描述統(tǒng)計學chap04

1、4-1,Chapter 4Basic Probability,Business Statistics,4-2,Chapter Goals,After completing this chapter, you should be able to: Explain basic probability concepts and definitionsUse contingency tables to view a sample sp

2、aceApply common rules of probabilityCompute conditional probabilitiesDetermine whether events are statistically independentUse Bayes’ Theorem for conditional probabilities,4-3,4-4,Basic Probability Concepts,Probabili

3、ty – the chance that an uncertain 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)

4、,4-5,Assessing Probability,There are three approaches to assessing the probability of an uncertain event:1. a priori -- based on prior knowledge of the process2. empirical probability3. subjective probability

5、,,,based on a combination of an individual’s past experience, personal opinion, and analysis of a particular situation,,Assuming all outcomes are equally likely,,,probability of occurrence,probability of occurrence,4-6,E

6、xample of a priori probability,Find the probability of selecting a face card (Jack, Queen, or King) from a standard deck of 52 cards.,Example of a priori probability,When randomly selecting a day from the year 2015 what

7、is the probability the day is in January?,4-8,Example of empirical probability,Find the probability of selecting a male taking statistics from the population described in the following table:,Probability of male taking s

8、tats,Subjective probability,Subjective 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 opt

9、imistic and assigns a 40% of success to the same campaignThe assignment of a subjective probability is based on a person’s experiences, opinions, and analysis of a particular situationSubjective probability is useful i

10、n situations when an empirical or a priori probability cannot be computed,4-10,Probability,Probability is the numerical measure of the likelihood that an event will occurThe probability of any event must be between 0 an

11、d 1, inclusively,Certain,Impossible,.5,1,0,,,,,,0 ≤ P(A) ≤ 1 For any event A,4-11,Contingency table 4.2(table of cross-classifications),4-12,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,Sample Space,,,,,The Sample Space is the collec

12、tion of all possible eventse.g. All 6 faces of a die:e.g. All 52 cards of a bridge deck:,4-13,Events,Simple eventAn outcome from a sample space with one characteristice.g., A red card from a deck of cardsComple

13、ment of an event A (denoted A’)All outcomes that are not part of event Ae.g., All cards that are not diamondsJoint eventInvolves two or more characteristics simultaneouslye.g., An ace that is also red from a deck o

14、f cards,4-14,Visualizing Events,Venn DiagramsLet A = acesLet B = red cards,,,A,B,,,,A ∩ B = ace and red,,A U B = ace or red,4-15,DefinitionsSimple vs. Joint Probability,Simple Probability refers to the probability of

15、a simple event.ex. P(King)ex. P(Spade)Joint Probability refers to the probability of an occurrence of two or more events (joint event).ex. P(King and Spade),4-16,,,,,Visualizing Events,Contingency Tables Tree Di

16、agrams,,Red 2 24 26,Black 2 24 26,,,Total 4 48 52,Ace Not Ace Total,Full Deck of 52 Cards,Re

17、d Card,Black Card,,,Not an Ace,Ace,Ace,Not an Ace,,,,,,,,Sample Space,,Sample Space,,,,224224,Organizing & Visualizing Events,Venn Diagram For All Days In 2015,,,,,Sample Space (All Days In 2015),,January Days,,We

18、dnesdays,,Days That Are In January and Are Wednesdays,,,,Organizing & Visualizing Events,Contingency Tables -- For All Days in 2015Decision Trees,All Days In 2015,Not Jan.,Jan.,,,Not Wed.,Wed.,Wed.,Not Wed.,,

19、,,,Sample Space,,TotalNumberOfSampleSpaceOutcomes,,4 27 48286,(continued),Definition: Simple Probability,Simple Probability refers to the probability of a simple event.ex. P(Jan.)ex. P(Wed.),P(Jan.) = 31 / 365

20、,P(Wed.) = 52 / 365,,,,Not Wed. 27 286 313,Wed. 4 48 52,,,Total 31 334 365,Jan. Not Jan. Total,,,,,Definit

21、ion: Joint Probability,Joint Probability 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 / 365,P(Not Jan. and Not We

22、d.)= 286 / 365,,,,Not Wed. 27 286 313,Wed. 4 48 52,,,Total 31 334 365,Jan. Not Jan. Total,,,,,,Mutually e

23、xclusive eventsEvents that cannot occur simultaneouslyExample: Randomly choosing a day from 2015 A = day in January; B = day in FebruaryEvents A and B are mutually exclusive,Mutually Exclusive Events,,Collectiv

24、ely Exhaustive Events,Collectively exhaustive eventsOne of the events must occur The set of events covers the entire sample spaceExample: Randomly choose a day from 2015 A = Weekday; B = Weekend; C = Jan

25、uary; 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 exclusive,4-23,Cont

