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1、Basic Probability,Chapter 4,Basic ProbabilityChapter 4,Objectives,The 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 rules,ObjectivesThe objectives for t,
2、Basic Probability Concepts,Probability 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),Basic Probability ConceptsProb,Assessing Probability,
3、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,based on a combination of an individuals past experience, personal opinion, and analysis of a particular situation,Assu
4、ming all outcomes are equally likely,probability of occurrence,probability of occurrence,Assessing ProbabilityThere are,Example of a priori probability,When randomly selecting a day from the year 2015 what is the probability the day is in January?,Example of a priori probabilit,Example of empirical
5、probability,Find the probability of selecting a male taking statistics from the population described in the following table:,Probability of male taking stats,Example of empirical probabili,Subjective probability,Subjective probability may differ from person to personA media development team assigns
6、a 60% probability of success to its new ad campaign.The chief media officer of the company is less optimistic and assigns a 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 prob
7、ability is useful in situations when an empirical or a priori probability cannot be computed,Subjective probabilitySubjecti,Events,Each possible outcome of a variable is an event.Simple eventAn event described by a single characteristice.g., A day in January from all days in 2015Joint eventAn event
8、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 that are not part of event Ae.g., All days from 2015 that are not in January,EventsEach possible outcome of,Sample Space,The Sample Space is the
9、collection of all possible eventse.g. All 6 faces of a die:e.g. All 52 cards of a bridge deck:,Sample SpaceThe Sample Space i,Organizing & Visualizing Events,Venn Diagram For All Days In 2015,Sample Space (All Days In 2015),January Days,Wednesdays,Days That Are In January and Are Wednesdays,Organizi
10、ng & Visualizing Event,Organizing & Visualizing Events,Contingency Tables - For All Days in 2015Decision Trees,All Days In 2015,Not Jan.,Jan.,Not Wed.,Wed.,Wed.,Not Wed.,Sample Space,TotalNumberOfSampleSpaceOutcomes,4 27 48286,(continued),Organizing & Visualizing Event,Definition: Simple Probability
11、,Simple Probability refers to the probability of a simple event.ex. P(Jan.)ex. P(Wed.),P(Jan.) = 31 / 365,P(Wed.) = 52 / 365,Not Wed. 27 286 313,Wed. 4 48 52,Total 31 334 365,Jan. Not Jan. Total,Definition: Simple Probabilit,Definition: Joint Probability,Joint Probability refers to the probability o
12、f 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 Wed.)= 286 / 365,Not Wed. 27 286 313,Wed. 4 48 52,Total 31 334 365,Jan. Not Jan. Total,Definition: Joint Probability,Mutually exclusive eventsEvents that
13、 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,Mutually exclusive eventsMutua,Collectively Exhaustive Events,Collectively exhaustive eventsOne of the events must occur The s
14、et of events covers the entire sample spaceExample: 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 a
15、lso mutually exclusive,Collectively Exhaustive Events,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 mutually exclusive and collectively exhaustive events,Computing Joint and Marginal,Joint Pr
16、obability Example,P(Jan. and Wed.),Joint Probability ExampleP(Jan,Marginal Probability Example,Marginal Probability ExampleP(,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 Probabilities,Marginal (Simple) Probabiliti
17、es,A1,A2,B1,B2,P(B1),P(B2),P(A2 and B2),P(A2),P(A1 and B2)P(A1)Total,Probability Summary So Far,Probability 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 c
18、ollectively exhaustive events is 1,Certain,Impossible,0.5,1,0,0 P(A) 1 For any event A,If A, B, and C are mutually exclusive and collectively exhaustive,Probability Summary So FarProb,General Addition Rule,P(A or B) = P(A) + P(B) - P(A and B),General Addition Rule:,If A and B are mutually exclusive,
19、 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,General Addition RuleP(A or B),General Addition Rule Example,P(Jan. or Wed.) = P(Jan.) + P(Wed.) - P(Jan. and Wed.),= 31/365 + 52/365 - 4/365 = 79/365,Dont count the four Wednesdays in
20、January twice!,General Addition Rule ExampleP,Computing Conditional Probabilities,A conditional 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 or simple probability of AP(B) = marginal or simple prob
21、ability of B,The conditional probability of A given that B has occurred,The conditional probability of B given that A has occurred,Computing Conditional Probabil,What 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 Example,Of t
