Probability, Statistic & Random Process - Lecture 2 - Counting.pdf
- 1. Counting techniques 1 / 45
- 2. Why need counting • Suppose (Ω, P) has equally likely outcomes. • To calculate P(E), we need to count the number of elements in E and Ω. • Need to learn counting technique. 2 / 45
- 3. Why need counting • Suppose (Ω, P) has equally likely outcomes. • To calculate P(E), we need to count the number of elements in E and Ω. • Need to learn counting technique. 2 / 45
- 4. Why need counting • Suppose (Ω, P) has equally likely outcomes. • To calculate P(E), we need to count the number of elements in E and Ω. • Need to learn counting technique. 2 / 45
- 5. Table of contents 1 A Fundamental Principle of Count- ing 2 Venn Diagrams and Counting 3 The Multiplication Principle 4 Permutations and Combinations 3 / 45
- 6. Notation • Denote number of elements in A by n(A) Example If A = {1, 3, 9} then n(A) = 3 • if A = ∅ then n(A) = 0 4 / 45
- 7. Notation • Denote number of elements in A by n(A) Example If A = {1, 3, 9} then n(A) = 3 • if A = ∅ then n(A) = 0 4 / 45
- 8. Notation • Denote number of elements in A by n(A) Example If A = {1, 3, 9} then n(A) = 3 • if A = ∅ then n(A) = 0 4 / 45
- 9. Inclusion–Exclusion Principle n(A ∪ B) = n(A) + n(B) − n(A ∩ B) 5 / 45
- 10. Proof • n(A ∪ B): the number of elements in either A or B • In n(A) + n(B), an element lies in both A and B, it is counted twice • To make up for this double counting, we must subtract the number of elements counted twice, namely, n(A ∩ B) • So n(A ∪ B) = n(A) + n(B) − n(A ∩ B)6 / 45
- 11. Example In the year 2012, Executive magazine surveyed the presidents of the 500 largest corporations in the United States. Of these 500 people, 310 had degrees (of any sort) in business, 238 had under- graduate degrees in business, and 184 had graduate degrees in business. How many presidents had both undergrad- uate and graduate degrees in business 7 / 45
- 12. Solution • A = {presidents with an undergraduate degree in business} • B ={presidents with a graduate degree in business}. then A ∪ B = {presidents with at least one degree in business} and A ∩ B = {presidents had both undergraduate and graduate degrees in business} 8 / 45
- 13. We have n(A) = 238, n(B) = 184, n(A∪B) = 310 By inclusion - exclusion principle, we have n(A ∪ B) = n(A) + n(B) − n(A ∩ B) So n(A∩B) = n(A)+n(B)−n(A∪B) = 238+1 9 / 45
- 14. Table of contents 1 A Fundamental Principle of Count- ing 2 Venn Diagrams and Counting 3 The Multiplication Principle 4 Permutations and Combinations 10 / 45
- 15. Basic regions Each Venn diagram divides the sam- ple space U into a certain number of regions. 11 / 45
- 16. Example Let • U = {Nobel laureates during the period 1901–2011} • A = American Nobel laureates during the period 1901–2011 • C = Chemistry Nobel laureates during the period 1901–2011 • P = Nobel Peace Prize laureates during the period 1901–2011 12 / 45
- 17. How many Americans received a No- bel Prize during the period 1901–2011? 13 / 45
- 18. Solution The number of Americans who received a Nobel Prize is the total contained in the circle A , which is 236+26+1+60 = 323. 14 / 45
- 19. Practice 1 How many Americans received Nobel Prizes in fields other than chemistry and peace during this period? 2 How many Americans received the Nobel Peace Prize during this period? 15 / 45
- 20. Practice Among 170 clients using media, there are 115 use television (T), 100 use radio (R), 130 use magazines (M), 75 use television and radio, 95 use radio and magazines, 85 use television and magazines, 70 use all three Complete the Venn diagram to display the clients’ use of mass media. 16 / 45
- 21. Table of contents 1 A Fundamental Principle of Count- ing 2 Venn Diagrams and Counting 3 The Multiplication Principle 4 Permutations and Combinations 17 / 45
- 22. Example - Routes between Cities Suppose that there are three different routes from city A to city B and five different routes from city B to city C. Count the number of different routes from A to C that pass through B 18 / 45
