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Chapter 4 Introduction to Probability I. Basic Definitionsp.150 II. Identify Sample Space with Counting Rules p.151 III. Probability of Outcomep.155 IV. Relationship and Probabilities of Events p.164 V. Conditional Probabilityp.171 VI. Bayes’ Theoremp.178

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I. Basic Definitionsp.150 Experiment: a process that generates well-defined outcomes (how many, what, mutually exclusive). (p.150) Example: Toss a coin, roll a die, select a part for inspection. Sample Space (U for Universe): all possible outcomes for an experiment.(p.150) Example: Toss a coin: {Head, Tail} Sample Point: a particular outcome in the sample space.(p.150) Example: Toss a coin: two sample points.

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II. Identify Sample Space with Counting Rules 1. Counting Rule for Multiple-Step Experiment p.151 It is a sequence of k independent sub-experiments (steps). Given the number of outcomes for each step (n 1, n 2, …, n k ), the number of outcomes for overall experiment = ?(n 1 )( n 2 ) …( n k ) Example: Toss coin twice. Example: Three sales calls (Yes, No).

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II. Identify Sample Space with Counting Rules 2. Combinationp.154 The experiment is to select n objects from a set of N different objects. Each combination is an outcome. The number of outcomes = ? Factorial: N!=N(N-1)…(2)(1)n!=n(n-1)…(2)(1) 0! = 1 Example: Elect 2 committee members from 3 professors. How many election results can be?

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II. Identify Sample Space with Counting Rules 3. Permutationp.154 The experiment is to select n objects from a set of N different objects where the order of selection is important. Each permutation is an outcome. The number of outcomes = ? Example: Elect one committee chair and one member from 3 professors. How many election results can be? Homework: p.158 #1, #2, #3

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III. Probability Definition: A measure of “chance”, 0 P 1, P(U) = 1. 1. Probability of an outcomep.155 Classical Method: outcomes are equally likely to occur.P(A) = 1/n n: the # of all possible outcomes A: Outcome A. Relative Frequency Method: empirical probability is a frequency obtained from a sample or historical data.P(A) = x/n n: sample size, the # of elements. x: Outcome A occurred x times. Subjective Method

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III. Probability 1. Probability of an outcome (Examples) (1) “Equally likely" Example: Toss a coin.P(Head) = ? Example: Roll a die.P(1) = ? (2) Relative frequency Example: A sample of 40 students and 5 students got “A”. (1) P(A) = ? (What is the probability that a student will get an A in this class?) (2) What is the probability that a student will not have an A? Homework: p.160 #13

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III. Probability 2. Probability of an event p.160 Event: a collection of sample points (outcomes). (p.160) As previous definition, all outcomes are mutually exclusive. For Event A,(p.161) (1) Generally, P(A) = (Sample point i in event A) (2) If all sample points (outcomes) in the sample space are equally likely to occur, then P(A) = x/n n: the # of all sample points (outcomes) in sample space. x: the number of sample points for the event A.

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III. Probability (continued) 2. Probability of an event (Examples) Example: Roll a die. What is the probability that the experiment ends with 3 points or less? Example: p.164 #20 Homework: p.162 #14, p.163 #18.

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IV. Relationship of Events and Their Probabilities 1.Contingency table for a sample with two variables 2.Relationship of events - Given probabilities, find other probabilities (1) Relationships between events Complement Intersection Union (2) Rules Probability of complement Addition law for “AND” Conditional probability and multiplication law for “OR”

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Relationships between events Complement of event A: A c consists all sample points that are not in event A. (p.164) Intersection of events A AND B: A B consists of all sample points belonging to both A and B. (p.166) Union of events A and B: A B consists of all sample points belonging to A OR B OR both. (p.165) Example: p.169 #23(for revised questions) Find events B c, A B, and A B.

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1. Contingency Table Approach Example: p.171 Table 4.4 Given: A sample of 1200 police officers. Two characteristics: Gender, Promotion. (1) Use characteristics to define events. (2) Find probabilities for events: P(M), P(M c ), P(W A), P(A A c ), P(W A), P(M W). Men (M)Women (W)Totals Promoted(A) 288 36 324 Not Promoted(A c ) 672 204 876 Totals 960 2401200

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Summary: contingency table for empirical probability (1) Use characteristics to define events in contingency table: Simple events: events defined by one characteristic. Based on one characteristic, all elements are assigned to mutually exclusive events. Joint events: events defined by two characteristics. (2) Find probabilities for events: P(M), P(M c ), P(W A), P(A A c ), P(W A), P(M W). Simple probability (Marginal Probability) and Joint Probability Homework: p.176 #33 a, b

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2. Relationship of probabilities Probability of complement (p.165) P(A c ) = 1 - P(A) Probability of intersection Multiplication Law (coming soon) Probability of union and the Addition Law Addition Law - General format: (p.166) P(A B) = P(A) + P(B) - P(A B) - If A and B are mutually exclusive, (p.168) P(A B) = P(A) + P(B) Mutually exclusive P(A B) = 0

