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Chapter 4 Introduction to Probability

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## Chapter 4 Introduction to Probability

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**I. Basic Definitions p.150**II. Identify Sample Space with Counting Rules p.151 III. Probability of Outcome p.155 IV. Relationship and Probabilities of Events p.164 V. Conditional Probability p.171 VI. Bayes’ Theorem p.178 Chapter 4 Introduction to Probability**I. Basic Definitions p.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.**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 (n1, n2, …, nk), the number of outcomes for overall experiment = ? (n1)( n2) …( nk) Example: Toss coin twice. Example: Three sales calls (Yes, No).**II. Identify Sample Space with Counting Rules**2. Combination p.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?**II. Identify Sample Space with Counting Rules**3. Permutation p.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**III. Probability**• Definition: A measure of “chance”, 0 P 1, P(U) = 1. • 1. Probability of an outcome p.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**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**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.**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.**IV. Relationship of Events and Their Probabilities**• Contingency table for a sample with two variables • 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”**Relationships between events**• Complement of event A: Ac consists all sample points that are not in event A. (p.164) • Intersection of events A AND B: AB consists of all sample points belonging to both A and B. (p.166) • Union of events A and B: AB consists of all sample points belonging to A OR B OR both. (p.165) • Example: p.169 #23 (for revised questions) • Find events Bc, AB, and AB.**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(Mc), P(WA), P(AAc), P(WA), P(MW). Men (M) Women (W) Totals Promoted (A) 288 36 324 Not Promoted (Ac) 672 204 876 Totals 960 240 1200**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(Mc), P(WA), P(AAc), P(WA), P(MW). Simple probability (Marginal Probability) and Joint Probability Homework: p.176 #33 a, b**2. Relationship of probabilities**• Probability of complement (p.165) • P(Ac) = 1 - P(A) • Probability of intersection • Multiplication Law (coming soon) • Probability of union and the Addition Law • Addition Law • - General format: (p.166) • P(AB) = P(A) + P(B) - P(AB) • - If A and B are mutually exclusive, (p.168) • P(AB) = P(A) + P(B) • Mutually exclusive P(AB) = 0**Example: p.169 #22**Example: p.170 #26 Homework: p.169 #23, p.170 #28**V. Conditional Probability p.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)**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(AB) = P(A)P(B|A) P(BA) = ? • If events A and B are independent, (p.175) • P(AB) = P(A)P(B) • Jordan is A and Jerry is A? Jordan is the first and Jerry is the second (no tie)?**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**(2) Rules: given probabilities, find other probabilities.**Example: p.177 #36 What probabilities are given? Make the first shot: A P(A) = .89 Make the second shot: B P(B) = .89 Assume events A and B are independent. Find: a. Assume events A and B are independent, P(AB)=? b. P(AB) = ? c. P((AB)c) = 1 - P(AB) Answer: a. Assume events A and B are independent, P(AB)=P(A)P(B)=(.89)(.89)=.7921 b. P(AB) = P(A)+P(B)-P(AB)=.89+.89-.7921=.9879 c. P((AB)c) = 1 - P(AB)=1-.9879 = .0121 Homework: p.175 #30, P.176 #31**VI. Bayes’ Theorem (Two-Event Case) p.181**• When to use: • Prior Probability New info Update (Posterior) • for events Ai for event B for events Ai • P(A1) P(B|A1) P(A1|B) • P(A2) P(B|A2) P(A2|B) • Two events A1 and A2 that are mutually exclusive (P(A1A2)=0), and A1 and A2 are collectively exhaustive (P(A1)+P(A2)=1). • We often have two variables to define events on one sample space. One variable defines two events A1 and A2. Another variable defined event B.**VI. Bayes’ Theorem (Two-Event Case Formulas) p.181**• Bayes’ Theorem: • If A1 and A2 are mutually exclusive (P(A1A2)=0), and A1 and A2 are collectively exhaustive (P(A1)+P(A2)=1), then • A useful formula: P(B) = ? Why?**VI. Bayes’ Theorem (Examples)**Example: p.183 #39 Think: Why Bayes’s theorem can work for this problem? Prior Probability New info Update (Posterior) for events Ai for event B for events Ai P(A1)=.4 P(B|A1)=.2 P(A1|B)=? P(A2)=.6 P(B|A2)=.05 P(A2|B)=? P(B) = ? Two events A1 and A2 that are mutually exclusive (P(A1A2)=0), and A1 and A2 are collectively exhaustive (P(A1)+P(A2)=1). Answer: c. & d. P(B) = (.4)(.2)+(.6)(.05) = .11 P(A1|B)= (.4)(.2)/.11 = .7273 P(A2|B)= (.6)(.05)/.11 = .2727**VI. Bayes’ Theorem (Examples)**Example: p.183 #42 Think: Why Bayes’s theorem can work for this problem? Prior Probability New info Update (Posterior) for events Ai for event B for events Ai P(A1)=.05 (default) P(B|A1)=1 P(A1|B)=? P(A2)=? (not default) P(B|A2)=.20 P(A2|B)=? B: miss payments. P(B) =? We have two variables to define events on one sample space. One variable defines two events A1 and A2 that are mutually exclusive (P(A1A2)=0), and A1 and A2 are collectively exhaustive (P(A1)+P(A2)=1). Another variable defined event B. Answer: a. P(?) = .2083. Follow-up: Probability that a cardholder will not miss any payment? P(Bc)**VI. Bayes’ Theorem (General Case Formulas) p.181**• Bayes’ Theorem: • If A1, A2, …, An are mutually exclusive, and A1, A2, …, Anare collectively exhaustive, then • …... • A useful formula: P(B) = ? Why? • Homework: p.183 #41, p.184 #43