Warm up

1. Warm up • The Leafs have won 45% of their games this season. When Phil Kessel scores, the Leafs win 30% of the time. What is the probability that Phil Kessel scored last night, given that the Leafs won?

2. Solution The probability that Phil Kessel scored given that the Leafs won is 0.67.

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

4. 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 Toss 1 Toss 2

5. 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)

6. 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 3/12 or ¼ • P(odd roll, head) = ¼

7. 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

8. 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 have thrown an even die? • these are independent events, so knowing the outcome of the second does not change the probability of the first

9. 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)

10. 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

11. 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

12. 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

13. 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

14. 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

15. MSIP / Homework • Read the examples on pages 239-244 • Complete pp. 245 – 249 #2, 3, 5, 7, 9, 12, 13a, 14

16. Warm up • How many different outcomes are there in a Dungeons and Dragons game where a 20-sided die is rolled, then a spinner with 5 sections is spun? • 20 x 5 = 100

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

18. 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 3 blocks of different colours? • How many for 4 blocks? • How many for 5 blocks? • How many for 6 blocks? • What is the pattern?

19. 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)

20. Factorial Notation • The notation is called factorial • n! (n factorial) 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 • NOTE: 0! = 1 • If we have 10 books to place on a shelf, how many possible ways are there to arrange them? • 10! = 3 628 800 ways

21. 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

22. 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

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

24. 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?

25. 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

26. 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?

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

28. Warm up • Canada’s 2010 Olympic Team has 13 forwards. If head coach Mike Babcock randomly selects his lines, what is the probability that the three San Jose Sharks, Dany Heatley, Joe Thornton and Patrick Marleau, play together on the first line.

29. Solution • There are 3! = 6 different ways to slot the 3 Sharks on the first line. • There are P(13, 3) = 13! ÷ (13-3)! = 1 716 possible line combinations. • So the probability is 6÷1 716 = 0.0035 or 0.35%. • It’s a good thing they are playing so well together in San Jose!

30. 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

31. Permutations and Probability • If you have 10 different coloured socks in a drawer, what is the probability of picking the red, green and blue socks? • 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

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

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

34. 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.

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

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

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

38. 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

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

40. 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

41. 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!

42. 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

43. 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

44. 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

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., Presidency • In Combinations, order doesn’t matter • e.g., Committee

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

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