26、ingency table 4.3(table of cross-classifications),,Computing Joint and Marginal Probabilities,The probability of a joint event, A and B:Computing a marginal (or simple) probability:Where B1, B2, …, Bk are k mutu

27、ally exclusive and collectively exhaustive events,Marginal Probability Example,,,,,P(A1 and B2),P(A1),Total,Event,Marginal & Joint Probabilities In A Contingency Table,P(A2 and B1),P(A1 and B1),Event,Total,1,Joint Pr

28、obabilities,Marginal (Simple) Probabilities,A1,A2,B1,B2,P(B1),P(B2),P(A2 and B2),P(A2),,,,,,,,,,,,,,,,4-27,,Mutually Exclusive Events,Mutually exclusive eventsEvents that cannot occur togetherexample: A = queen of

29、diamonds; B = queen of clubsEvents A and B are mutually exclusive,4-28,Collectively Exhaustive Events,Collectively exhaustive events A set of mutually exclusive events is collectively exhaustive if one of th

30、e events must occur.One of the events must occur The set of events covers the entire sample space,4-29,Example:,Let A = aces; B = black cards; C = diamonds; D = hearts; E=queensEvents B and D are mutuall

31、y exclusive Events B, C and D are collectively exhaustive and also mutually exclusiveEvent A and E are mutually exclusive,4-30,,Probability,Probability is the numerical measure of the likelihood that an event will oc

32、curThe probability of any event must be between 0 and 1, inclusivelyThe sum of the probabilities of all mutually exclusive and collectively exhaustive events is 1,Certain,Impossible,.5,1,0,,,,,,0 ≤ P(A) ≤ 1 For any

33、event A,If A, B, and C are mutually exclusive and collectively exhaustive,4-31,,Computing Joint and Marginal Probabilities,The probability of a joint event, A and B:Computing a marginal (or simple) probability:Wh

34、ere B1, B2, …, Bk are k mutually exclusive and collectively exhaustive events,4-32,,,,,Joint Probability Example,P(Red and Ace),Black,,Color,Type,Red,Total,Ace,2,,2,4,Non-Ace,24,24,48,Total,26,26,52,,,,,,,,,,,,4-33,,,,,M

35、arginal Probability Example,P(Ace),Black,,Color,Type,Red,Total,Ace,2,,2,4,Non-Ace,24,24,48,Total,26,26,52,,,,,,,,,,,,4-34,,,,,P(A1 and B2),P(A1),Total,Event,Joint Probabilities Using Contingency Table,P(A2 and B1),P(A1 a

36、nd B1),Event,Total,1,Joint Probabilities,Marginal (Simple) Probabilities,A1,A2,B1,B2,P(B1),P(B2),P(A2 and B2),P(A2),,,,,,,,,,,,,,,,4-35,General Addition Rule,P(A or B) = P(A) + P(B) - P(A and B),General Addition Rule:,If

37、 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 B,4-36,General Addition Rule Example,,,,,,,P(Red or Ace) = P(Red) +P(Ace)

38、 - P(Red and Ace),= 26/52 + 4/52 - 2/52 = 28/52,Don’t count the two red aces twice!,,Black,,Color,Type,Red,Total,Ace,2,,2,4,Non-Ace,24,24,48,Total,26,26,52,,,,,,,,,,,,,4-37,Computing Conditional Probabilities,A condit

39、ional probability 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 probability of AP(B) = marginal probability of B,The conditi

40、onal probability of A given that B has occurred,,The conditional probability of B given that A has occurred,,4-38,What is the probability that a car has a CD player, given that it has AC ?i.e., we want to find P(CD

41、 | AC),Conditional Probability Example,Of 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.,4-39,,,,Conditional Probability Example,No CD,,CD,Total,AC,.

42、2,,.5,.7,No AC,.2,.1,.3,Total,.4,.6,1.0,,,,,,,,,Of 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.,,,,(continued),4-40,,Conditional Probability Exam

43、ple,No CD,,CD,Total,AC,.2,,.5,.7,No AC,.2,.1,.3,Total,.4,.6,1.0,,,,,,,,,Given AC, we only consider the top row (70% of the cars). Of these, 20% have a CD player. 20% of 70% is about 28.57%.,,,,,,(continued),4-41,,,,,Usi

44、ng Decision Trees,,,,,,,Has AC,Does not have AC,Has CD,Does not have CD,Has CD,Does not have CD,P(AC)= .7,P(AC’)= .3,P(AC and CD) = .2,P(AC and CD’) = .5,P(AC’ and CD’) = .1,P(AC’ and CD) = .2,AllCars,Given AC or no AC:

45、,4-42,,,,,Using Decision Trees,,,,,,,Has CD,Does not have CD,Has AC,Does not have AC,Has AC,Does not have AC,P(CD)= .4,P(CD’)= .6,P(CD and AC) = .2,P(CD and AC’) = .2,P(CD’ and AC’) = .1,P(CD’ and AC) = .5,AllCars,Given

