Bài giảng Statistical Techniques in Business and Economics - Chapter 5 A Survey of Probability Concepts

Chapter 5A Survey ofProbabilityConcepts1When you have completed this chapter, you will be able to: Explain the terms random experiment, outcome, sample space, permutations,  and combinations. Define probability.Describe the classical, empirical, and subjective approaches to probability.and...Chapter Goals Explain and calculate conditional probability and joint probability. 12342Use a tree diagram to organize and  compute probabilities. Calculate probability using the rules of addition and rules of multiplication.Chapter GoalsCalculate a probability using Bayes’ theorem.5673Types of StatisticsMethods of collecting organizing presenting and analyzing dataDescriptiveScience of making inferences about a population, based on sample information.InferentialRecallEmphasis now to be on this!4TerminologyProbabilityis a measure of the likelihood that an event in the future will happen! It can only assume a value between 0 and 1. A value near zero means the event is not likely happen; near one means it is likely.. There are three definitions of probability: classical, empirical, and subjectiveMore5TerminologyRandom Experimentis a process repetitive in nature the outcome of any trial is uncertain well-defined set of possible outcomes each outcome has a probability associated with itMore6Outcomeis a particular result of a random experiment.Sample space... is the collection or set of all the possible outcomes of a random experiment.TerminologyEventis the collection of one or more outcomes of an experiment.More7Approaches to Assigning ProbabilitySubjectiveprobability is based on whatever information is availableObjectiveClassical Probability is based on the assumption that the outcomes of an experiment are equally likelyProbability of an Event= NUMBER of favourable outcomesTotal NUMBER of possible outcomesEmpirical Probability applies when the number of times the event happens is divided by the number of observationsExamples8. refers to the chance of occurrence assigned to an event by a particular individualIt is not computed objectively, i.e., not from prior knowledge or from actual data S ubjectiveProbabilityExample that the Toronto Maple Leafs will win the Stanley Cup next season!that you will arrive to class on time tomorrow!9Students measure the contents of their soft drink cans 10 cans are underfilled, 32 are filled correctly and 8 are overfilled When the contents of the next can is measured, what is the probability that it is (a) filled correctly?P(C) = 32 / 50 = 64%(b) not filled correctly?P(~C) = 1 – P(C) = 1 - .64 = 36%This is called the Complement of CAE mpiricalProbability10Random ExperimentThe experiment is rolling the die...once!The possible outcomes are the numbers1 2 3 4 5 6An event is the occurrence of an even numberi.e. we collect the outcomes 2, 4, and 6.11 Consider the random experiment of flipping a coin twice.Tree Diagrams This is a useful device to show all the possible outcomes of the experimentand their corresponding probabilities121.00Tree DiagramsOriginFirst FlipHTHTHTHHHTTTTHSimple EventsP(HH)= 0.25P(HT)= 0.25P(TH)= 0.25P(TT)= 0.25SecondFlipNewExpressed as:13Tree Diagrams Menu Appetizer: Soup or JuiceEntrée:Beef Turkey FishDessert:PieIce CreamOriginAppetizerEntrée DessertSoupJuiceBeefTurkeyFishBeefTurkeyFishPieIce CreamPieIce CreamPieIce CreamPieIce CreamPieIce CreamPieIce Cream14Tree DiagramsHow many complete dinners are there?1215Tree DiagramsHow many dinners include beef?41.2.3.4.16Tree DiagramsWhat is the probability that a complete dinner will includeJuice?Turkey?Both beef and soup?6/124/122/12See next slide 17 If one thing can be done in M ways, and if after this is done, something else can be done in N ways, then both things can be done in a total of M*N different ways in that stated order!Refer back to tree diagram example: # different meals = 2 * 3 * 2 = 12 # meals with beef = 2 * 1 * 2 = 4 # meals with juice = 1 * 3 * 2 = 6M * N Rule TheAppetizerEntrée DessertLegend:18 3 * 2 * 5 = 30 When getting dressed, you have a choice between wearing one of:3 pairs of shoes2 pairs of pants5 shirtsFind the number of different “outfits” possibleExample A19What is the probability of drawing a red Ace from a deck of well-shuffled cards?P( Red Ace) = 