobability,fractionorapercentage.

Weusuallyuseprobabilitiesbecausetheyareeasytocompare.

____________

Combiningprobabilities

Thisisveryimportantandfrequentlycomesupinexams.

and____x___

or____+____

6

=

=________

=________

pg11-12

7

MultiplicationPrinciple

1.Withdifferentitems:egJeremiahhas6shirts,2tiesand4pairsoftrousersInhowmanywayscanhebedressed?

2.Whereitemsaresimilar(andallitemsareavailableforpositions)butnorepeatsareallowed:

eg10peoplearerunningarace-inhowmanywayscanthefirst3placesbefilled.

3.Whereitemsaresimilar,butrepeatsareallowed.egnumberofNewZealandlicenseplates

FactorialsHowmanydifferentwayscan4peoplesitonabench?

Howmanydifferentwayscan12peoplesitonabench?

Challenge:Howmanydifferentwayscan4peoplesitaroundacirculartable?

Pg15-16

8

Methodsofdisplay

Howdowefindthisvalue?

Thetableneedstoaddto__________

What’sthedifferencebetweenaprobabilitytableandfrequencytable?________________________________________pg18-20

Howdowefindthisnumber?

9

ProbabilityNotation

P(A)

P(A’)

P(AÇB)

P(AÇB’)

P(A∪B)

P(A’∪B)

P(A∪B)’orP(A’∩B’)

10

2x2VennDiagrams

P(A)

P(A’)

P(A∩B)

P(A∪B)

P(A∪B)’orP(A’∩B’)

11

Onyourformulasheet

Questionsfor2x2VennorTwowaycontingencytables

Thedatamaybegiveninfractions,decimals,probabilitiesor

frequencies(ienumbersof).

A A’

B

B’

Checkthatalltheprobabilitiesaddto_____________

FindP(A∪B)=

12

A A’

B

B’

pg24-27

13

3x3VennDiagrams

pg30

14

3x3Frequencytable

pg31-35

15

Probabilities from Tables of Counts

pg37-39

16

TreeDiagrams

Thesecanbeusedasanalternativetovenndiagramsandtables,theyareparticularlyusefulwhenthereisasequenceofevents.

1. Firstidentifytheeventsareandwhichordertheygoin

vaccinatedandflu

Whichislikelytohappenfirst?vaccinated

2. Addtheeventsandtheprobabilitiestothetreediagrambelow

Nowanswerthequestions

pg42-48

17

ComplementaryEvents(AandA’)If A is an event, then A’ (not A) is the complementary event. If A doesn’thappenthenA’doeshappen.

P(A)+P(A’)=1

Example:Pickingaredcardandpickingablackcardfromapack(nojokers)

P(R)+P(B)=1

InGeneral:

P(A)+P(A')=P(AÈA')=1

P(AÇA')=0

Thinkofsomeexamplesofacomplementaryevents:

• _____________________________________________________

• _____________________________________________________

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Minustheintersection,howeverits0

MutuallyExclusiveEvents

Mutuallyexclusiveeventscannotbothoccur.InaVennDiagramthecirclesdon’toverlap.

P(A∪B)=P(A)+P(B)

Example:Heartsandblackcardsaremutuallyexclusive.

Findtheprobabilityofpickingaheartorablackcard.

P(H)+P(B)=¼+½=0.75

B B' H 0 .25 .25H' .5 .25 .75 .5 .5 1.0

InGeneral:

P(AÈB)=P(A)+P(B)

and

P(AÇB)=Ø

19

Onyourformulasheet

Thisboxwouldbe0formutuallyexclusiveevents

EventswhichareNOTMutuallyExclusive

EventswhichareNOTmutuallyexclusivehave

overlappingsets

A B

P(AÈB)=P(A)+P(B)–P(AÇB)

Youcan’tcounttheintersectiontwice.

Example:Theprobabilityofdrawingaheartora6fromapackofcards.

Hearts 6 of Hearts

Sixes

P(Hand6)=¼x1/13=0.019

6 6’ H 0.019 0.231 0.25H’ 0.058 0.692 0.75 0.077 0.923 1.0

InGeneral:

P(AÈB)=P(A)+P(B)–P(AÇB)

Complementaryeventsaremutuallyexclusive

Howevermutuallyexclusivearenotnecessarilycomplementary.

