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

  9. Example • if you draw a card, replace it and draw another, what is the probability of two aces? • 4/52 x 4/52 • independent events • if you draw a card and then draw a second card (no 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. 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 + 0.3 = 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. Counting Techniques and Probability Strategies - Permutations Chapter 4.6 – Introduction to Probability Mathematics of Data Management (Nelson) MDM 4U

  15. 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 colors • how many for 4 blocks • how many for 5 blocks • how many for 6 blocks • what is the pattern?

  16. 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)

  17. Factorial Notation • the notation is called factorial • 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! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3 628 800 ways

  18. Permutations • Suppose we have a group of 10 people and we want to choose 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

  19. 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

  20. Picking 3 people from 10… • we get 720 possible arrangements

  21. Permutations When Some Objects Are Alike • suppose you are creating arrangements and some objects are alike? • for example, the word ear has 6 or 3! 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?

  22. Permutations With Some Objects 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

  23. 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?

  24. Peter’s Example • How many arrangements are there of the letters in the word BOOKKEEPER?

  25. 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

  26. Permutations and Probability • if you have 10 different colored 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 red, green and blue socks by the total number of ways that 3 socks can be pulled from the drawer

  27. The Answer • so we have 1 chance in 120 or 0.833% probability

  28. Application of 4.6 - Circular Permutations • How many arrangements are there of 6 old chaps around a table?

  29. 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.

  30. 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

  31. 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.

  32. Exercises • p. 255-257 #1-7, 11, 13, 14, 16

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

  34. 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 line of 5 players • 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

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

  36. An Example of a Restriction on a Combination • suppose that one of Mr. Russell’s players is the superintendent’s son, 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 • so now there are 126 possible combinations for the starting lineup

  37. 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

  38. 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 • how many combinations of team lineups are there? • you need to take into account team combinations with 3, 4, or 5 women

  39. 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

  40. 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

  41. Finding Probabilities Using Combinations • what is the probability of drawing 10-J-Q-K-A from the same suit in 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

  42. 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

  43. 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)

  44. Exercises • p. 262 – 265 # 1, 2, 3, 5, 7, 9, 18, 23

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

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