商业统计学英文课件：ch04 Basic Probability

1、Chap 4-1Chapter 4Basic ProbabilityStatistics For Managers Using Microsoft Excel6th EditionChap 4-2Learning ObjectivesIn this chapter, you learn: nBasic probability concepts and definitionsnConditional probability nTo use Bayes Theorem to revise probabilitiesnVarious counting rulesChap 4-3Important T

2、ermsnProbability the chance that an uncertain event will occur (always between 0 and 1)nEvent Each possible outcome of a variablenSimple Event an event that can be described by a single characteristicnSample Space the collection of all possible eventsChap 4-4Assessing ProbabilitynThere are three app

3、roaches to assessing the probability of an uncertain event:1. a priori classical probability2. empirical classical probability3. subjective probability an individual judgment or opinion about the probability of occurrenceoutcomeselementaryofnumbertotaloccurcaneventthewaysofnumberTXoccurrenceofyproba

4、bilitobservedoutcomesofnumbertotalobservedoutcomesfavorableofnumberoccurrenceofyprobabilitChap 4-5Sample SpaceThe Sample Space is the collection of all possible eventse.g. All 6 faces of a die:e.g. All 52 cards of a bridge deck:Chap 4-6EventsnSimple eventnAn outcome from a sample space with one char

5、acteristicne.g., A red card from a deck of cardsnComplement of an event A (denoted A)nAll outcomes that are not part of event Ane.g., All cards that are not diamondsnJoint eventnInvolves two or more characteristics simultaneouslyne.g., An ace that is also red from a deck of cardsChap 4-7Visualizing

6、EventsnContingency TablesnTree Diagrams Red 2 24 26 Black 2 24 26Total 4 48 52 Ace Not Ace TotalFull Deck of 52 CardsRed CardBlack CardNot an AceAceAceNot an Ace Sample SpaceSample Space224224Chap 4-8Visualizing EventsnVenn DiagramsnLet A = acesnLet B = red cardsABA B = ace and redA U B = ace or red

7、Chap 4-9Mutually Exclusive EventsnMutually exclusive eventsnEvents that cannot occur togetherexample: A = queen of diamonds; B = queen of clubsnEvents A and B are mutually exclusiveChap 4-10Collectively Exhaustive EventsnCollectively exhaustive eventsnOne of the events must occur nThe set of events

8、covers the entire sample spaceexample: A = aces; B = black cards; C = diamonds; D = heartsnEvents A, B, C and D are collectively exhaustive (but not mutually exclusive an ace may also be a heart)nEvents B, C and D are collectively exhaustive and also mutually exclusiveChap 4-11ProbabilitynProbabilit

9、y is the numerical measure of the likelihood that an event will occurnThe probability of any event must be between 0 and 1, inclusivelynThe sum of the probabilities of all mutually exclusive and collectively exhaustive events is 1CertainImpossible0.5100 P(A) 1 For any event A1P(C)P(B)P(A)If A, B, an

10、d C are mutually exclusive and collectively exhaustiveChap 4-12Computing Joint and Marginal ProbabilitiesnThe probability of a joint event, A and B:nComputing a marginal (or simple) probability:nWhere B1, B2, , Bk are k mutually exclusive and collectively exhaustive eventsoutcomeselementaryofnumbert

11、otalBandAsatisfyingoutcomesofnumber)BandA(P)BdanP(A)BandP(A)BandP(AP(A)k21Chap 4-13Joint Probability ExampleP(Red and Ace)BlackColorTypeRedTotalAce224Non-Ace242448Total262652522cards of number totalace and red are that cards of numberChap 4-14Marginal Probability ExampleP(Ace)BlackColorTypeRedTotalA

12、ce224Non-Ace242448Total262652524522522)BlackandAce(P)dReandAce(PChap 4-15 P(A1 and B2)P(A1)TotalEventJoint Probabilities Using Contingency TableP(A2 and B1)P(A1 and B1)EventTotal1Joint ProbabilitiesMarginal (Simple) Probabilities A1 A2B1B2 P(B1) P(B2)P(A2 and B2)P(A2)Chap 4-16General Addition RuleP(

13、A or B) = P(A) + P(B) - P(A and B)General Addition Rule:If A and B are mutually exclusive, then P(A and B) = 0, so the rule can be simplified:P(A or B) = P(A) + P(B) For mutually exclusive events A and BChap 4-17General Addition Rule ExampleP(Red or Ace) = P(Red) +P(Ace) - P(Red and Ace) = 26/52 + 4

14、/52 - 2/52 = 28/52Dont count the two red aces twice!BlackColorTypeRedTotalAce224Non-Ace242448Total262652Chap 4-18Computing Conditional ProbabilitiesnA conditional probability is the probability of one event, given that another event has occurred:P(B)B)andP(AB)|P(AP(A)B)andP(AA)|P(BWhere P(A and B) =

15、 joint probability of A and B P(A) = marginal probability of AP(B) = marginal probability of BThe conditional probability of A given that B has occurredThe conditional probability of B given that A has occurredChap 4-19nWhat is the probability that a car has a CD player, given that it has AC ?i.e.,

