Finding Probability Using Tree Diagrams and Outcome Tables

1
Finding Probability Using Tree Diagrams and
Outcome Tables

• Chapter 4.5 Introduction to Probability
• Mathematics of Data Management (Nelson)
• MDM 4U

2
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 H HH
H T HT
T H TH
T T TT
T
3
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)
4
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 in 12 or ¼
• P(odd roll, head) ¼

5
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

6
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
  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

7
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)

8
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

9
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

10
Example 3 - Predicting Outcomes

• Mr. Lieff is playing Texas HoldEm
• 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

11
Tree Diagram
P0.6 x 0.5 0.3
Win pot
0.5
bluff
0.6
0.5
P0.6 x 0.5 0.3
Lose pot
Win pot
0.7
P0.4 x 0.7 0.28
0.4
no bluff
P0.4 x 0.3 0.12
Lose pot
0.3
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

13
Exercises

• Read the examples on pages 239-244
• Try pp. 245 249 2, 3, 5, 7, 9, 12, 13a, 14
• Check each answer with the back of the book to
  make sure that you are understanding these
  concepts

14
Warm up

• How many different outcomes are there if a
  20-sided die is rolled then a spinner with 5
  sections is spun?
• 20 x 5 100

15
Counting Techniques and Probability Strategies -
Permutations

• Chapter 4.6 Introduction to Probability
• Mathematics of Data Management (Nelson)
• MDM 4U

16
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 colours?
• How many for 4 blocks?
• How many for 5 blocks?
• How many for 6 blocks?
• What is the pattern?

17
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)

18
Factorial Notation

• The notation is called factorial
• n! 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
• If we have 10 books to place on a shelf, how many
  possible ways are there to arrange them?
• 10! 3 628 800 ways

19
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

20
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

21
Picking 3 people from 10

• We get 720 possible arrangements

22
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?

23
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

24
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?

25
Another Example

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

26
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

27
Permutations and Probability

• If you have 10 different coloured 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 the red, green and blue socks by
  the total number of ways that 3 socks can be
  pulled from the drawer

28
The Answer

• so we have 1 chance in 120 or 0.0083 probability

29
Circular Permutations

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

30
Circular Permutations

• There are 6! different orderings of 6 people
• Number the people in the original photo
• 1-2-3-4-5-6.
• 2-3-4-5-6-1 is the permutation when everyone
  moves one seat to the left.
• This is not considered to be a different
  arrangement, since the people are seated in the
  same order, but we have counted it 6 times.
• This is true of every arrangement - we have
  counted them 6 times, so we must divide the total
  by 6.

31
Circular Permutations

• So the number of orderings of 6 people around a
  table is
• 6! 5!
• 6
• Therefore the number of arrangements of n people
  in a circle is
• n! (n-1)!
• n

32
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.

33
MSIP / Homework

• p. 255-257 1-7, 11, 13, 14, 16

34
Warm up

• How many ways can 8 children be placed on a
  8-horse Merry-Go-Round?
• 7! 5 040
• What if Simone insisted on riding the red horse?
• Here we are only arranging 7 children on 7
  horses, so 6! 720

35
Counting Techniques and Probability Strategies -
Combinations

• Chapter 4.7 Introduction to Probability
• Mathematics of Data Management (Nelson)
• MDM 4U

36
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

37
Combinations

• A combination of 5 players from 10 is calculated
  the following way, giving 252 ways for Mr.
  Russell to choose his starting lineup

38
An Example of a Restriction on a Combination

• Suppose that one of Mr. Russells players is the
  superintendents 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

39
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!

40
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

41
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

42
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

43
Warm up

• For your favourite sport
• How many players on the roster?
• How many players in the starting lineup?
• How many different groups of players can be put
  in the starting lineup? (no assigned positions)
• How many ways can the coach set the starting
  lineup? (assigned positions)

44
Hockey

• 12F, 6D, 2G 20 players
• 3F, 2D, 1G start
• There are
• C(20,6) 38 760 groups of 6 players
• P(12, 3) x P(6, 2) x P(2, 1) 1320 x 30 x 2 79
  200 starting lineups

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 / 649 740
• 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 58)

48
Combinatorics Summary

• In Permutations, order matters
• e.g., President
• In Combinations, order doesnt matter
• e.g., Committee

49
MSIP / Homework

• p. 262 265 1, 2, 3, 5, 7, 9, 18, 23

50
References

• Wikipedia (2004). Online Encyclopedia. Retrieved
  September 1, 2004 from http//en.wikipedia.org/wik
  i/Main_Page

51
The Birthday Paradox

• I will bet any takers 1 that two people in the
  room have the same birthday (I didnt check
  beforehand).
• In the long run, will I win money? lose money?
  break even?

52
The Birthday Paradox

• Assume all birthdays are equally likely
• Find the probability that no two birthdays are
  the same
• because the second person cannot have the same
  birthday as the first (364/365), the third cannot
  have the same birthday as the first two
  (363/365), etc.

53
The Birthday Paradox

• The event of at least two of the n persons having
  the same birthday is complementary to all n
  birthdays being different. Therefore, its
  probability p(n) is
• The approximate probability that no two people
  share a birthday in a group of n people.
• This probability surpasses 1/2 for n 23 (with
  value about 50.7).

54
The Birthday Paradox
n p(n)
10 11.7
20 41.1
23 50.7
50 70.6
53 97.0
57 99.0
100 99.99997
366 100