warm up #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?

Solution

83.060.0

50.0

)(

)()|(

WP

WKPWKP

The probability that Phil Kessel scored, given that the Leafs won is 0.83.

Let W be the event that the Leafs won.Let K be the event that Phil Kessel scored.

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

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.

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

Tree Diagram 1 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

How to Draw a Tree Diagram

Draw the branch for the first event Add a branch to each for the second event Repeat for additional events

Follow each branch to list the outcomes

Tree Diagram 2 if you roll 1 die and then flip a coin there are

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)

Tree Diagram 2 (cont’d) The sample space is 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

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

Examples Pt 1 How many outcomes are there when

rolling 2 dice? 6 x 6 = 36 How many pairs of initials are there? 26 x 26 = 676 What if letters cannot be repeated? This adds a restriction on the 2nd letter

26 x 25 = 650

Examples Pt 2 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

Examples Pt 3 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? (1st letter K, L, M, N or P)

5 x 10 x 26 x 10 x 26 x 10 = 13 520 000

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

)()|(,2

1)(

2

1

6

312

3

)(

)()|(

headsPevenheadsPsaycanweheadsPas

evenP

evenheadsPevenheadsP

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)

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

R

R

R

R

B

B

B

BG

G

G

G

9

1

3

1

3

1

)()(

)(

redPredP

redandredP

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

Example 3 - Predicting Outcomes What if the outcomes are not equally likely? 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 / Home Learning

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

13a, 14

Steps to Solving Problems

What information am I given? What are the key words?

Number/probability, and/or, given 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?

Minds on!

Counting Techniques and Probability Strategies - Permutations


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

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

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?

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

3 x 2 x 1 = 6 There is a mathematical notation for this (and

your calculator has it)

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 e.g.,

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

Circular Permutations

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

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? How does it connect to n!?

Circular Permutations

There are 6! ways to arrange the 6 old chaps 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.

Recap of yesterday

n! is read “n-factorial”

n! = n(n-1)(n-2)…(2)(1) is the number of ways of arranging n unique objects

n! = (n-1)! is the number of ways of arrangingn n objects in a circle

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?

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 identical

n is the number of objects a, b, c are the number of

objects that occur more than once

!...!!

!

cba

n

nsPermutatio

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?

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

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

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

!!

),(rn

nrnP

Picking 3 people from 10…

We get 720 possible arrangements

72089101234567

12345678910

!7

!10

)!310(

!10)3,10(

P

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

10 different coloured socks in a drawer What is the probability of picking green, red

and blue in any order?

The Answer so we have 1 chance in 120 or 0.0083

probability

0083.0120

1

720

6

8910

!3

!7!10!3

)!310(!10

!3

)3,10(

!3)(

P

RGBP


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

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?

Drawing a diagram

LW C RW

Line 1 3 2 1

Line 2 X X X

Line 3 X X X

Line 4 X X X

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

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%

Minds on! The Coca-Cola freestyle

machine boasts 100+ drink choices (1-2 flavours).

How many different drinks are possible if there are 20 flavours?

P(20, 2) + 20= 380 + 20

= 400 Does order matter?

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?

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

Counting Techniques and Probability Strategies - Combinations


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

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

Combinations A combination of 5 students from 29 is calculated

the following way, giving 118 755 ways for Mr. Lieff to choose his committee.

118755)!5!24(

!29

!5)!529(

!29

5

29)5,29(

!)!(

!),(

C

rrn

n

r

nrnC

An Example of a Restriction on a Combination Suppose that one of Mr. Lieff’s students is

the superintendent’s daughter, so she must be one of the 5 committee members

Here there are really only 4 choices from 28 students

So the calculation is C(28,4) = 20 475 Now there are 20 475 possible combinations

for the committee

How many combos are possible?

ColaDiet ColaRoot BeerOrangeWater

Combinations from Complex Sets Combinations of mains = C(4,2) = 6 Combinations of sides = C(2,1) = 2 Combinations of desserts = C(2,1) = 2 Combinations of drinks = C(5,1) = 5 Possible combinations =

C(4,2) x C(2,1) x C(2,1) x C(5, 1) = 6 x 2 x 2 x 5 = 120 You have 120 possible meals, so you had better make

a run for the border! What if you can order 2 of the same main? There are 4 ways to do this

So Combinations of mains = 6 + 4 = 10 Possible combinations = 10 x 2 x 2 x 5 = 200

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

Permutationsorder matters

e.g., Presidency

Combinationsorder doesn’t matter

e.g., Committee!)!(

!),(

rrn

nr

nrnC

!!

),(rn

nrnP

MSIP / Home Learning pp. 262 – 265 # 1, 2, 3, 5, 7, 9, 18


pp. 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

warm up #1

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