Warm up #1

1. Warm up #1 Entering the Olympic Break last season, the Toronto Maple Leafs had won 60% of their games. When Phil Kessel scored a goal, the Leafs won 50% of the time. What is the probability that Phil Kessel scored a goal in their last game, given that the Leafs won?

2. Solution Let W be the event that the Leafs won. Let K be the event that Phil Kessel scored. The probability that Phil Kessel scored, given that the Leafs won is 0.83.

3. Warm up #2 • Find the probability that two hearts are dealt consecutively from a standard deck (no replacement). • Let H1 be the event that the first card is a heart • Let H2 be the event that the second card is a heart • P(H1 ∩ H2) = P(H2 | H1) x P(H1) = 12/51 x 13/52 = 156/2652 = 0.06

4. Application of Warm-Up #2 • In a Texas Hold ‘Em game, how often will your hand (2 cards) be suited? • 4 x0.06 = 0.24 or almost ¼ of the time.

5. Finding Probability Using Tree Diagrams and Outcome Tables Chapter 4.5 – Introduction to Probability Learning goal: Calculate probabilities using tree diagrams and outcome tables Questions? pp. 235 – 238 #1, 2, 4, 6, 7, 9, 10, 19 MSIP / Home Learning: pp. 245 – 249 #2, 3, 5, 7, 9, 12, 13a, 14

6. Tree diagrams and Outcome tables • e.g., Flipping 2 coins • Draw the branch for the first event • Add a branch to each for the second event • Follow each branch to list the outcomes

7. 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

8. 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)

9. Sample Space • All ordered pairs in the form (d,c) • d represents the roll of a die • c represents the flip of a coin • e.g., (4,T) represents rolling a 4 then flipping a tail • There are 6 x 2 = 12 possible outcomes • What is P(odd roll,head) = ? • There are 3 possible outcomes for an odd die and a head (1, H), (3, H) and (5, H) • So the probability is 3/12 or ¼

10. 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

11. Examples • How many outcomes are there when rolling 2 dice? • 6 x 6 = 36 • How many sets of initials are there? • 26 x 26 = 676 • What if the letters must be unique? • 26 x 25 = 650

12. Examples • Airport codes are arrangements of 3 letters. • e.g., YOW=Ottawa, YYZ=Toronto • How many are possible? • 26 choices for each of the 3 positions • 26 x 26 x 26 = 17 576 • What if no letters can be repeated? • 26 x 25 x 24 = 15 600

13. Examples • How many possible postal codes are there in Canada? LDL DLD • 26 x 10 x 26 x 10 x 26 x 10 = 17 576 000 • 263 x 103 • How many are in Ontario? (K, L, N or P) • 4 x 10 x 26 x 10 x 26 x 10 = 2 704 000

14. Independent and Dependent Events • Independent events: the occurence of one event does not change the probability of the other • Ex. What is the probability of getting heads when you have thrown an even die? • These are independent events, because knowing the outcome of the first does not change the probability of the second

15. Multiplicative Principle for Probability of Independent Events • If we know that A and B are independent events, then… • P(B | A) = P(B) • We can also prove that if two events are independent the probability of both occurring is… • P(A and B) = P(A) × P(B)

16. 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 the red sock both times? • these are independent events

17. Example 2 • If you draw a card, replace it and draw another, what is the probability of getting two aces? • 4/52 x 4/52 = 16/2704 = 1/169 = 0.006 • These are independent events

18. 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

19. 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

20. 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

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

22. Steps to Solving Problems • What information am I given? • What are the key words? • Which formula is best? • How did I solve a similar problem? • Can I create a diagram / table / list? • How can I split the information into variables / events / sets?

23. 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

24. Counting Techniques and Probability Strategies - Permutations Chapter 4.6 – Introduction to Probability Learning goal: count arrangements of objects when order matters Questions: pp. 245 – 249 #2, 3, 5, 7, 9, 12, 13a, 14 MSIP / Home Learning: pp. 255-257 #1-7, 11, 13, 14, 16

25. Arranging blocks when order matters • Activity 1a – Arranging unique blocks in a line • Activity 1b – Arranging unique blocks in a circle • Activity 1c – Arranging blocks in a line when some are identical

26. Activity 1a – Arranging unique blocks in a line • Start with 3 cube-a-links of different colours • How many ways can you arrange them in a line on your desk? • Record the number in the first column. • How about 4 blocks? • 5? • 6? • What is the pattern?

27. Selecting When Order Matters • There are fewer choices for later places • For 3 blocks: • First block - 3 choices • Second block - 2 choices left • Third block - 1 choice left • Number of arrangements for 3 blocks is 3x2x1=6 • There is a mathematical notation for this (and your calculator has it)

28. Factorial Notation (n!) • n! is read “n-factorial” • n! = n x (n – 1) x (n – 2) x … x 2 x 1 • n! is the number of ways of arranging n unique objects when order matters • Ex: • 3! = 3 x 2 x 1 = 6 • 5! = 5 x 4 x 3 x 2 x 1 = 120 • NOTE: 0! = 1 • Ex. 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 … x 2 x 1 = 3 628 800 ways

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

30. Activity 1b – Arranging Unique Blocks in a Circle • Start with 3 blocks of different colours • Arrange them in a circle • Find the number of different arrangements / orders • Repeat for 4 • Try it for 5 • Do you see a pattern?

31. Circular Permutations • There are 6! ways to arrange the 6 old chaps around a table • However, if everyone shifts one seat, 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. Activity 1c – Arranging blocks in a line when some are identical • Start with 3 blocks where 2 are the same colour • How many ways can you arrange them in a line? • How does this number compare to the number of ways of arranging 3 unique blocks? • Repeat for 4 blocks where 2 are the same • Try it for 5 where 2 are the same • Do you see a pattern?

33. 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?

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

35. So back to our problem • Arrangements of the letters in the word eel • What is the number of arrangements of 8 socks if 3 are red, 2 are blue, 1 is black, one is white and one is green?

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

37. Permutations of SOME objects • 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

38. Permutation Notation • A permutation is an ordered arrangement of objects selected from a set • Written P(n,r) or nPr • The number of possible arrangements of r objects from a set of n objects

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

40. 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

41. Permutations and Probability • 10 different coloured socks in a drawer • What is the probability of picking green, red and blue in any order?

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

43. MSIP / Home Learning • pp. 255-257 #1-7, 11, 13, 14, 16

44. Warm up • At the 2013 NHL All Star Game, Team Alfredsson featured all 3 Ottawa Senators forwards. If head coach John Tortorella randomly selected his lines from 12 forwards, what is the probability that the three Senators played together on the first line?

45. Drawing a diagram • On the first line, there are: • 3 choices for the first position • 2 choices for the second position • 1 choice for the third position LW C RW

46. Solution • There are 3! = 6 different ways to slot the 3 Sens on the first line. • There are P(12, 3) = (12)! ÷ (12-3)! = 1320 possible line combinations. • So the probability is 6÷1320 = 0.0045 or 0.45%. • What is the probability they play together on ANY of the 4 lines? • 4 x 0.45% = 1.8%

47. 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?

48. Warm up - Solution • 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

49. Counting Techniques and Probability Strategies - Combinations Chapter 4.7 – Introduction to Probability Learning goal: Count arrangements when order doesn’t matter Questions? pp. 255-257 #1-7, 11, 13, 14, 16 MSIP / Home Learning: pp. 262 – 265 # 1, 2, 3, 5, 7, 9, 18

50. 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. Lieff has 29 MDM4U students and must choose a Data Fair Committee of 5 • Order 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