20

QuestionOne

Onaparticularday,P(rain)=0.27,P(heavywind)=0.24,P(neither)=0.64.

Aretheseeventsmutuallyexclusive?

I I' B .15 .09 .24B' .12 .64 .76 .27 .73 1.0

QuestionTwo

Theprobabilityofapersonenjoyinggardeningis0.43andtheprobabilityofapersonenjoyingplayingcomputergamesis0.32.Theprobabilitythatapersonenjoysneithergardeningnorcomputergamesis0.38.Aretheseeventsmutuallyexclusive?

QuestionThree

16peoplestudyFrench,21studySpanishandthereare30altogether.AretheeventsstudyingFrenchandSpanishmutuallyexclusive?

HowmanystudyonlySpanish?

QuestionFour

Inasampleof135shoppers,29boughtapples,34boughtcolaand72boughtneither.Aretheevents‘boughtapples’and‘boughtcola’mutuallyexclusive?

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ConditionalProbabilityP(A/B)

Conditional probabilities are those that are influenced by the occurrence or non-occurrence of previous events.P(A/B)=InwordsthismeanstheprobabilitythatAhasoccurredgiventhatB

hasoccurred

P(B/A)=InwordsthismeanstheprobabilitythatBhasoccurredgiventhatAhasoccurred

Theformulaforcalculatingthiscanbeobtainedfromaprobabilitytree:

P(BÇA)

Onyourformulasheet

23

pg55-58

24

Ifthisistruethetwoeventsareindependent

IfthisisthecasethenthetwoeventsareNOT

independent

IndependentEvents

Independenteventsarewhentheoccurrenceononehasnoeffectontheother.

P(AÇB)=P(A)xP(B)

P(AÇB)≠P(A)xP(B)

Example:

P(havingabirthdayinDecember) P(becomingaprefect)

Wewouldexpectthesetwoeventstobeindependent.

InGeneral:

KnowingthatBhasoccurredmakesnodifferencetotheprobabilityofAoccurring.

25

pg61-64

26

Exercises on Probability Tables

1. P(A) = 0.4, P(B) = 0.5 P(A∩B) = 0.1

A A’

B 0.1 0.5

B’

0.4 1

Complete the table and answer the following questions.

(a) What is the probability that A or B occurs?

(b) What is the probability that neither A nor B occurs?

(c) Are A and B independent?

(d) Find P(A|B)

(e) Are A and B mutually exclusive?

2. P(A) is 0.4, P(B) is 0.6, and A and B are independent events.

A A’

B 0.6

B’

0.4 1



(b) What is P(B|A’)?

(c) Are A and B mutually exclusive?

27

3. P(A) is 0.6, P(A∩B) = 0.3, and A and B are independent events.

A A’

B 0.3

B’

0.6 1


(a) What is the probability that of A or B occurs?


(c) What is P(A|B)?

4. P(A) = 0.45, P(B) = 0.6, P(AÈB) = 0.9

A A’

B

B’


(a) Are A and B mutually exclusive?


(c) What is P(A|B)?

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5. P(AUB) = 0.7, P(A) = 0.5, P(B) = 0.4

A A’

B

B’


(a) What is the probability that both of A and B occur?


(c) What is P(A|B)?

(d) Are A and B independent events?

6. P(A) = 0.4, P(B) = 0.5, P(A|B) = 0.5

A A’

B

B’




(c) What is P(B|A)?

(d) Are A and B independent?

29

AbsoluteRisk(Risk)Riskorabsoluteriskisthechanceofaeventoccuring.Thisisusuallyexpressedasadecimalorpercentage.

Riskisusuallyassociatedwithsomethingbad(cancer,death,foodpoisoningetc).

Riskcanbearatio

EgTheriskofbeingstruckbylighteningis4.5per1,000,000people

Thereforetheriskis0.0000045or0.00045%

AsitisaprobabilityRiskwillalwaysbebetween0and1

What is the absolute risk of getting heart disease in each group?

Risk of heart disease in an overweight

male

___________________________________

___________________________________

Risk of heart disease in an smoking

male

___________________________________

___________________________________

Risk of heart disease in a non-

overweight male

___________________________________

___________________________________

Risk of heart disease in a non-smoking

male

___________________________________

___________________________________

What conclusion can you come to from this?