- 23. Solution All possible routes is (1, 4) (1, 5) (1, 6) (1, 7) (1, 8) (2, 4) (2, 5) (2, 6) (2, 7) (2, 8) (3, 4) (3, 5) (3, 6) (3, 7) (3, 8) The number of routes is 3 × 5 = 15 19 / 45
- 24. Example - Experiment in Two Parts Consider an experiment that has the following two characteristics: i The experiment is performed in two parts. ii The first part of the experiment has m possible outcomes x1, . . . , xm, and, regardless of which one of these outcomes xi occurs, the second part of the experiment has n possible outcomes y1, . . . , yn. 20 / 45
- 25. Each outcome in the sample space Ω of such an experiment will therefore be a pair having the form (xi, yj) Ω = (x1, y1) (x1, y2) . . . (x1, yn) (x2, y1) (x2, y2) . . . (x2, yn) . . . (xm, y1) (xm, y2) . . . (xm, yn) Total number of outcomes in Ω is n(Ω) = mn 21 / 45
- 26. Tree diagram in which end-nodes represent outcomes 22 / 45
- 27. Multiplication rule • Suppose there is a job that has 2 steps • There are m ways to do step 1 • There are n ways to do step 2 • There are m × n ways to do the job. 23 / 45
- 28. Multiplication rule • Suppose there is a job that has 2 steps • There are m ways to do step 1 • There are n ways to do step 2 • There are m × n ways to do the job. 23 / 45
- 29. Multiplication rule • Suppose there is a job that has 2 steps • There are m ways to do step 1 • There are n ways to do step 2 • There are m × n ways to do the job. 23 / 45
- 30. Multiplication rule • Suppose there is a job that has 2 steps • There are m ways to do step 1 • There are n ways to do step 2 • There are m × n ways to do the job. 23 / 45
- 31. General multiplication rule Suppose that a task consists of k steps performed consecutively. Suppose that step 1 can be performed in m1 ways; for each of these, step 2 in m2 ways; for each of these, step 3 in m3 ways; and so forth. Then the task can be performed in m = m1m2 . . . mk ways 24 / 45
- 32. Example A corporation has a board of direc- tors consisting of 10 members. The board must select from among its mem- bers a chairperson, vice chairperson, and secretary. In how many ways can this be done? 25 / 45
- 33. Solution • The task Select the three officers can be divided into three consecutive steps Select chairperson → Select vice chairperson → Select secretary • Step 1: there are 10 ways to select chairperson • After select chairperson, there are 9 directors left • Step 2: three are 9 ways to select vice chairperson 26 / 45
- 34. • After this step, there are 8 directors who are possible candidates for secretary • Step 3: there are 8 ways to select secretary • the number of possible ways to perform the sequence of three choices equals 10 .9 . 8, or 720 • So the officers of the board can be selected in 720 ways. 27 / 45
- 35. Sequential selection process Number of ways is 10 × 9 × 8 = 720 28 / 45
- 36. Example There are 6 balls of different colors. How many ways to arrange them on a straight line? 29 / 45
- 37. Solution • Think of a line with 6 positions • 6 ways to choose ball for 1st position • 5 ways for 2nd position, 4 for 3rd ... • Total 6 · 5 · 4 . . . 2 · 1 = 6! = 720 ways 30 / 45
- 38. Solution • Think of a line with 6 positions • 6 ways to choose ball for 1st position • 5 ways for 2nd position, 4 for 3rd ... • Total 6 · 5 · 4 . . . 2 · 1 = 6! = 720 ways 30 / 45
- 39. Solution • Think of a line with 6 positions • 6 ways to choose ball for 1st position • 5 ways for 2nd position, 4 for 3rd ... • Total 6 · 5 · 4 . . . 2 · 1 = 6! = 720 ways 30 / 45
- 40. Solution • Think of a line with 6 positions • 6 ways to choose ball for 1st position • 5 ways for 2nd position, 4 for 3rd ... • Total 6 · 5 · 4 . . . 2 · 1 = 6! = 720 ways 30 / 45
- 41. Sequential selection process Number of ways is 6 · 5 · 4 . . . 2 · 1 = 6! = 720 31 / 45
- 42. Practice A telecommunications channel trans- mits digits that are either 0 or 1. A se- quence of four digits is transmitted. 32 / 45
