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Finding Probability Using Tree Diagrams and Outcome Tables

Finding Probability Using Tree Diagrams and Outcome Tables. Chapter 4.5 – Introduction to Probability Mathematics of Data Management (Nelson) MDM 4U. H. H. T. H. T. Tree Diagrams.

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Finding Probability Using Tree Diagrams and Outcome Tables

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  1. Finding Probability Using Tree Diagrams and Outcome Tables Chapter 4.5 – Introduction to Probability Mathematics of Data Management (Nelson) MDM 4U

  2. H H T H T Tree Diagrams • if you flip a coin twice, you can model the possible outcomes using a tree diagram or an outcome table resulting in 4 possible outcomes T

  3. H H H H H H T T T T T T Tree Diagrams Continued • if you rolled 1 die and then flipped a coin you have 12 possible outcomes (1,H) 1 (1,T) (2,H) 2 (2,T) (3,H) 3 (3,T) (4,H) (4,T) 4 (5,H) 5 (5,T) (6,H) 6 (6,T)

  4. Sample Space • the sample space for the last experiment would be all the ordered pairs in the form (d,c), where d represents the roll of a die and c represents the flip of a coin • clearly there are 12 possible outcomes (6 x 2) • P(odd roll,head) = ? • there are 3 possible outcomes for an odd die and a head • so the probability is 3 in 12 or ¼ • P(odd roll, head) = ¼

  5. Multiplicative Principle for Counting • The total number of outcomes is the product of the number of possible outcomes at each step in the sequence • if a is selected from A, and b selected from B… • n (a,b) = n(A) x n(B) • (this assumes that each outcome has no influence on the next outcome) • How many possible three letter ‘words’ are there? • you can choose 26 letters for each of the three positions, so there are 26 x 26 x 26 = 17576 • how many possible postal codes are there in Canada? • 26 x 10 x 26 x 10 x 26 x 10 =17 576 000

  6. Independent and Dependent Events • two events are independent of each other if an occurence of one event does not change the probability of the occurrence of the other • what is the probability of getting heads when you know in advance that you will throw an even die? • these are independent events, so knowing the outcome of the second does not change the probability of the first

  7. Multiplicative Principle for Probability of Independent Events • If we know that if A and B are independent events, then… • P(B | A) = P(B) • if this is not true, then the events are dependent • we can also prove that if two events are independent the probability of both occurring is… • P(A and B) = P(A) × P(B)

  8. R R B G R B B G R G B G Example 1 • a sock drawer has a red, a green and a blue sock • you pull out one sock, replace it and pull another out a) draw a tree diagram representing the possible outcomes b) what is the probability of drawing 2 red socks? • these are independent events

  9. Example 2 • a) If you draw a card, replace it and draw another, what is the probability of getting two aces? • 4/52 x 4/52 • These are independent events • b) If you draw an ace and then draw a second card (“without replacement”), what is the probability of two aces? • 4/52 x 3/51 • second event depends on first event • the sample space is reduced by the first event

  10. Example 3 - Predicting Outcomes • Mr. Lieff is playing Texas Hold’Em • He finds that he wins 70% of the pots when he does not bluff • He also finds that he wins 50% of the pots when he does bluff • If there is a 60% chance that Mr. Lieff will bluff on his next hand, what are his chances of winning the pot? • We will start by creating a tree diagram

  11. Tree Diagram P=0.6 x 0.5 = 0.3 Win pot 0.5 bluff 0.6 0.5 P=0.6 x 0.5 = 0.3 Lose pot Win pot 0.7 P=0.4 x 0.7 = 0.28 0.4 no bluff P=0.4 x 0.3 = 0.12 0.3 Lose pot

  12. Continued… • P(no bluff, win) = P(no bluff) x P(win | no bluff) • = 0.4 x 0.7 = 0.28 • P(bluff, win) = P(bluff) x P(win | bluff) • = 0.6 x 0.5 = 0.30 • Probability of a win: 0.28 + 0.30 = 0.58 • So Mr. Lieff has a 58% chance of winning the next pot

  13. Exercises • Read the examples on pages 239-244 • Try pp. 245 – 249 #2, 3, 5, 7, 9, 12, 13a, 14 • Check each answer with the back of the book to make sure that you are understanding these concepts

  14. Warm up • How many different outcomes are there if a 20-sided die is rolled then a spinner with 5 sections is spun? • 20 x 5 = 100

  15. Counting Techniques and Probability Strategies - Permutations Chapter 4.6 – Introduction to Probability Mathematics of Data Management (Nelson) MDM 4U

  16. Arrangements of objects • Suppose you have three people in a line • How many different arrangements are there? • It turns out that there are 6 • How many arrangements are there for blocks of different colours? • How many for 4 blocks? • How many for 5 blocks? • How many for 6 blocks? • What is the pattern?