2/52Probability202.1.Determine.the Outcomes that Meet Our ConditionList.all Possible OutcomesKey steps 4 SuitsHeartsDiamondsClubsSpadesDeck13 cards in each= 52 CardsProbabilityUsing AnalysisProbability21P = probabilityof getting four(4) aces= 52 Cards(the Population)DeckData13 cards4 Suitsx 4 SuitsHeartsDiamondsClubsSpades13 cards in eachProbability22‘Honours’ cardsEach Suit has a.ProbabilityScenarios= 52 Cards 4 Suits (13 cards in each)HeartsDiamondsClubsSpadesDeck2352Scenarios1. Draw an Ace Condition Outcomes All Possible Outcomes 4 2. Draw a Black Ace Condition Outcomes All Possible Outcomes 2 523. Draw a Red Card Condition Outcomes All Possible Outcomes26 52= 52 Cards 4 Suits (13 cards in each)HeartsDiamondsClubsSpadesDeck244. Drawinga Red Card or a Queen Condition Outcomes All Possible Outcomes2652+ 2 522852=Scenarios= 52 Cards 4 Suits (13 cards in each)HeartsDiamondsClubsSpadesDeck-or- P(Red) + P(Queen) - P (Red Queen)= 26 + 4 - 2 52 2852=25Scenarios= 52 Cards 4 Suits (13 cards in each)HeartsDiamondsClubsSpadesDeckWhat is the probability of drawing a Jack or a King from a deck of well-shuffled cards?= 4/52= 4/52= 8/52P( Jack or King)= 4/52+ 4/5226ScenariosWhat is the probability of drawing one card that is both a Jack and a King from a deck of well-shuffled cards?These are MUTUALLY EXCLUSIVE events, i.e. they can’t both happen at the same time!Note= 0P( Jack and King)= 52 Cards 4 Suits (13 cards in each)HeartsDiamondsClubsSpadesDeck27Scenarios=2/52P( Black and King)What is the probability of drawing one card that is both BLACK and a King from a deck of well-shuffled cards?= 52 Cards 4 Suits (13 cards in each)HeartsDiamondsClubsSpadesDeck28Scenarios= 52 Cards 4 Suits (13 cards in each)HeartsDiamondsClubsSpadesDeckWhat is the probability of drawing a card that is either BLACK or a King from a deck of well-shuffled cards?Formula P(A or B) == 28/52P( Black or King)= 26/52+ 4/52- 2/52NoteThis is called the Addition RuleP (A) + P(B) –P(Both)29Scenarios= 52 Cards 4 Suits (13 cards in each)HeartsDiamondsClubsSpadesDeckWhat is the probability of drawing a King given that you have drawn a BLACK card?= 2/26P(King|Black )NoteThis is called a CONDITIONAL probabilityOur sample space is now just the BLACK cardsAlternate solution30= 52 Cards 4 Suits (13 cards in each)HeartsDiamondsClubsSpadesDeckScenariosWhat is the probability of drawing a King given that you have drawn a BLACK card?Formula P(A|B) =P(Given)P (Both) = 2/26= (2/52) / (26/52)= (2/52) * (52/26)31Scenarios= 52 Cards 4 Suits (13 cards in each)HeartsDiamondsClubsSpadesDeckScenarios= 52 Cards 4 Suits (13 cards in each)HeartsDiamondsClubsSpadesDeckWhat is the probability of drawing a King of Clubs given that you have drawn a BLACK card?P(King of Clubs|Black )= 1/26= (1/52) / (26/52)= (1/52) * (52/26)P(A|B) =P(Given)P (Both) Formula 32Scenarios= 52 Cards 4 Suits (13 cards in each)HeartsDiamondsClubsSpadesDeckWhat is the probability of drawing a King of Clubs given that you have drawn a CLUB?P(King of Clubs given Club)= 1/13= (1/52) * (52/13)= P(1/52) / (13/52) = P(King of Clubs|Club)33 Reading Probabilities from a Table34What is the Probability of selecting a female student?400/750 = 53.33%AA survey of undergraduate students in the School of Business Management at Eton College revealed the following regarding the gender and majors of the students: Gender Accounting International HR TOTAL Male 150 150 50 350 Female 175 160 65 400 325 310 115 750MoreReading Probabilities from a Table35What is the Probability of selecting a Human Resources or International major?A Gender Accounting International HR TOTAL Male 150 150 50 350 Female 175 160 65 400 325 310 115 750More= 115/750 + 310/750 = 425/750= 56.67%P(HR or I) = P(HR) + P(I) 310 115Reading Probabilities from a Table36What is the Probability of selecting a Female or International major?A Gender Accounting International HR TOTAL Male 150 150 50 350 Female 175 160 65 400 325 310 115 750= 400/750 P(F or I) = P(F) + P(I) – P(F and I)MoreReading Probabilities from a Table+ 310/750– 160/750= 550/750 = 73.33%37What is the Probability of selecting a Female Accounting student?A Gender Accounting International HR TOTAL Male 150 150 50 350 Female 175 160 65 400 325 310 115 750= 175/750 = 23.33%P(F and A)MoreReading Probabilities from a Table38What is the Probability of selecting a