商务统计学课件英文版BSFC7e_CH04

1、Basic ProbabilityChapter 4ObjectivesThe objectives for this chapter are: To understand basic probability concepts.To understand conditional probability To be able to use Bayes Theorem to revise probabilitiesTo learn various counting rulesBasic Probability ConceptsProbability the chance that an uncer

2、tain event will occur (always between 0 and 1)Impossible Event an event that has no chance of occurring (probability = 0)Certain Event an event that is sure to occur (probability = 1)Assessing ProbabilityThere are three approaches to assessing the probability of an uncertain event:1. a priori - base

3、d on prior knowledge of the process2. empirical probability3. subjective probabilitybased on a combination of an individuals past experience, personal opinion, and analysis of a particular situation Assuming all outcomes are equally likelyprobability of occurrenceprobability of occurrenceExample of

4、a priori probabilityWhen randomly selecting a day from the year 2015 what is the probability the day is in January?Example of empirical probabilityTaking StatsNot Taking StatsTotalMale 84145229Female 76134210Total160279439Find the probability of selecting a male taking statistics from the population

5、 described in the following table:Probability of male taking statsSubjective probabilitySubjective probability may differ from person to personA media development team assigns a 60% probability of success to its new ad campaign.The chief media officer of the company is less optimistic and assigns a

6、40% of success to the same campaignThe assignment of a subjective probability is based on a persons experiences, opinions, and analysis of a particular situationSubjective probability is useful in situations when an empirical or a priori probability cannot be computedEventsEach possible outcome of a

7、 variable is an event.Simple eventAn event described by a single characteristice.g., A day in January from all days in 2015Joint eventAn event described by two or more characteristicse.g. A day in January that is also a Wednesday from all days in 2015Complement of an event A (denoted A)All events th

8、at are not part of event Ae.g., All days from 2015 that are not in JanuarySample 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:Organizing & Visualizing EventsVenn Diagram For All Days In 2015Sample Space (All Days In 2015)J

9、anuary DaysWednesdaysDays That Are In January and Are WednesdaysOrganizing & Visualizing EventsContingency Tables - For All Days in 2015Decision TreesAll Days In 2015Not Jan.Jan.Not Wed.Wed.Wed.Not Wed.Sample SpaceTotalNumberOfSampleSpaceOutcomesNot Wed. 27 286 313 Wed. 4 48 52Total 31 334 365 Jan.

10、Not Jan. Total 4 27 48286(continued)Definition: Simple ProbabilitySimple Probability refers to the probability of a simple event.ex. P(Jan.)ex. P(Wed.)P(Jan.) = 31 / 365P(Wed.) = 52 / 365Not Wed. 27 286 313 Wed. 4 48 52Total 31 334 365 Jan. Not Jan. TotalDefinition: Joint ProbabilityJoint Probabilit

11、y refers to the probability of an occurrence of two or more events (joint event).ex. P(Jan. and Wed.)ex. P(Not Jan. and Not Wed.)P(Jan. and Wed.) = 4 / 365P(Not Jan. and Not Wed.)= 286 / 365Not Wed. 27 286 313 Wed. 4 48 52Total 31 334 365 Jan. Not Jan. TotalMutually exclusive eventsEvents that canno

12、t occur simultaneouslyExample: Randomly choosing a day from 2015 A = day in January; B = day in FebruaryEvents A and B are mutually exclusiveMutually Exclusive EventsCollectively Exhaustive EventsCollectively exhaustive eventsOne of the events must occur The set of events covers the entire sample sp

13、aceExample: Randomly choose a day from 2015 A = Weekday; B = Weekend; C = January; D = Spring;Events A, B, C and D are collectively exhaustive (but not mutually exclusive a weekday can be in January or in Spring)Events A and B are collectively exhaustive and also mutually exclusiveComputing Joint an

14、d Marginal ProbabilitiesThe probability of a joint event, A and B:Computing a marginal (or simple) probability:Where B1, B2, , Bk are k mutually exclusive and collectively exhaustive eventsJoint Probability ExampleP(Jan. and Wed.)Not Wed. 27 286 313 Wed. 4 48 52Total 31 334 365 Jan. Not Jan. TotalMa

15、rginal Probability ExampleP(Wed.)Not Wed. 27 286 313 Wed. 4 48 52Total 31 334 365 Jan. Not Jan. Total P(A1 and B2)P(A1)TotalEventMarginal & Joint Probabilities In A Contingency TableP(A2 and B1)P(A1 and B1)EventTotal1Joint ProbabilitiesMarginal (Simple) Probabilities A1 A2B1B2 P(B1) P(B2)P(A2 and B2