30

RelativeRiskRelativeRiskisusedtocomparetheriskfortwogroups.Itgivesameasureoftheimpactofthebehaviourortreatmentgroupisexposedto.

Usuallythebaselinegroup(denominator)isthenon-treatmentgroup(thiswillbeinthewordingofthequestion).

RelativeRisk = Riskfortreatmentgroup

Riskfornontreatmentgroup

Relativeriskdoesnothavetobebetween0and1

RRof2.5means=youare2.5timesmorelikelyto...(ora150%increase)

RRof1means=youareequallylikelyto......

RRof0.8means=thereis0.8timesthechanceof....(adecreasedriskegexerciseonthechanceofaheartattack)(Thisis20%LESSriskforthosethatexercise)

or0.8timesaslikely.....

What is the Relative Risk of heart

disease in an overweight male

compared with a non-overweight male.

___________________________________

___________________________________

What can we conclude?

___________________________________

___________________________________

___________________________________

___________________________________

31

What is the Relative Risk of heart

disease in an smoking male compared

with a non-smoking male.

___________________________________

___________________________________

What can we conclude?

___________________________________

___________________________________

___________________________________

___________________________________

RelativerisktellsnothingofACTUALrisk

Struckbylightning

NotStruck Total

Hat 3 1000000

NoHat 1 2000000

Whatistherelativeriskofbeingstruckbylightningwhilewearingahatcomparedwithnotwearingahat?

Whatdoyounowconclude?

32

pg67-72

34

TheoreticalvsExperimentalvsActualProbability

TheoreticalProbability

TheoreticalProbabilityalsoknownasmodelprobabilityiswhenweuseatooltofindoutthetheoreticalprobabilityofaneventoccurring

Example:

Thetheoreticalprobabilityofrollinga3onadiceis56or

0.16666or16%

ExperimentalProbability

Experimentalprobabilityisbasedonthenumberoftimestheeventoccursoutofthetotalnumberoftrials.

Example:

Samrolledadice50times.A3appeared10times.

Thentheexperimentalprobabilityofrollinga3is10outof50or20%.Inprobabilityanexperimentisoneormoretrialsofaprobabilitysituation.

ActualProbability

Alsoknownastrueprobabilityisthe(almostalways)unknownactualprobabilitythatan

eventwilloccurinagivensituation.

Example:

Theactualprobabilityofacoinlandingheadsupisaffectedbythepositionfromwhichitistossed,theasymmetryofthetwofacesofthecoinetc,soisnotexactly0.5,thoughtheprobabilityofafaircoinlandingheadswillbeverycloseto0.5.

35

NZQA Sample Paper Question Emma was given a set of six different keys to her new house. One of the keys opened the dead lock on the front door, and a different key opened the door lock on the front door. Emma did not know which keys were the correct keys for each lock, and was able to open both locks after four key attempts. Emma’s friend Sene thought this was a low number of key attempts, and wondered what process Emma used to find the correct keys. Sene designed and carried out a simulation to estimate how many key attempts it would take before both locks were open, using a ‘trial and error’ process, to see if this might have been the process Emma used.

For the ‘trial and error’ process, Sene assumed:

• that a key was selected at random to try to open the dead lock

• once Emma had tried a key for the dead lock, she did not try it again

• once Emma found the correct key for the dead lock, she removed this from the set of keys and tried the same process with the door lock.

The results of Sene’s simulation are shown below:

Numberofkeyattemptsbefore

bothlocksareopen Frequency

2 29

3 70

4 103

5 136

6 164

7 176

8 134

9 74

10 76

11 38

36

(a) Calculate the theoretical probability of a person using a 'trial and error’ process taking more than two key attempts before both locks are open. Compare this probability with the results from Sene’s simulation and discuss any differences.

(b) Sene used the central 90% of her simulation results to check if the number of key attempts Emma

took (four) was likely if the ‘trial and error’ process was used, and concluded it was. Discuss if the results of Sene’s simulation support this conclusion.

(c) Calculate the theoretical probability that a person using Emma’s process takes four key attempts before both locks are open.

37

Randomness

ActivityOne

Plot3'randompoints'onpaper

Plot5'randompoints'onpaper

Howdotheycompare?

ActivityTwopg74

Ineedtopickwhichoneisthefakeandwhichisreal.

3.13 Probability workbook 2018 - Mathematics with Ms Walker · 2018-02-04 · Theoretical...

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