Female, given that the person selected is an International major?A Gender Accounting International HR TOTAL Male 150 150 50 350 Female 175 160 65 400 325 310 115 750160/310 = 51.6%P(F|I) =Alternative SolutionReading Probabilities from a Table39What is the Probability of selecting a Female, given that the person selected is an International major?A= (160/750) / (310/750)P(F|I) = P(F and I) / P(I)P(A|B) =Formula P(Both)P(Given)Reading Probabilities from a Table= 51.6%= 160/31040What is the Probability of selecting an International major, given that the person selected is a Female?A Gender Accounting International HR TOTAL Male 150 150 50 350 Female 175 160 65 400 325 310 115 750160/400 = 40%P(I|F) =MoreReading Probabilities from a Table41Notice the significant difference:Reading Probabilities from a Tablebetween F given I I given F!and42Each flip is independent of the other!Flip onceFlip twiceTerminologyEvents are independent if the occurrence of one event does not affect the probability of the other Consider the random experiment of flipping a coin twice.Independent EventsFind the probability of flipping 2 Heads in a row P(2H) = .5*.5 = .25 or 25%A43TerminologyIndependent EventsEach draw is independent of the otherDraw three cards with replacement i.e., draw one card, look at it, put it back, and repeat twice more.Find the probability of drawing 3 Queens in a row:P(3Q) = 4/52 * 4/52 *4/52= 0.00046 = most unlikely!44Independent EventsConsider 2 events: Drawing a RED card from a deck of cards Drawing a HEART from a deck of cardsAre these two events considered to be independent?If two events, A and B are independent, then P(A|B) = P(A)NoteP(Red) =P(Red|Heart) = 26/52 = 1/213/13 = 1Therefore these are NOT independent events!45TBayes’heorem46is a method for revising a probability given additional information!Formula ExampleTBayes’heoremP(A1|B) =P(A1 )P(B|A1 )P(A1 )P(B|A1)+P(A2 )P(B|A2 )47TBayes’heoremDuff Cola Company recently received several complaints that their bottles are under-filled. A complaint was received today but the production manager is unable to identify which of the two Springfield plants (A or B) filled this bottle. What is the probability that the under-filled bottle came from plant A?% of Total Production% of Underfilled BottlesA553B45448TBayes’heorem% of Total Production% of Underfilled BottlesA553B454What is the probability that the under-filled bottle came from plant A?1List the Probabilities given2Input values into formula and computeP(plant A) = .55P(plant B) = .45P(Underfilled -A) = .03P(Underfilled -B) = .0449TBayes’heoremWhat is the probability that the under-filled bottle came from plant A?1List the Probabilities given2Input values into formula and computeP(plant A) = .55P(plant B) = .45P(Underfilled/A) = .03P(Underfilled/B) = .04P(A1 |B) =P(A1 )P(B|A1 )P(A1 )P(B|A1 )+P(A2 )P(B|A2 )= .55(.03).55(.03) +.45(.04)= .4783 The likelihood that the underfilled bottle came from Plant A has been reduced from 55% to 47.83%50CountingRules51Factorials! this is just a shorthand notation that is sometimes used to save time!Examples:5! Means 5*4*3*2*1 = 1204! Means 4*3*2*1 = 24By definition, 1! =1 and 0! =1Note52is a counting technique that is used when order is important!is a counting technique that is used when order is NOT important!PermutationCombinationn Pr =n!(n – r)!n Cr =n!r!(n – r)!53How many ways can you arrange n things, taking r at a time, when order is important?You are assigned the task of choosing 2 of your 6 classmates to serve on a task force. One will act as the Chair of the task force, and the other will be the Secretary. In how many ways can you make this assignment?Permutationn Pr =n!(n – r)!Example:6P2 =6! / (6-2)!= 6! / 4!= 6*5 = 3054You are assigned the task of choosing 2 of your 6 classmates to serve on a task force. Responsibilities are evenly shared. In how many ways can you make this assignment?Example:6C2 =6! / (2!(6-2)!)= 6! /2!4!= (6*5)/2 = 15is a counting technique that is used when order is NOT important!Combinationn Cr =n!r(n – r)!Using55UsingTexas Instruments BAII PLUSi 15 30 CombinationPermutation2nd+nCr621562nd230nPr56Test your learning www.mcgrawhill.ca/college/lindClick onOnline Learning Centrefor quizzesextra contentdata setssearchable glossaryaccess to Statistics Canada’s E-Stat dataand much more!57This completes Chapter 558