  17. Selecting When Order Matters • When order matters, we have fewer choices for later places in the arrangements • For the problem of 3 people: • For person 1 we have 3 choices • For person 2 we have 2 choices left • For person 3 we have one choice left • The number of possible arrangements for 3 people is 3 x 2 x 1 = 6 • There is a mathematical notation for this (and your calculator has it)

  18. Factorial Notation • The notation is called factorial • n! is the number of ways of arranging n unique objects when order matters • n! = n x (n – 1) x (n – 2) x … x 2 x 1 • for example: • 3! = 3 x 2 x 1 = 6 • 5! = 5 x 4 x 3 x 2 x 1 = 120 • If we have 10 books to place on a shelf, how many possible ways are there to arrange them? • 10! = 3 628 800 ways

  19. Permutations • Suppose we have a group of 10 people. How many ways are there to pick a president, vice-president and treasurer? • In this case we are selecting people for a particular order • However, we are only selecting 3 of the 10 • For the first person, we can select from 10 • For the second person, we can select from 9 • For the third person, we can select from 8 • So there are 10 x 9 x 8 = 720 ways

  20. Permutation Notation • a permutation is an ordered arrangement of objects selected from a set • written P(n,r) or nPr • it is the number of possible permutations of r objects from a set of n objects

  21. Picking 3 people from 10… • We get 720 possible arrangements

  22. Permutations When Some Objects Are Alike • Suppose you are creating arrangements and some objects are alike • For example, the word ear has 3! or 6 arrangements (aer, are, ear, era, rea, rae) • But the word eel has repeating letters and only 3 arrangements (eel, ele, lee) • How do we calculate arrangements in these cases?

  23. Permutations When Some Objects Are Alike • To perform this calculation we divide the number of possible arrangements by the arrangements of objects that are similar • n is the number of objects • a, b, c are objects that occur more than once

  24. So back to our problem • Arrangements of the letters in the word eel • What would be the possible arrangements of 8 socks if 3 were red, 2 were blue, 1 black, one white and one green?

  25. Another Example • How many arrangements are there of the letters in the word BOOKKEEPER?

  26. Arrangements With Replacement • Suppose you were looking at arrangements where you replaced the object after you had chosen it • If you draw two cards from the deck, you have 52 x 51 possible arrangements • If you draw a card, replace it and then draw another card, you have 52 x 52 possible arrangements • Replacement increases the possible arrangements

  27. Permutations and Probability • If you have 10 different coloured socks in a drawer, what is the probability of choosing a red, green and blue sock? • Probability is the number of possible outcomes you want divided by the total number of possible outcomes • You need to divide the number of possible arrangements of the red, green and blue socks by the total number of ways that 3 socks can be pulled from the drawer

  28. The Answer • so we have 1 chance in 120 or 0.0083 probability

  29. Circular Permutations • How many arrangements are there of 6 old chaps around a table?

  30. Circular Permutations • There are 6! different orderings of 6 people • Number the people in the original photo 1-2-3-4-5-6. • 2-3-4-5-6-1 is the permutation when everyone moves one seat to the left. • This is not considered to be a different arrangement, since the people are seated in the same order, but we have counted it 6 times. • This is true of every arrangement - we have counted them 6 times, so we must divide the total by 6.

  31. Circular Permutations • So the number of orderings of 6 people around a table is: 6! = 5! 6 • Therefore the number of arrangements of n people in a circle is: n! = (n-1)! n

  32. Circular Permutations • There are 6! ways to arrange 6 the old chaps around a table • However, if everyone shifts one seat to the left, the arrangement is the same • This can be repeated 4 more times (6 total) • Therefore 6 of each arrangement are identical • So the number of DIFFERENT arrangements is 6! / 6 = 5! • In general, there are (n-1)! ways to arrange n objects in a circle.

