1 / 38

10.1 – Counting by Systematic Listing

10.1 – Counting by Systematic Listing. One-Part Tasks. The results for simple, one-part tasks can often be listed easily. . Tossing a fair coin:. Heads or tails. Rolling a single fair die. 1, 2, 3, 4, 5, 6. Consider a club N with four members:.

chinara
Télécharger la présentation

10.1 – Counting by Systematic Listing

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. 10.1 – Counting by Systematic Listing One-Part Tasks The results for simple, one-part tasks can often be listed easily. Tossing a fair coin: Heads or tails Rolling a single fair die 1, 2, 3, 4, 5, 6 Consider a club N with four members: N = {Mike, Adam, Ted, Helen} or N = {M, A, T, H} In how many ways can this group select a president? There are four possible results: M, A, T, and H.

  2. 10.1 – Counting by Systematic Listing Product Tables for Two-Part Tasks Determine the number of two-digit numbers that can be written using the digits from the set {2, 4, 6}. The task consists of two parts: 1. Choose a first digit 2. Choose a second digit The results for a two-part task can be pictured in a product table. 2 4 6 2 26 24 22 4 9 possible numbers 42 44 46 6 62 64 66

  3. 10.1 – Counting by Systematic Listing Product Tables for Two-Part Tasks What are the possible outcomes of rolling two fair die?

  4. 10.1 – Counting by Systematic Listing Product Tables for Two-Part Tasks Find the number of ways club N can elect a president and secretary. N = {Mike, Adam, Ted, Helen} or N = {M, A, T, H} The task consists of two parts: 1. Choose a president 2. Choose a secretary MA MM MT MH 12 outcomes AA AT AM AH TH TM TA TT HM HA HT HH

  5. 10.1 – Counting by Systematic Listing Product Tables for Two-Part Tasks Find the number of ways club N can elect a two member committee. N = {Mike, Adam, Ted, Helen} or N = {M, A, T, H} 6 committees MA MM MT MH AA AT AH AM TM TA TH TT HM HA HT HH

  6. 10.1 – Counting by Systematic Listing Tree Diagrams for Multiple-Part Tasks A task that has more than two parts is not easy to analyze with a product table. Another helpful device is a tree diagram. Find the number of three digit numbers that can be written using the digits from the set {2, 4, 6} assuming repeated digits are not allowed. A product table will not work for more than two digits. Generating a list could be time consuming and disorganized.

  7. 10.1 – Counting by Systematic Listing Tree Diagrams for Multiple-Part Tasks Find the number of three digit numbers that can be written using the digits from the set {2, 4, 6} assuming repeated digits are not allowed. 1st # 2nd # 3rd # 246 4 6 2 6 4 264 2 6 426 6 possibilities 4 6 2 462 2 4 624 6 4 2 642

  8. 10.1 – Counting by Systematic Listing Other Systematic Listing Methods There are additional systematic ways to produce complete listings of possible results besides product tables and tree diagrams. How many triangles (of any size) are in the figure below? D One systematic approach is begin with A, and proceed in alphabetical order to write all 3-letter combinations (like ABC, ABD, …), then cross out ones that are not triangles and those that repeat. E C F B A Another approach is to “chunk” the figure to smaller, more manageable figures. There are 12 triangles.

  9. 10.2 – Using the Fundamental Counting Principle Uniformity Criterion for Multiple-Part Tasks: A multiple part task is said to satisfy the uniformity criterion if the number of choices for any particular part is the same no matter which choices were selected for previous parts. Uniformity exists: Find the number of three letter combinations that can be written using the letters from the set {a, b, c} assuming repeated letters are not allowed. 2 dimes and one six-sided die numbered from 1 to 6 are tossed. Generate a list of the possible outcomes by drawing a tree diagram. Uniformity does not exists: A computer printer allows for optional settings with a panel of three on-off switches. Set up a tree diagram that will show how many setting are possible so that no two adjacent switches can be on?

