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. 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
Flip 1
Flip 2
Simple
Event H HH T HT TH TT H T T
Toss Toss 2

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

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. 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 R B G

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 0.5 Win pot P=0.6 x 0.5 = 0.3 bluff 0.6 0.5 Lose pot
0.7
P=0.4 x 0.7 = 0.28
0.4
no bluff
0.3
Lose pot
P=0.4 x 0.3 = 0.12

14. Continued… P(no bluff, win) = P(no bluff) x P(win | no bluff)
P(bluff, win) = P(bluff) x P(win | bluff)
= 0.6 x 0.5 = 0.30
Probability of a win: = 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! = 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)! = possible line combinations.
So the probability is 6÷1 716 = 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 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 #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 /
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