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Finding Probability Using Tree
Diagrams and Outcome Tables
Chapter 4.5 – Introduction to Probability
Mathematics 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
H
Flip 1
Flip 2
H
H
Simple
Event
HH
H
T
HT
T
H
TH
T
T
TT
H
T
H
T
T
Tree Diagrams Continued

if you rolled 1 die and then flipped a coin you
have 12 possible outcomes
1
H
T
2
H
T
(1,H)
(1,T)
(2,H)
(2,T)
3
H
T
(3,H)
(3,T)
4
H
T
(4,H)
(4,T)
5
H
T
(5,H)
(5,T)
H
T
(6,H)
(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)
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)


How many possible three letter words are there?


(this assumes that each outcome has no influence on the next
outcome)
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
 3 
 
P (heads  even)  12  1
P (heads | even) 


P (even)
2
3
 
6
1
as P (heads)  , we can say P (heads | even)  P ( heads)
2
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
B
G
B
G
R
B
G
R
B
G
P(red and red )
 P(red )  P(red )
1 1 1
  
3 3 9
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
0.5
fish dinner
P=0.3
0.5
bean dinner
P=0.3
west
0.6
0.7
0.4
fish dinner
P=0.28
east
0.3
bean dinner
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 Permutations
Chapter 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, vicepresident 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
n!
P(n, r ) 
n  r !
Choosing 3 people from 10…

we get 720 possible arrangements
10!
10 !
P(10 ,3) 


(10  3)! 7!
10  9  8  7  6  5  4  3  2 1
7  6  5  4  3  2 1
 10  9  8  720
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
Permutations
n!

a!b!c!...
So back to our problem

combinations of the
letters in the word eel

what would be the
possible arrangements
8! 8  7  6  5  4  3  2 1
of 8 socks if 3 were

3!2!
3  2  1 2  1
red, 2 were blue, 1
black, one white and
 3360
one green?
3! 3  2  1

3
2!
2 1
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
3!
P

P(10,3)
3!
10!
(10  3)!
3!
3!


10! 10  9  8
7!
6
1


720 120
Exercises

Page 255 #1, 4, 5, 7, 11, 12
Counting Techniques and
Probability Strategies Combinations
Chapter 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
n
n!
C (n, r )    
 r  (n  r )! r!
10 
10!
C (10 ,5)    
 5  (10  5)!5!
10!
 252

5!5!
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
 4  5   4  5   4  5 
           
 3  3   2  4   1  5 
 4  10  6  5  4  1
 40  30  4  74
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
 9   4  5 
      
 6   4  2 
 84  1 10
 84  10  74
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
 4  48 
13  
4  1 
1

P

4165
 52 
 
5
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