  10. 10.2 – Using the Fundamental Counting Principle Uniformity Find the number of three letter combinations that can be written using the letters from the set {a, b, c} assuming repeated letters are not allowed. 1st letter 2nd letter 3rd letter abc b c a c b acb a c bac 6 possibilities b c a bca a b cab c b a cba

  11. 10.2 – Using the Fundamental Counting Principle Uniformity 2 dimes and one six-sided die numbered from 1 to 6 are tossed. Generate a list of the possible outcomes by drawing a tree diagram. Die # Dime 1 d1 d1 d1 d1 d1 d1 d1 1 1 d2 d2 d2 d2 d2 d2 d2 2 d1 2 2 d2 3 d1 3 3 d2 12 possibilities 4 d1 4 4 d2 5 d1 5 5d2 6 d1 6 6 d2

  12. 10.2 – Using the Fundamental Counting Principle Uniformity does not exist A computer printer is designed for optional settings with a panel of three on-off switches. Set up a tree diagram that will show how many setting are possible so that no two adjacent switches can be on? (o = on, f = off) 1st switch 2nd switch 3rd switch o o f o o f f o o f f o f f

  13. 10.2 – Using the Fundamental Counting Principle Fundamental Counting Principle The principle which states that all possible outcomes in a sample space can be found by multiplying the number of ways each event can occur. Example: At a firehouse fundraiser dinner, one can choose from 2 proteins (beef and fish), 4 vegetables (beans, broccoli, carrots, and corn), and 2 breads (rolls and biscuits). How many different protein-vegetable-bread selections can she make for dinner? Proteins Vegetables Breads 2  4  2 = 16 possible selections

  14. 10.2 – Using the Fundamental Counting Principle Example At the local sub shop, customers have a choice of the following: 3 breads (white, wheat, rye), 4 meats (turkey, ham, chicken, bologna), 6 condiments (none, brown mustard, spicy mustard, honey mustard, ketchup, mayo), and 3 cheeses (none, Swiss, American). How many different sandwiches are possible? Breads Meats Condiments Cheeses  6  3 3  4 = 216 possible sandwiches

  15. 10.2 – Using the Fundamental Counting Principle Example: Consider the set of digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. (a) How many two digit numbers can be formed if repetitions are allowed? 1st digit 2nd digit 10 90 9  = (b) How many two digit numbers can be formed if no repetitions are allowed? 1st digit 2nd digit 9 81 9  = (c) How many three digit numbers can be formed if no repetitions are allowed? 1st digit 2nd digit 3rd digit 9  9  8 = 648

  16. 10.2 – Using the Fundamental Counting Principle Example: (a) How many five-digit codes are possible if the first two digits are letters and the last three digits are numerical? 1st digit 2nd digit 3rd digit 4th digit 5th digit  26 10 10 26    10 676000 possible five-digit codes (a) How many five-digit codes are possible if the first two digits are letters and the last three digits are numerical and repeats are not permitted? 1st digit 2nd digit 3rd digit 4th digit 5th digit  25 10 9 26    8 468000 possible five-digit codes

  17. 10.2 – Using the Fundamental Counting Principle Factorials For any counting number n, the product of all counting numbers from n down through 1 is called n factorial, and is denoted n!. For any counting number n, the quantity n factorial is calculated by: n! = n(n – 1)(n – 2)…(2)(1). Definition of Zero Factorial: 0! = 1 Examples: b) (4 – 1)! a) 4! c) 3! 4321 321 20 = 24 = 54 6

  18. 10.2 – Using the Fundamental Counting Principle Arrangements of Objects Factorials are used when finding the total number of ways to arrange a given number of distinct objects. The total number of different ways to arrange n distinct objects is n!. Example: How many ways can you line up 6 different books on a shelf?   1 5 4 3 6    2 720 possible arrangements

  19. 10.2 – Using the Fundamental Counting Principle Arrangements of n Objects Containing Look-Alikes The number of distinguishable arrangements of n objects, where one or more subsets consist of look-alikes (say n1 are of one kind, n2 are of another kind, …, and nk are of yet another kind), is given by Example: Determine the number of distinguishable arrangements of the letters of the word INITIALLY. 9 letters with 3 I’s and 2 L’s 9! 30240 possible arrangements  3! 2!

  20. 10.3 – Using Permutations and Combinations Permutation: The number of ways in which a subset of objects can be selected from a given set of objects, where order is important. Given the set of three letters, {A, B, C}, how many possibilities are there for selecting any two letters where order is important? (AB, AC, BC, BA, CA, CB) Combination: The number of ways in which a subset of objects can be selected from a given set of objects, where order is not important. Given the set of three letters, {A, B, C}, how many possibilities are there for selecting any two letters where order is not important? (AB, AC, BC).