46、 CD or no CD:,(continued),4-43,Statistical Independence,Two events are independent if and only if:Events A and B are independent when the probability of one event is not affected by the other event,4-44,Example,Suppo

47、se that of all individuals buying a certain digital camera, 60% include an optional memory card in their purchase, 40% include an extra battery, and 30% include both a card and battery. Consider randomly selecting a buye

48、r and let A={memory card purchased} and B={battery purchased}. Then P(A)=0.6, P(B)=0.4, and P(A∩B)=0.3. Given that the selected individual purchased an extra battery, what is the probability that an optional card was als

49、o purchased?,4-45,,Multiplication Rules,Multiplication rule for two events A and B:,Note: If A and B are independent, then,and the multiplication rule simplifies to,4-46,Example,It is known that 30% of a certain company’

50、s air conditioners require service while under warranty, whereas only 10% of its washing machines need such service. If someone purchases both a washer and air conditioner made by this company, what is the probability t

51、hat both machines need warranty service?,4-47,Marginal Probability(the Law of total Probability),Marginal probability for event A:Where B1, B2, …, Bk are k mutually exclusive and collectively exhaustive events,4-4

52、8,Example,Four individuals have responded to a request by a blood bank for blood donations. None of them has donated before, so their blood types are unknown. Suppose only type O is desired and only one of the four actua

53、lly has this type. If the potential donors are selected in random order for typing, what is the probability that at least three individuals must be typed to obtain the desired type?,4-49,Example,Each day, Monday through

54、Friday, a batch of components sent by a first supplier arrives at a certain inspection facility. Two days a week, a batch also arrives from a second supplier. Eighty percent of all supplier 1’s batches pass inspection, a

55、nd 90% of supplier 2’s do likewise. What is the probability that, on a randomly selected day, two batches pass inspection?,4-50,Bayes’ Theorem,where:Bi = ith event of k mutually exclusive and collectively exhausti

56、ve eventsA = new event that might impact P(Bi),4-51,Bayes’ Theorem Example,A drilling company has estimated a 40% chance of striking oil for their new well. A detailed test has been scheduled for more information. Hi

57、storically, 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 wel

58、l will be successful?,4-52,Let S = successful well U = unsuccessful wellP(S) = .4 , P(U) = .6 (prior probabilities)Define the detailed test event as DConditional probabilities:P(D|S) = .6 P(D|U)

59、= .2Goal is to find P(S|D),Bayes’ Theorem Example,(continued),4-53,Bayes’ Theorem Example,(continued),Apply Bayes’ Theorem:,So the revised probability of success, given that this well has been scheduled for a detailed

60、 test, is .667,,4-54,Given the detailed test, the revised probability of a successful well has risen to .667 from the original estimate of .4,Bayes’ Theorem Example,Sum = 0.36,,,,(continued),4-55,Example 4.10,Let event

61、 D=has disease event D’=does not have disease event T= test is positive event T’=test is negative,4-57,Bayes’ Theorem Example,A manufacturing firm that receives shipments of parts fr

62、om two different suppliers. Let A1 denote the event that a part is from supplier 1 and A2 denote the event that a part is form supplier 2. Currently, 65% of the parts purchased by the company are from supplier 1 and the

63、remaining 35% are from supplier 2. The quality of the purchased parts varies with the source of supply. As shown in the table,Historical quality levels of two suppliers,4-59,Suppose now that the parts from the two su

64、ppliers are used in the firm’s manufacturing process and that a machine breaks down because it attempts to process a bad part. Given the information that the part is bad, what is the probability that it came from supplie

65、r 1 and what is the probability that it came from supplier 2?,,,,,,,,,,,P(G|A1)=0.98,P(B|A1)=0.02,P(G|A2)=0.95,P(B|A2)=0.05,P(A1)=0.65,P(A2)=0.35,P= 0.6370,P=0.0130,P=0.0175,P=0.3325,,Step 1supplier,4-61,Example: Incide

66、nce of a rare disease,Only 1 in 1000 adults if afflicted with a rare disease for which a diagnostic test has been developed. The test is such that when an individual actually has the disease,a positive result will occur

67、99% of the time, whereas an individual without the disease will show a positive test result only 2% of the time. If a randomly selected individual is tested and the result is positive, what is the probability that the in

68、dividual has the disease?,4-62,Chapter Summary,Discussed basic probability conceptsSample spaces and events, contingency tables, simple probability, and joint probabilityExamined basic probability rulesGeneral additio

69、n rule, addition rule for mutually exclusive events, rule for collectively exhaustive eventsDefined conditional probabilityStatistical independence, marginal probability, decision trees, and the multiplication ruleDis