1. Review Chapter 1 - 3

2. Chapter 1 Combinatorial Analysis Basic principle of counting Permutation Combination 2

3. 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(EF  EF c ) = P(EF) + P(EF c ) 3

4. 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

5. 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

6. Combination The number of different groups of r objects that could be formed from a total of n objects. 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

7. Chapter 2 Axioms of Probability Sample space and events Axioms of probability Propositions Sample spaces having equally likely outcomes 7

8. 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

9. Axioms and Propositions Axioms 1-3 –0 ≤ P(E) ≤ 1 –P(S) = 1 –If E and F are mutually exclusive events, then Propositions –4.1: P(E c ) = 1-P(E) –4.2: –4.3: 9

10. Inclusion-exclusion identity Chap-2: Movie recommendation system, ex. 4a, 5l, HW 2.8, HW 2.9, HW 2.12, HW 2.14. 10

11. 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? 11

12. 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

13. Chapter 3. Conditional Probability and Independence Conditional probabilities Bayes’ formula Independent events Conditional probability is a probability 13

14. 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(EF  EF c ) = P(EF) + P(EF c ) = P(E|F)P(F) + P(E|F c )P(F c ) Bayeis Formula 14

15. Independent Events P(EF) = P(E)P(F) Proposition 4.1 If E and F are independent, then so are E and F c. –Chap-3: Ex 4f, 4g, 4h, 4i, 15

16. 16 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;

17. Conditional Probability is a Probability 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. A i, i = 1,2, be the event that he or she has had an accident in the i-th year. P(A 2 |A 1 ) = ? Conditioning on whether or not the policyholder is accident prone: P(A 2 |A 1 ) = P(A 2 |AA 1 )P(A|A 1 ) + P(A 2 |A c A 1 ) P(A c |A 1 ) 17

18. P(E)? P(E) = (number of outcomes in E) / (number of outcomes in S) –Sample space with equally likely events P(E) = P(EF  EF c ) = P(EF) + P(EF c ) = P(E|F)P(F) + P(E|F c )P(F c ) –Chap-3: ex 3a part 1, P(E) = 1 - P(E c ) 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(EFG  EFG c ) = P(EFG) + P(EFG c ) = P(EF|G)P(G) + P(EF|G c )P(G c ) 18

19. 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|E c )P(E c )) –Chap-3 ex. 3d, 3f, … 19