  21. 10.3 – Using Permutations and Combinations Factorial Formula for Permutations Factorial Formula for Combinations

  22. 10.3 – Using Permutations and Combinations Evaluate each problem. d) 6C6 b) 5C3 c) 6P6 a) 5P3 543 60 720 10 1

  23. 10.3 – Using Permutations and Combinations How many ways can you select two letters followed by three digits for an ID if repeats are not allowed? Two parts: 1. Determine the set of two letters. 2. Determine the set of three digits. 26P2 10P3 2625 1098 650 720 650720 468,000

  24. 10.3 – Using Permutations and Combinations A common form of poker involves hands (sets) of five cards each, dealt from a deck consisting of 52 different cards. How many different 5-card hands are possible? Hint: Repetitions are not allowed and order is not important. 52C5 2,598,960 5-card hands

  25. 10.3 – Using Permutations and Combinations Find the number of different subsets of size 3 in the set: {m, a, t, h, r, o, c, k, s}. Find the number of arrangements of size 3 in the set: {m, a, t, h, r, o, c, k, s}. 9C3 9P3 987 504 arrangements 84 Different subsets

  26. 10.3 – Using Permutations and Combinations Guidelines on Which Method to Use

  27. 11.1 – Probability – Basic Concepts Probability The study of the occurrence of random events or phenomena. It does not deal with guarantees, but with the likelihood of an occurrence of an event. Experiment: - Any observation or measurement of a random phenomenon. Outcomes: - The possible results of an experiment. Sample Space: - The set of all possible outcomes of an experiment. Event: - A particular collection of possible outcomes from a sample space.

  28. 11.1 – Probability – Basic Concepts Example: If a single fair coin is tossed, what is the probability that it will land heads up? Sample Space: S = {h, t} Event of Interest: E = {h} P(heads) = P(E) = 1/2 The probability obtained is theoretical as no coin was actually flipped Theoretical Probability: number of favorable outcomes n(E) = P(E) = total number of outcomes n(S)

  29. 11.1 – Probability – Basic Concepts Example: A cup is flipped 100 times. It lands on its side 84 times, on its bottom 6 times, and on its top 10 times. What is the probability that it lands on it top? 10 1 number of top outcomes = = P(top) = total number of flips 100 10 The probability obtained is experimental or empirical as the cup was actually flipped. Empirical or Experimental Probability: number of times event E occurs P(E) ͌ number of times the experiment was performed

  30. 11.1 – Probability – Basic Concepts Example: There are 2,598,960 possible five-card hand in poker. If there are 36 possible ways for a straight flush to occur, what is the probability of being dealt a straight flush? number of possible straight flushes P(straight flush) = total number of five-card hands 36 = = 0.0000139 2,598,960 This probability is theoretical as no cards were dealt.

  31. 11.1 – Probability – Basic Concepts Example: A school has 820 male students and 835 female students. If a student is selected at random, what is the probability that the student would be a female? number of possible female students P(female) = total number of students 835 167 835 = = = 331 1655 820 + 835 0.505 P(female) = This probability is theoretical as no experiment was performed.

  32. 11.1 – Probability – Basic Concepts The Law of Large Numbers As an experiment is repeated many times over, the experimental probability of the events will tend closer and closer to the theoretical probability of the events. Flipping a coin Spinner Rolling a die

  33. 11.1 – Probability – Basic Concepts Odds A comparison of the number of favorable outcomes to the number of unfavorable outcomes. Odds are used mainly in horse racing, dog racing, lotteries and other gambling games/events. Odds in Favor: number of favorable outcomes (A) to the number of unfavorable outcomes (B). A to B A : B Example: What are the odds in favor of rolling a 2 on a fair six-sided die? 1 : 5 What is the probability of rolling a 2 on a fair six-sided die? 1/6

  34. 11.1 – Probability – Basic Concepts Odds Odds against: number of unfavorable outcomes (B) to the number of favorable outcomes (A). B to A B : A Example: What are the odds against rolling a 2 on a fair six-sided die? 5 : 1 What is the probability against rolling a 2 on a fair six-sided die? 5/6

  35. 11.1 – Probability – Basic Concepts Odds Example: Two hundred tickets were sold for a drawing to win a new television. If you purchased 10 tickets, what are the odds in favor of you winning the television? 10 Favorable outcomes 200 – 10 = 190 Unfavorable outcomes 10 : 190 = 1 : 19 What is the probability of winning the television? 10/200 = 1/20 = 0.05

  36. 11.1 – Probability – Basic Concepts Converting Probability to Odds Example: The probability of rain today is 0.43. What are the odds of rain today? P(rain) = 0.43 Of the 100 total outcomes, 43 are favorable for rain. Unfavorable outcomes: 100 – 43 = 57 43 : 57 The odds for rain today: 57 : 43 The odds against rain today:

  37. 11.1 – Probability – Basic Concepts Converting Odds to Probability Example: The odds of completing a college English course are 16 to 9. What is the probability that a student will complete the course? 16 : 9 The odds for completing the course: Favorable outcomes + unfavorable outcomes = total outcomes 16 + 9 = 25 P(completing the course) = = 0.64

More Related