  33. MSIP / Homework • p. 255-257 #1-7, 11, 13, 14, 16

  34. Warm up • How many ways can 8 children be placed on a 8-horse Merry-Go-Round? • 7! = 5 040 • What if Simone insisted on riding the red horse? • Here we are only arranging 7 children on 7 horses, so 6! = 720

  35. Counting Techniques and Probability Strategies - Combinations Chapter 4.7 – Introduction to Probability Mathematics of Data Management (Nelson) MDM 4U

  36. When Order is Not Important • A combination is an unordered selection of elements from a set • There are many times when order is not important • Suppose Mr. Russell has 10 basketball players and must choose a starting lineup of 5 players (without specifying positions) • Order of players is not important • We use the notation C(n,r) or nCr where n is the number of elements in the set and r is the number we are choosing

  37. Combinations • A combination of 5 players from 10 is calculated the following way, giving 252 ways for Mr. Russell to choose his starting lineup

  38. An Example of a Restriction on a Combination • Suppose that one of Mr. Russell’s players is the superintendent’s daughter, and so must be one of the 5 starting players • Here there are really only 4 choices from 9 players • So the calculation is C(9,4) = 126 • Now there are 126 possible combinations for the starting lineup

  39. Combinations from Complex Sets • If you can choose of 1 of 3 entrees, 3 of 6 vegetables and 2 of 4 desserts for a meal, how many possible combinations are there? • Combinations of entrees = C(3,1) = 3 • Combinations of vegetables = C(6,3) = 20 • Combinations of desserts = C(4,2) = 6 • Possible combinations = • C(3,1) x C(6,3) x C(4,2) = 3 x 20 x 6 = 360 • You have 360 possible dinner combinations, so you had better get eating!

  40. Calculating the Number of Combinations • Suppose you are playing coed volleyball, with a team of 4 men and 5 women • The rules state that you must have at least 3 women on the floor at all times (6 players) • How many combinations of team lineups are there? • You need to take into account team combinations with 3, 4, or 5 women

  41. Solution 1: Direct Reasoning • In direct reasoning, you determine the number of possible combinations of suitable outcomes and add them • Find the combinations that have 3, 4 and 5 women and add them

  42. Solution 2: Indirect Reasoning • In indirect reasoning, you determine the total possible combinations of outcomes and subtract unsuitable combinations • Find the total combinations and subtract those with 2 women

  43. Warm up • For your favourite sport… • How many players on the roster? • How many players in the starting lineup? • How many different groups of players can be put in the starting lineup? (no assigned positions) • How many ways can the coach set the starting lineup? (assigned positions)

  44. Hockey • 12F, 6D, 2G = 20 players • 3F, 2D, 1G start • There are • C(20,6) = 38 760 groups of 6 players • P(12, 3) x P(6, 2) x P(2, 1) = 1320 x 30 x 2= 79 200 starting lineups

  45. Finding Probabilities Using Combinations • What is the probability of drawing a Royal Flush (10-J-Q-K-A from the same suit) from a deck of cards? • There are C(52,5) ways to draw 5 cards • There are 4 ways to draw a royal flush • P(Royal Flush) = 4 / C(52,5) = 1 / 649 740 • You will likely need to play a lot of poker to get one of these hands!

  46. Finding Probability Using Combinations • What is the probability of drawing 4 of a kind? • There are 13 different cards that can be used to make up the 4 of a kind, and the last card can be any other card remaining

  47. Probability and Odds • These two terms have different uses in math • Probability involves comparing the number of favorable outcomes with the total number of possible outcomes • If you have 5 green socks and 8 blue socks in a drawer the probability of drawing a green sock is 5/13 • Odds compare the number of favorable outcomes with the number of unfavorable • With 5 green and 8 blue socks, the odds of drawing a green sock is 5 to 8 (or 5:8)

  48. Combinatorics Summary • In Permutations, order matters • e.g., President • In Combinations, order doesn’t matter • e.g., Committee

  49. MSIP / Homework • p. 262 – 265 # 1, 2, 3, 5, 7, 9, 18, 23

  50. References • Wikipedia (2004). Online Encyclopedia. Retrieved September 1, 2004 from http://en.wikipedia.org/wiki/Main_Page

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