22、he cars on a used car lot, 70% have air conditioning (AC) and 40% have a GPS. 20% of the cars have both.,What is the probability that a,Conditional Probability Example,No GPS,GPS,Total,AC,0.2,0.5,0.7,No AC,0.2,0.1,0.3,Total,0.4,0.6,1.0,Of the cars on a used car lot, 70% have air conditioning (AC) an
23、d 40% have a GPS and 20% of the cars have both.,(continued),Conditional Probability Exampl,Conditional Probability Example,No GPS,GPS,Total,AC,0.2,0.5,0.7,No AC,0.2,0.1,0.3,Total,0.4,0.6,1.0,Given AC, we only consider the top row (70% of the cars). Of these, 20% have a GPS. 20% of 70% is about 28.57
24、%.,(continued),Conditional Probability Exampl,Using Decision Trees,Has AC,Does not have AC,Has GPS,Does not have GPS,Has GPS,Does not have GPS,P(AC)= 0.7,P(AC)= 0.3,P(AC and GPS) = 0.2,P(AC and GPS) = 0.5,P(AC and GPS) = 0.1,P(AC and GPS) = 0.2,AllCars,Given AC or no AC:,ConditionalProbabilities,Usi
25、ng Decision TreesHas ACDoes,Using Decision Trees,Has GPS,Does not have GPS,Has AC,Does not have AC,Has AC,Does not have AC,P(GPS)= 0.4,P(GPS)= 0.6,P(GPS and AC) = 0.2,P(GPS and AC) = 0.2,P(GPS and AC) = 0.1,P(GPS and AC) = 0.5,AllCars,Given GPS or no GPS:,(continued),ConditionalProbabilities,Using D
26、ecision TreesHas GPSDoe,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 fact that the other event has occurred,IndependenceTwo events are ind,Multiplication Rules,Multiplication rule for two events A and B
27、:,Note: If A and B are independent, then,and the multiplication rule simplifies to,Multiplication RulesMultiplica,Marginal Probability,Marginal probability for event A:Where B1, B2, , Bk are k mutually exclusive and collectively exhaustive events,Marginal ProbabilityMarginal p,Bayes Theorem,Bayes Th
28、eorem 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 probability.,Bayes TheoremBayes Theorem i,Bayes Theorem,where:Bi = ith event of k mutually exclusive and collectively exhaustive events
29、A = new event that might impact P(Bi),Bayes Theoremwhere:,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. Historically, 60% of successful wells have had detailed tests, and 20% of unsucce
30、ssful 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?,Bayes Theorem ExampleA drilli,Let S = successful well U = unsuccessful wellP(S) = 0.4 , P(U) = 0.6 (prior probabilities)Define the detailed test
31、 event as DConditional probabilities:P(D|S) = 0.6 P(D|U) = 0.2Goal is to find P(S|D),Bayes Theorem Example,(continued),Let S = successful well Baye,Bayes Theorem Example,(continued),Apply Bayes Theorem:,So the revised probability of success, given that this well has been scheduled for a detailed tes
32、t, is 0.667,Bayes Theorem Example(continu,Given the detailed test, the revised probability of a successful well has risen to 0.667 from the original estimate of 0.4,Bayes Theorem Example,Sum = 0.36,(continued),Given the detailed test, the r,Counting Rules Are Often Useful In Computing Probabilities,
33、In many cases, there are a large number of possible outcomes.Counting rules can be used in these cases to help compute probabilities.,Counting Rules Are Often Usefu,Counting Rules,Rules for counting the number of possible outcomesCounting Rule 1:If any one of k different mutually exclusive and colle
34、ctively 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 outcomes,kn,Counting RulesRules for counti,Counting Rules,Counting Rule 2:If there are k1 events on the first trial, k2 events on
35、 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 poss
36、ibilities,(k1)(k2)(kn),(continued),Counting RulesCounting Rule 2:,Counting Rules,Counting Rule 3:The number of ways that n items can be arranged in order isExample:You have five books to put on a bookshelf. How many different ways can these books be placed on the shelf?Answer: 5! = (5)(4)(3)(2)(1) =
37、 120 different possibilities,n! = (n)(n 1)(1),(continued),Counting RulesCounting Rule 3:,Counting Rules,Counting Rule 4:Permutations: The number of ways of arranging X objects selected from n objects in order isExample:You have five books and are going to put three on a bookshelf. How many different
38、 ways can the books be ordered on the bookshelf?Answer: different possibilities,(continued),Counting RulesCounting Rule 4:,Counting Rules,Counting Rule 5:Combinations: The number of ways of selecting X objects from n objects, irrespective of order, isExample:You have five books and are going to sele
39、ct three are to read. How many different combinations are there, ignoring the order in which they are selected?Answer: different possibilities,(continued),Counting RulesCounting Rule 5:,Chapter Summary,In this chapter we covered: Understanding basic probability concepts.Understanding conditional probability Using Bayes Theorem to revise probabilitiesVarious counting rules,Chapter SummaryIn this chapter,