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Example: p.169 #22 Example: p.170 #26 Homework: p.169 #23, p.170 #28

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V. Conditional Probabilityp.171 1. Conditional Probability Condition provides more information - “Given”, “If”, “of”, “among”. Condition may lead to a restricted sample space. Condition versus Intersection: restricted sample space and one event; two events. Example: Take one student from my class at random. (1) If the student is a junior, what is the probability for an A-student? (2) What is the probability of a student being a junior and an A-student? 2. Multiplication Law for Intersection (AND)

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V. Conditional Probability(Formulas) 1. Conditional Probability (p.173) P(B|A) = ? Independent events if P(A|B) = P(A)(p.174) - A card is King and a card is King given Club? 2. Multiplication Law for Intersection (AND)(p.174) P(A B) = P(A)P(B|A)P(B A) = ? If events A and B are independent, (p.175) P(A B) = P(A)P(B) Jordan is A and Jerry is A? Jordan is the first and Jerry is the second (no tie)?

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Two Approaches: (1) Contingency Table Approach Example: p.171 Table 4.4 Given: a sample in contingency table. Find: (1) Probability that an officer is a man. (.8) (2) Probability that an officer is a man who got promotion. (.24) (3) Probability that an officer is promoted given that the officer is a man. (.3) (4) Probability that an officer is a man given that the officer is not promoted.(.7671) (5) Are events “man” and “promotion” independent? Homework: p.176 #33

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(2) Rules: given probabilities, find other probabilities. Example: p.177 #36 What probabilities are given? Make the first shot: AP(A) =.89 Make the second shot: BP(B) =.89 Assume events A and B are independent. Find: a. Assume events A and B are independent, P(A B)=? b. P(A B) = ? c. P((A B) c ) = 1 - P(A B) Answer: a. Assume events A and B are independent, P(A B)=P(A)P(B)=(.89)(.89)=.7921 b. P(A B) = P(A)+P(B)-P(A B)=.89+.89-.7921=.9879 c. P((A B) c ) = 1 - P(A B)=1-.9879 =.0121 Homework: p.175 #30, P.176 #31

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VI. Bayes’ Theorem (Two-Event Case) p.181 When to use: Prior ProbabilityNew infoUpdate (Posterior) for events A i for event B for events A i P(A 1 ) P(B|A 1 ) P(A 1 |B) P(A 2 ) P(B|A 2 ) P(A 2 |B) Two events A 1 and A 2 that are mutually exclusive (P(A 1 A 2 )=0), and A 1 and A 2 are collectively exhaustive (P(A 1 )+P(A 2 )=1). We often have two variables to define events on one sample space. One variable defines two events A 1 and A 2. Another variable defined event B.

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VI. Bayes’ Theorem (Two-Event Case Formulas) p.181 Bayes’ Theorem: If A 1 and A 2 are mutually exclusive (P(A 1 A 2 )=0), and A 1 and A 2 are collectively exhaustive (P(A 1 )+P(A 2 )=1), then A useful formula: P(B) = ? Why?

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VI. Bayes’ Theorem (Examples) Example: p.183 #39 Think: Why Bayes’s theorem can work for this problem? Prior ProbabilityNew infoUpdate (Posterior) for events A i for event B for events A i P(A 1 )=.4 P(B|A 1 )=.2 P(A 1 |B)=? P(A 2 )=.6 P(B|A 2 )=.05 P(A 2 |B)=? P(B) = ? Two events A 1 and A 2 that are mutually exclusive (P(A 1 A 2 )=0), and A 1 and A 2 are collectively exhaustive (P(A 1 )+P(A 2 )=1). Answer: c. & d. P(B) = (.4)(.2)+(.6)(.05) =.11 P(A 1 |B)= (.4)(.2)/.11 =.7273 P(A 2 |B)= (.6)(.05)/.11 =.2727

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VI. Bayes’ Theorem (Examples) Example: p.183 #42 Think: Why Bayes’s theorem can work for this problem? Prior ProbabilityNew infoUpdate (Posterior) for events A i for event B for events A i P(A 1 )=.05 (default) P(B|A 1 )=1 P(A 1 |B)=? P(A 2 )=? (not default) P(B|A 2 )=.20 P(A 2 |B)=? B: miss payments. P(B) =? We have two variables to define events on one sample space. One variable defines two events A 1 and A 2 that are mutually exclusive (P(A 1 A 2 )=0), and A 1 and A 2 are collectively exhaustive (P(A 1 )+P(A 2 )=1). Another variable defined event B. Answer: a. P(?) =.2083. Follow-up: Probability that a cardholder will not miss any payment? P(B c )

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VI. Bayes’ Theorem (General Case Formulas) p.181 Bayes’ Theorem: If A 1, A 2, …, A n are mutually exclusive, and A 1, A 2, …, A n are collectively exhaustive, then …... A useful formula: P(B) = ? Why? Homework: p.183 #41, p.184 #43

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