16、)P(A2)Probability Summary So FarProbability is the numerical measure of the likelihood that an event will occurThe probability of any event must be between 0 and 1, inclusivelyThe sum of the probabilities of all mutually exclusive and collectively exhaustive events is 1CertainImpossible0.5100 P(A) 1

17、 For any event AIf A, B, and C are mutually exclusive and collectively exhaustiveGeneral Addition RuleP(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

18、A and BGeneral Addition Rule ExampleP(Jan. or Wed.) = P(Jan.) + P(Wed.) - P(Jan. and Wed.)= 31/365 + 52/365 - 4/365 = 79/365Dont count the four Wednesdays in January twice!Not Wed. 27 286 313 Wed. 4 48 52Total 31 334 365 Jan. Not Jan. TotalComputing Conditional ProbabilitiesA conditional probability

19、 is the probability of one event, given that another event has occurred:Where P(A and B) = joint probability of A and B P(A) = marginal or simple probability of AP(B) = marginal or simple probability of BThe conditional probability of A given that B has occurredThe conditional probability of B given

20、 that A has occurredWhat is the probability that a car has a GPS, given that it has AC ?i.e., we want to find P(GPS | AC)Conditional Probability ExampleOf the cars on a used car lot, 70% have air conditioning (AC) and 40% have a GPS. 20% of the cars have both.Conditional Probability ExampleNo GPSGPS

21、TotalAC0.20.50.7No AC0.20.10.3Total0.40.6 1.0Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a GPS and 20% of the cars have both.(continued)Conditional Probability ExampleNo GPSGPSTotalAC0.20.50.7No AC0.20.10.3Total0.40.6 1.0Given AC, we only consider the top row (70% of t

22、he cars). Of these, 20% have a GPS. 20% of 70% is about 28.57%.(continued)Using Decision TreesHas ACDoes not have ACHas GPSDoes not have GPSHas GPSDoes not have GPSP(AC)= 0.7P(AC)= 0.3P(AC and GPS) = 0.2P(AC and GPS) = 0.5P(AC and GPS) = 0.1P(AC and GPS) = 0.2AllCarsGiven AC or no AC:ConditionalProb

23、abilitiesUsing Decision TreesHas GPSDoes not have GPSHas ACDoes not have ACHas ACDoes not have ACP(GPS)= 0.4P(GPS)= 0.6P(GPS and AC) = 0.2P(GPS and AC) = 0.2P(GPS and AC) = 0.1P(GPS and AC) = 0.5AllCarsGiven GPS or no GPS:(continued)ConditionalProbabilitiesIndependenceTwo events are independent if a

24、nd only if:Events A and B are independent when the probability of one event is not affected by the fact that the other event has occurredMultiplication RulesMultiplication rule for two events A and B:Note: If A and B are independent, thenand the multiplication rule simplifies toMarginal ProbabilityM

25、arginal probability for event A:Where B1, B2, , Bk are k mutually exclusive and collectively exhaustive eventsBayes TheoremBayes Theorem is used to revise previously calculated probabilities based on new information.Developed by Thomas Bayes in the 18th Century.It is an extension of conditional prob

26、ability.Bayes Theoremwhere:Bi = ith event of k mutually exclusive and collectively exhaustive eventsA = new event that might impact P(Bi)Bayes Theorem ExampleA drilling company has estimated a 40% chance of striking oil for their new well. A detailed test has been scheduled for more information. His

27、torically, 60% of successful wells have had detailed tests, and 20% of unsuccessful wells have had detailed tests. Given that this well has been scheduled for a detailed test, what is the probability that the well will be successful?Let S = successful well U = unsuccessful wellP(S) = 0.4 , P(U) = 0.

28、6 (prior probabilities)Define the detailed test event as DConditional probabilities:P(D|S) = 0.6 P(D|U) = 0.2Goal is to find P(S|D)Bayes Theorem Example(continued)Bayes Theorem Example(continued)Apply Bayes Theorem:So the revised probability of success, given that this well has been scheduled for a

29、detailed test, is 0.667Given the detailed test, the revised probability of a successful 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

30、)(0.2) = 0.120.12/0.36 = 0.333Sum = 0.36(continued)Counting Rules Are Often Useful In Computing ProbabilitiesIn many cases, there are a large number of possible outcomes.Counting rules can be used in these cases to help compute probabilities.Counting RulesRules for counting the number of possible ou

31、tcomesCounting Rule 1:If 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 toExampleIf you roll a fair die 3 times then there are 63 = 216 possible outcomesknCounting RulesCounting Rule 2:If there are

32、k1 events on the first trial, k2 events on the second trial, and kn events on the nth trial, the number of possible outcomes isExample:You 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?Answer: (3)(4)(6) = 72 different possibilities(k1)(k2)(kn)(continued)Counting RulesCounting Rule