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Review Chapter 1 - 3 Chapter 1 Combinatorial Analysis • Basic principle of counting • Permutation • Combination 2 Basic Principle of Counting • Suppose that two experiments are to be performed. Then if experiment 1 can result in any one of m possible outcomes and if for each outcome of experiment 1 there are n possible outcomes of experiment 2, then together there are mn possible outcomes of the two experiments. • If experiment 1 has two possible outcomes, for the first outcome there are m possible outcomes for experiment 2, and for the second outcome, there are n possible outcomes for experiment 2, then there are m+n total possible outcomes for the two experiments. • Useful when computing probabilities. – P(EF) = P(E)P(F) – P(E) = P(EFEFc) = P(EF) + P(EFc) 3 Permutations • Possible linear ordering of n items. – n! – A class in probability theory consists of 6 men and 4 women with 2 graduate and 8 undergraduate students. An examination is given and the students are ranked according to their performance. Assume that no two students obtain the same score. How many different rankings are possible? – 10! 4 Variations of Permutation Problems • Ordering subset of students separately – If the graduate students are ranked just among themselves and undergraduate students among themselves, how many different rankings are possible? • 2!8! • Some students are indistinguishable – If we just list the gender of the students how many different rankings are possible? • 10!/(6!4!) • Some students need to be placed together – If we know that student X and Y rank next to each other, how many different rankings are possible? • 9!2! • Some students cannot be placed together – If we know that student X and Y are not ranked next to each other, how many different rankings are possible? • 10! - 9!2! 5 Combination • The number of different groups of r objects that could be formed from a total of n objects. n n! r (n r )! r! • A group of 20 people, from which a committee of 5 is to be formed. How many different committees are possible? – 20!/(15!5!) • Multiple groups – If two committee of 5 and 4, respectively, are to be formed from this group of 20 people, how many different committees are possible? • 20!/(5!4!11!) 6 Chapter 2 Axioms of Probability • Sample space and events • Axioms of probability • Propositions • Sample spaces having equally likely outcomes 7 Sample Space and Events • Sample space and events are important concepts. – List all the outcomes in a sample space or an event. – HW 2.3, 2.5, 2.7 • Venn diagram and operators of events such as intersection, union, and complement. – Chap-2 slides page 13-22, 32-33, 38 8 Axioms and Propositions • Axioms 1-3 – 0 ≤ P(E) ≤ 1 – P(S) = 1 – If E and F are mutually exclusive events, then P( E F ) P( E ) P( F ) • Propositions – 4.1: P(Ec) = 1-P(E) – 4.2: If E F , then P( E ) P( F ). – 4.3: P( E F ) P( E ) P( F ) P( EF ) 9 Inclusion-exclusion identity P( E F ) P( E ) P( F ) P( EF ) P( E F G ) P( E ) P( F ) P(G ) P( EF ) P( EG ) P( FG ) P( EFG ) • Chap-2: Movie recommendation system, ex. 4a, 5l, • HW 2.8, HW 2.9, HW 2.12, HW 2.14. 10 Sample Space Having Equally Likely Outcomes • If in an experiment, all outcomes in the sample space are equally likely to occur – P(E) = (number of outcomes in E) / (number of outcomes in S) • Most problems use the counting techniques we learned in Chapter 1. – A committee of 8 is to be selected from a group of people with 9 men and 11 women. If the selection is made randomly, what is the probability that the committee consists of 3 men and 5 women? 9 11 3 5 20 8 11 Sample Space Having Equally Likely Outcomes • P(E) = (number of outcomes in E) / (number of outcomes in S) • Chap-2: ex. 5a, 5b, 5d, 5e, 5f, 5g, 5h, 5i, 5j • HW 2.31, 2.37, 2.41 12 Chapter 3. Conditional Probability and Independence • Conditional probabilities • Bayes’ formula • Independent events • Conditional probability is a probability 13 Conditional Probabilities • Conditional probabilities are very important concept. – P(E|F) = P(EF)/P(F) – P(EF) = P(E|F)P(F) – P(E) = P(EFEFc) = P(EF) + P(EFc) = P(E|F)P(F) + P(E|Fc)P(Fc) • Bayeis Formula P( AB) P( A | B) P( B) P( B | A) P( A) P( A | B) P( B) P( A | B c ) P( B c ) P( F j | E ) P( EFj ) P( E ) P( E | F j ) P( F j ) n P( E | F ) P( F ) i 1 i i 14 Independent Events • P(EF) = P(E)P(F) • Proposition 4.1 If E and F are independent, then so are E and Fc. – Chap-3: Ex 4f, 4g, 4h, 4i, 15 • Ex 4f. An infinite sequence of independent trials is to be performed. Each trial results in a success with probability p and a failure with probability 1-p. What is the probability that (a) all trials result in successes in the first n trials? (b) at least 1 success occurs in the first n trials; (c) exactly k successes occur in the first n trials; 16 Conditional Probability is a Probability P( E1 | F ) P( E1 | E2 F ) P( E2 | F ) P( E1 | E2c F ) P( E2c | F ) • Ex 5a. Insurance company problem. • What is the conditional probability that a new policyholder will have an accident in his or her second year of policy ownership, given that the policyholder has had an accident in the first year? • A: the event that the policyholder is accident prone. • Ai, i = 1,2, be the event that he or she has had an accident in the i-th year. • P(A2|A1) = ? • Conditioning on whether or not the policyholder is accident prone: • P(A2|A1) = P(A2|AA1)P(A|A1) + P(A2|AcA1) P(Ac|A1) 17 P(E)? • P(E) = (number of outcomes in E) / (number of outcomes in S) – Sample space with equally likely events • P(E) = P(EFEFc) = P(EF) + P(EFc) = P(E|F)P(F) + P(E|Fc)P(Fc) – Chap-3: ex 3a part 1, • P(E) = 1 - P(Ec) • Conditioning on a particular event – Chap-3: ex 4h, HW 3.74 • P(EF)? – P(EF) = P(E|F)P(F) = P(F|E)P(E) – Chap-3: ex 2e – P(EF) = P(EFGEFGc) = P(EFG) + P(EFGc) = P(EF|G)P(G) + P(EF|Gc)P(Gc) 18 P(E|F)? • P(E|F) = P(EF) / P(F) • = P(F|E)P(E) / P(F) – Chap-3 ex. 3a part 2 • = P(F|E)P(E) / (P(F|E)P(E) + P(F|Ec)P(Ec)) – Chap-3 ex. 3d, 3f, … P( E1 | F ) P( E1 | E2 F ) P( E2 | F ) P( E1 | E2c F ) P( E2c | F ) 19