16、we want to find P(CD | AC)Conditional Probability ExamplenOf the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both.Chap 4-20Conditional Probability ExampleNo CDCDTotalAC0.20.50.7No AC0.20.10.3Total0.40.6 1.0nOf the cars on a used car lot,

17、 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both.0.28570.70.2P(AC)AC)andP(CDAC)|P(CD(continued)Chap 4-21Conditional Probability ExampleNo CDCDTotalAC0.20.50.7No AC0.20.10.3Total0.40.6 1.0nGiven AC, we only consider the top row (70% of the cars). Of these, 20%

18、have a CD player. 20% of 70% is about 28.57%.0.28570.70.2P(AC)AC)andP(CDAC)|P(CD(continued)Chap 4-22Using Decision TreesHas ACDoes not have ACHas CDDoes not have CDHas CDDoes not have CDP(AC)= 0.7P(AC)= 0.3P(AC and CD) = 0.2P(AC and CD) = 0.5P(AC and CD) = 0.1P(AC and CD) = 0.27.5.3.2.3.1.AllCars7.2

19、.Given AC or no AC:Chap 4-23Using Decision TreesHas CDDoes not have CDHas ACDoes not have ACHas ACDoes not have ACP(CD)= 0.4P(CD)= 0.6P(CD and AC) = 0.2P(CD and AC) = 0.2P(CD and AC) = 0.1P(CD and AC) = 0.54.2.6.5.6.1.AllCars4.2.Given CD or no CD:(continued)Chap 4-24Statistical IndependencenTwo even

20、ts are independent if and only if:nEvents A and B are independent when the probability of one event is not affected by the other eventP(A)B)|P(AChap 4-25Multiplication RulesnMultiplication rule for two events A and B:P(B)B)|P(AB)andP(A P(A)B)|P(ANote: If A and B are independent, thenand the multipli

21、cation rule simplifies toP(B)P(A)B)andP(A Chap 4-26Marginal ProbabilitynMarginal probability for event A:nWhere B1, B2, , Bk are k mutually exclusive and collectively exhaustive events)P(B)B|P(A)P(B)B|P(A)P(B)B|P(A P(A)kk2211Chap 4-27Bayes Theoremnwhere:Bi = ith event of k mutually exclusive and col

22、lectively exhaustive eventsA = new event that might impact P(Bi)P(BB|P(A)P(BB|P(A)P(BB|P(A)P(BB|P(AA)|P(Bk k 2 2 1 1 i i i Bayes theorem is used to revise previously calculated probabilities after new information is obtainedChap 4-28Bayes Theorem ExamplenA drilling company has estimated a 40% chance

23、 of striking oil for their new well. nA detailed test has been scheduled for more information. Historically, 60% of successful wells have had detailed tests, and 20% of unsuccessful wells have had detailed tests. nGiven that this well has been scheduled for a detailed test, what is the probability t

24、hat the well will be successful?Chap 4-29nLet S = successful well U = unsuccessful wellnP(S) = 0.4 , P(U) = 0.6 (prior probabilities)nDefine the detailed test event as DnConditional probabilities:P(D|S) = 0.6 P(D|U) = 0.2nGoal is to find P(S|D)Bayes Theorem Example(continued)Chap 4-300.6670.120.240.

25、24(0.2)(0.6)(0.6)(0.4)(0.6)(0.4)U)P(U)|P(DS)P(S)|P(DS)P(S)|P(DD)|P(SBayes Theorem Example(continued)Apply Bayes Theorem:So the revised probability of success, given that this well has been scheduled for a detailed test, is 0.667Chap 4-31nGiven the detailed test, the revised probability of a successf

26、ul well has risen to 0.667 from the original estimate of 0.4Bayes Theorem ExampleEventPriorProb.Conditional Prob.JointProb.RevisedProb.S (successful)0.40.6(0.4)(0.6) = 0.240.24/0.36 = 0.667U (unsuccessful)0.60.2(0.6)(0.2) = 0.120.12/0.36 = 0.333Sum = 0.36(continued)Chap 4-32Counting RulesnRules for

27、counting the number of possible outcomesnCounting Rule 1:nIf any one of k different mutually exclusive and collectively exhaustive events can occur on each of n trials, the number of possible outcomes is equal toknChap 4-33Counting RulesnCounting Rule 2:nIf there are k1 events on the first trial, k2

28、 events on the second trial, and kn events on the nth trial, the number of possible outcomes isnExample:nYou want to go to a park, eat at a restaurant, and see a movie. There are 3 parks, 4 restaurants, and 6 movie choices. How many different possible combinations are there?nAnswer: (3)(4)(6) = 72 d

29、ifferent possibilities(k1)(k2)(kn)(continued)Chap 4-34Counting RulesnCounting Rule 3:nThe number of ways that n items can be arranged in order isnExample:nYour restaurant has five menu choices for lunch. How many ways can you order them on your menu?nAnswer: 5! = (5)(4)(3)(2)(1) = 120 different possibilitiesn! = (n)(n 1)(1)(continued)Chap 4-35Counting RulesnCounting Rule 4:nPermutations: The number of ways of arranging X objects selected from n objects in order isnExample:nYour restaurant has five menu choices, and three are selected for daily specials. How many diff