Finding Probability Using Tree Diagrams and Outcome Tables

Chapter 4.5 – Introduction to ProbabilityMathematics of Data Management (Nelson)MDM 4U Author: G. Greer (ideas from K. Myers)

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 TT

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) there are 3 possible outcomes for an odd die

and a head so the probability is 3 in 12 or ¼ P(odd roll, head) = ¼

Multiplicative Principle for Counting The total number of outcomes is the product of the 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 license plates are there in Ontario?

26 x 26 x 26 x 26 x 10 x 10 x 10

Independent and Dependent Events two events are independent of each other if an

occurence in one event does not change the probability of an occurrence in the other

what is the probability of getting heads when you know in advance that you will throw 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 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 a sock drawer has a red, a green and a blue sock you pull out one sock, replace it and pull another out draw a tree diagram representing the possible outcomes 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 if you draw a card, replace it and draw

another, what is the probability of two aces? 4/52 x 4/52 independent events

if you draw a card and then draw a second card (no 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

Predicting Outcomes Mr. Greer is going fishing he finds that he catches fish 70% of the time

when the wind is out of the east he also finds that he catches fish 50% of the

time when the wind is out of the west if there is a 60% chance of a west wind

today, what are his chances of having fish for dinner?

we will start by creating a tree diagram

Tree Diagram

west

east

fish dinner

fish dinner

bean dinner

bean dinner

0.6

0.4 0.7

0.3

0.5

0.5

P=0.3

P=0.3

P=0.28

P=0.12

Continued… P(east, catch) = P(east) x P(catch | east) = 0.4 x 0.7 = 0.28 P(west, catch) = P(west) x P(catch | west) = 0.6 x 0.5 = 0.30 Probability of a fish dinner: 0.28 + 0.3 = 0.58 So Mr. Greer has a 58% chance of catching a

fish for dinner

Exercises

read the examples on pages 239-244 try page 245# 1,2,4,7,8,10

check each answer with the back of the book to make sure that you are understanding these concepts

Counting Techniques and Probability Strategies - PermutationsChapter 4.6 – Introduction to Probability

Mathematics of Data Management (Nelson)MDM 4U Author: G. Greer (ideas from K. Myers)

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 blocks of

different colors 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 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

If we have 10 books to place on a shelf, how many possible ways are there to arrange them?

10! = 3628800 ways

Permutations Suppose we have a group of 10 people and

we want to choose a president, vice-president and treasurer?

in this case we are choosing people for a particular order

however, we are only choosing 3 of the 10 for the first person, we can choose among 10 for the second person, we can choose among 9 for the third person, we can choose among 8 so there are 10 x 9 x 8 ways

Permutation Notation a permutation is an ordered arrangement of

objects selected from a set written P(n,r) or sometimes nPr

it is the number of possible permutations of r objects from a set of n objects

!!

),(rn

nrnP

Choosing 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 6 or 3!

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 With Some Objects Alike to perform this

calculation we divide the number of possible combinations by the combinations 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

combinations 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

336012123

12345678

!2!3

!8

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 colored socks in a

drawer, what is the probability of choosing a red, green and blue sock?

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 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 0.833%

probability

120

1

720

6

8910

!3

!7!10!3

)!310(!10!3

)3,10(

!3

P

P

Exercises

Page 255 #1, 4, 5, 7, 11, 12

Counting Techniques and Probability Strategies - CombinationsChapter 4.7 – Introduction to Probability

Mathematics of Data Management (Nelson)MDM 4U Author: G. Greer (ideas from K. Myers)

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. Greer has 10 players and must choose

a starting line of 5 players for his basketball team order of players is not important we use the notation C(n,r) 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. Greer 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. Greer’s players is the

superintendent’s son, 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 so now there are 126 possible combinations

for the starting lineup

Combinations from Complex Sets if you can choose of 1 of 3 meat dishes, 3 of

6 vegetables and 2 of 4 desserts for a meal, how many possible combinations are there?

combinations of meat dishes = 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 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 0, 1, or 2 women 741084

10184

2

5

4

4

6

9

Finding Probabilities Using Combinations what is the probability of drawing 10-J-Q-K-A

from the same suit in a deck of cards? there are C(52,5) ways to draw 5 cards there are 4 ways to draw this type of straight P = 4 / C(52,5) = 1/649740 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)

Exercises

Page 262 # 3, 5, 7, 9, 13, 14

References

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