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?

Solution

67.045.0

3.0

)(

)()|(

WP

WKPWKP

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

Finding Probability Using Tree Diagrams and Outcome Tables

Chapter 4.5 – Introduction to Probability

Mathematics of Data Management (Nelson)

MDM 4U

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

H

T

H

H

T

Flip 1 Flip 2 Simple

Event

H H HH

H T HT

T H TH

T T TTToss 1 Toss 2

Tree Diagrams Continued if you rolled 1 die and then flipped a coin you

have 12 possible outcomes

H

T

H

T

H

T

H

T

H

T

H

T

1

2

3

4

5

6

(2,H)

(1,H)

(3,H)

(4,H)

(5,H)

(6,H)

(2,T)

(1,T)

(3,T)

(4,T)

(5,T)

(6,T)

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

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

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

)()|(,2

1)(

2

1

6

312

3

)(

)()|(

headsPevenheadsPsaycanweheadsPas

evenP

evenheadsPevenheadsP

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)

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

R

R

R

B

B

B

BG

G

G

G

9

1

3

1

3

1

)()(

)(

redPredP

redandredP

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

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

Tree Diagram

bluff

no bluff

Win pot

Win pot

Lose pot

Lose pot

0.6

0.4 0.7

0.3

0.5

0.5

P=0.6 x 0.5 = 0.3

P=0.6 x 0.5 = 0.3

P=0.4 x 0.7 = 0.28

P=0.4 x 0.3 = 0.12

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

MSIP / Homework

Read the examples on pages 239-244 Complete pp. 245 – 249 #2, 3, 5, 7, 9, 12,

13a, 14

Warm up

o 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?

o 20 x 5 = 100

Counting Techniques and Probability Strategies - Permutations

Chapter 4.6 – Introduction to Probability

Mathematics of Data Management (Nelson)

MDM 4U

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?

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)

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

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

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

!!

),(rn

nrnP

Picking 3 people from 10…

We get 720 possible arrangements

72089101234567

12345678910

!7

!10

)!310(

!10)3,10(

P

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?

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

!...!!

!

cba

n

nsPermutatio

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?

312

123

!2

!3

3360

)12()123(

12345678

!2!3

!8

Another Example

How many arrangements are there of the letters in the word BOOKKEEPER?

200151

56789101231212

12345678910

!3!2!2

!10

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.

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!

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

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

The Answer so we have 1 chance in 120 or 0.0083

probability

120

1

720

6

8910

!3

!7!10!3

)!310(!10

!3

)3,10(

!3)(

P

RGBP

Circular Permutations

How many arrangements are there of 6 old chaps around a table?

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.

MSIP / Homework

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

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

Counting Techniques and Probability Strategies - Combinations

Chapter 4.7 – Introduction to Probability

Mathematics of Data Management (Nelson)

MDM 4U

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

Combinations A combination of 5 players from 10 is calculated

the following way, giving 252 ways for Mr. Russell to choose his starting lineup

252!5!5

!10

!5)!510(

!10

5

10)5,10(

!)!(

!),(

C

rrn

n

r

nrnC

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

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!

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

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

7443040

1456104

5

5

1

4

4

5

2

4

3

5

3

4

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 741084

10184

2

5

4

4

6

9

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!

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

4165

1

5

52

1

48

4

413

P

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)

Combinatorics Summary

In Permutations, order matters

e.g., Presidency

In Combinations, order doesn’t matter

e.g., Committee!)!(

!),(

rrn

nr

nrnC

!!

),(rn

nrnP

MSIP / Homework

p. 262 – 265 # 1, 2, 3, 5, 7, 9, 18

References

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