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Discrete Structures Chapter 4 Counting and Probability

Discrete Structures Chapter 4 Counting and Probability. Nurul Amelina Nasharuddin Multimedia Department. Outline. Rules of Sum and Product Permutations Combinations: The Binomial Theorem Combinations with Repetition: Distribution Probability. Combinations with Repetition. Example:

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Discrete Structures Chapter 4 Counting and Probability

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  1. Discrete StructuresChapter 4 Counting and Probability Nurul Amelina Nasharuddin Multimedia Department

  2. Outline • Rules of Sum and Product • Permutations • Combinations: The Binomial Theorem • Combinations with Repetition: Distribution • Probability

  3. Combinations with Repetition Example: How many ways are there to select 4 pieces of fruits from a bowl containing apples, oranges, and pears if the order does not matter, only the type of fruit matters, and there are at least 4 pieces of each type of fruit in the bowl

  4. Answer • Some of the results:

  5. Answer The number of ways to select 4 pieces of fruit = The number of ways to arrange 4 X’s and 2 |’s, which is given by = 6! / 4!(6-4)! = C(6,4) = 15 ways.

  6. Combinations with Repetition • In general, when we wish to select, with repetition, r of n distinct elements, we are considering all arrangements of r X’s and n-1 |’s and that their number is

  7. Combinations with Repetition • An r-combination of a set of n elements is an unordered selection of r elements from the set, with repetition is:

  8. Example (1) A person throwing a party wants to set out 15 assorted cans of drinks for his guests. He shops at a store that sells five different types of soft drinks. How many different selections of 15 cans can he make? (Here n = 5, r = 15)

  9. Example (1) • 4 |’s (to separate the categories of soft drinks) • 15 X’s (to represent the cans selected) = 19! / 15!(19-15)! = C(19,15) = 3876 ways.

  10. Example (2) A donut shop offers 20 kinds of donuts. Assuming that there are at least a dozen of each kind when we enter the shop, we can select a dozen donuts in (Here n = 20, r = 12). = C(31, 12) = 141,120,525 ways.

  11. Example (3) A restaurant offers 4 kinds of food. In how many ways can we choose six of the food? C(6 + 4 - 1, 6) = C(9, 6) = C(9, 3) = 9! = 84 ways. 3! 6!

  12. Which formula to use? Different ways of choosing k elements from n

  13. Counting and Probability

  14. Discrete Probability • The probability of an event is the likelihood that event will occur. • “Probability 1” means that it must happen while “probability 0” means that it cannot happen • Eg: The probability of… • “Manchester United defeat Liverpool this season” is 1 • “Liverpool win the Premier League this season” is 0 • Events which may or may not occur are assigned a number between 0 and 1.

  15. Discrete Probability Consider the following problems: • What’s the probability of tossing a coin 3 times and getting all heads or all tails? • What’s the probability that a list consisting of n distinct numbers will not be sorted?

  16. Discrete Probability • An experiment is a process that yields an outcome • A sample space is the set of all possible outcomes of a random process • An event is an outcome or combination of outcomes from an experiment • An event is a subset of a sample space • Examples of experiments: - Rolling a six-sided die - Tossing a coin

  17. Example Experiment 1: Tossing a coin. • Sample space: S = {Head or Tail} or we could write: S = {0, 1} where 0 represents a tail and 1 represents a head. Experiment 2: Tossing a coin twice • S = {HH, TT, HT, TH} where some events: • E1 = {Head}, • E2 = {Tail}, • E3 = {All heads}

  18. Definition of Probability • Suppose an event E can happen in r ways out of a total of n possible equally likely ways. • Then the probability of occurrence of the event (called its success) is denoted by • The probability of non-occurrence of the event (called its failure) is denoted by • Thus,

  19. Definition of Probabilityusing Sample Spaces • If S is a finite sample space in which all outcomes are equally likely and E is an event in S, then the probability of E, P(E), is • where N(E) is the number of outcomes in E N(S) is the total number of outcomes in S

  20. Example (1) What’s the probability of tossing a coin 3 times and getting all heads or all tails? Can consider set of ways of tossing coin 3 times: Sample space, S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Next, consider set of ways of tossing all heads or all tails: Event, E = {HHH, TTT} Assuming all outcomes equally likely to occur  P(E) = 2/8 = 0.25

  21. Example (2) • Five microprocessors are randomly selected from a lot of 1000 microprocessors among which 20 are defective. Find the probability of obtaining no defective microprocessors. There are C(1000,5) ways to select 5 micros. There are C(980,5) ways to select 5 good micros. The prob. of obtaining no defective micros is C(980,5)/C(1000,5) = 0.904

  22. Probability of Combinations of Events • Theorem: Let E1 and E2 be events in the sample space S. Then P(E1  E2) = P(E1) + P(E2) – P(E1  E2) • Eg: What is the probability that a positive integer selected at random from the set of positive integers not greater than 100 is divisible by either 2 or 5 E1: Event that the integer selected is divisible by 2 E2: Event that the integer selected is divisible by 5 P(E1  E2) = 50/100 + 20/100 – 10/100 = 3/5

  23. Exercise • If any seven digits could be used to form a telephone number, how many seven-digits telephone numbers would not have repeated digits? • How many seven-digit telephone numbers would have at least one repeated digit? • What is the probability that a randomly chosen seven-digit telephone number would have at least one repeated digit?

  24. Answer • 10 x 9 x 8 x 7 x 6 x 5 x 4 = 604800 • [no of PN with at least one digit] = [total no of PN] – [no of PN with no repeated digit] = 107 – 604800 = 9395200 • 9395200 / 107 = 0.93952

  25. Counting Elements of Sets The Principle of Inclusion/Exclusion Rule for Two or Three Sets If A, B, and C are finite sets, then N(AB) = N(A) + N(B) – N(A  B) and N(ABC) = N(A) + N(B) + N(C) – N(AB) – N(AC) – N(BC) + N(ABC)

  26. Example (1) • In a class of 50 college freshmen, 30 are studying BASIC, 25 studying PASCAL, and 10 are studying both. How many freshmen are not studying either computer language? A: set of freshmen study BASIC B: set of freshmen study PASCAL N(AB) = N(A)+N(B)-N(AB) = 30 + 25 – 10 = 45 • Not studying either: 50 – 45 =5 10 20 10 15

  27. Example (2) • A professor takes a survey to determine how many students know certain computer languages. The finding is that out of a total of 50 students in the class, 30 know Java; 18 know C++; 26 know SQL; 9 know both Java and C++; 16 know both Java and SQL; 8 know both C++ and SQL; 47 know at least one of the 3 languages.

  28. Example (2) a. How many students know none of the three languages? b. How many students know all three languages? c. How many students know Java and C++ but not SQL? How many students know Java but neither C++ nor SQL Answer: • 50 – 47 = 3 • ? • ?

  29. Example (2) • J = the set of students who know Java • C = the set of students who know C++ • S = the set of students who know SQL • Use Inclusion/Exclusion rule.

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