- 43. Assuming that the outcomes are equiprob able. Find the probability that 1 all transmitted digits are the same? 2 no two consecutive digits are the same? 3 there are three successive 1’s transmitted? 4 there are exactly three successive 1’s transmitted? 33 / 45
- 44. Table of contents 1 A Fundamental Principle of Count- ing 2 Venn Diagrams and Counting 3 The Multiplication Principle 4 Permutations and Combinations 34 / 45
- 45. Example Problem A How many words (by which we mean strings of letters) of two distinct letters can be formed from the letters a, b, c? Problem B Problem B A construction crew has three members. A team of two must be chosen for a particular job. In 35 / 45
- 46. Solution Enumerate all possibilities Problem A There are six possible words ab ac ba bc ca cb Problem B Designate the three crew members by a, b, and c. Then there are three possible two-person teams 36 / 45
- 47. Both problems are concerned with count- ing the numbers of arrangements of the elements of the set {a, b, c}, taken two at a time, without allowing repe- tition. (For example, aa was not al- lowed.) However, in Problem A, the order of the arrangement mattered, whereas in Problem B it did not. Ar- rangements of the sort considered in Problem A are called permutations, whereas those in Problem B are called combinations. 37 / 45
- 48. Permutation Given a set of n distinguishable ob- jects. Then a permutation of n ob- jects taken r at a time is an arrange- ment of r of the n objects in a specific order. Combination A combination of n objects taken r at a time is a selection of r objects from among the n, with order disre- garded. 38 / 45
- 49. Permutation formula The number of permutations of n ob- jects taken r at a time is P(n, r) = n(n − 1) . . . (n − r + 1) 39 / 45
- 50. Combination formula The number of combinations of n ob- jects taken r at a time is C(n, r) = P(n, r) r! = n . . . (n − r + 1) r(r − 1) . . . 1 Another convenient formula C(n, r) = n! r!(n − r)! Alternative notation of C(n, r) is n r read ”n choose r” 40 / 45
- 51. Example The board of directors of a corpora- tion has 10 members. In how many ways can they choose a committee of three board members to negotiate a merger? 41 / 45
- 52. Example The board of directors of a corpora- tion has 10 members. In how many ways can they choose a committee of three board members to negotiate a merger? 41 / 45
- 53. Solution Since the committee of three involves no ordering of its members, we are concerned here with combinations. The number of combinations of 10 people taken 3 at a time is C(10, 3) = 120 42 / 45
- 54. Example Eight horses are entered in a race in which a first, second, and third prize will be awarded. Assuming no ties, how many different outcomes are pos- sible? 42 / 45
- 55. Example Eight horses are entered in a race in which a first, second, and third prize will be awarded. Assuming no ties, how many different outcomes are pos- sible? 42 / 45
- 56. Solution we are considering ordered arrangements of three horses, so we are dealing with permutations. The number of permutations of eight horses taken three at a time is P(8, 3) = 8.7.6 = 336 43 / 45
- 57. Practice A club has 10 members. In how many ways can they choose a slate of four officers, consisting of a president, vice president, secretary, and treasurer? 43 / 45
- 58. Practice Suppose that an experiment consists of tossing a coin 10 times and observ- ing the sequence of heads and tails. 1 How many different outcomes are possible? 2 How many different outcomes have exactly four heads? 44 / 45
- 59. Practice An urn contains 25 numbered balls, of which 15 are red and 10 are white. A sample of 5 balls is to be selected. 1 How many different samples are possible? 2 How many samples contain all red balls? 3 How many samples contain 3 red balls and 2 white balls? 4 How many samples contain at least 4 red balls? 45 / 45