Download Overview of Unit 1

MDM4U
Unit 1 Master
Overview of Unit 1: Introduction to Probability (Chapter # 4 – Nelson)
Lesson 1:
Lesson 2:
Lesson 3:
Lesson 4:
Lesson 5:
Lesson 6:
Lesson 7:
Lesson 8:
Lesson 9:
Lesson 10:
Lesson 11:
Lesson 12:
Lesson 13:
Sec.4.1 An Introduction to Simulations
Sec.4.2 Theoretical Probability
Sec.4.3a. Finding Probability Using Sets
Sec.4.3b. Finding Probability Using Venn Diagrams
Sec.4.4 Conditional Probability
Sec.4.5 Finding Probability Using Tree Diagrams
Sec.4.6a Permutations
Sec.4.6b Probability & Permutations
Sec.4.7a Combinations
Sec.4.7b Probability & Odds
Sec.4.7c Problem Solving with Combinations
Unit Review Day # 1
Unit Review Day # 2
Unit 1: Homework
Topic
Homework
Simulations
p.210 # 4, 5, 8, 10, 11, 14, 19
Theoretical Probability
p.218 # 1 , 2b , 3 – 6 , 8 , 10 , 11, 14 , 18
Probability Using Sets
p.228 # 1abcdf , 6a , 8a
Probability Using Venn Diagrams
p.228 # 2 , 3 , 4 , 6b , 7 , 8b , 9 , 12 , 13 , 14
Conditional Probability
p.236 # 1 , 2 , 5 , 8 , 9 , 14ab
Tree Diagrams
p.245 # 3 , 4ab , 5abc (ii,iii,iv) , 6ab , 9ab
Permutations
p.255 # 1bde , 2b , 3c , 4abc , 5 , 6ab , 9a , 11 , 13 , 15 - 17
Probability & Permutations
YES
Combinations
p.262 # 1 - 3 , 6 , 7 omit g
Odds
YES
Problem Solving with Combinations YES
Page 1 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Probability & Odds
Homework:
1.
a)
Given:
P  A 
7
36
Find: odds in favour of A
b)
Given:
P  A ' 
2
6
Find: odds in favour of A
c)
Given:
(odds against A) = 5 : 4
d)
Given:
(odds in favour of A) = 10 : 8
(odds against B) = 5 : 1
A & B are mutually exclusive
Find: (i)
(ii)
e)
Find: P(A)
P A  B
odds in favour of either A or B occurring
Given:
P(A) = 0.5
P(B) = 0.3
A & B are independent
Find:
odds in favour that both A & B occur
[ 7:29 , 2:1 , 4/9 , 13/18 , 13:5 , 3:17 ]
2.
h
h
Show that if (odds in favour of A) =
then P A 
k
hk
3.
What are the odds in favour of rolling a pair, followed by two even numbers in two successive rolls of a
pair of dice?
4.
Over time, Natalie has experienced that if she washer her car, it rains the next day 80% of the time.
What are the odds in favour of rain tomorrow if Natalie is washing her car today?
[ 1:23 ]
[ 4:1 ]
5. Statistics show that in 75% of the fatal accidents involving two cars, at least one of the drivers is
impaired. If you have the misfortune to witness such an accident, what are the odds in favour of one of
the drivers being impaired?
[ 3:1 ]
Page 2 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Probability & Permutations
Homework:
1.
2.
3.
Match each expression on the left with an equivalent expression on the right.
14 !
a)
i)
10 100
13!
52 !
b)
ii)
6!
51!
101!
c)
iii)
52
99!
d)
20  19!
iv)
10!
e)
90 x 8!
v)
14
f)
30 x 4!
vi)
20!
Find the value for each expression.
a)
8!
5!
b)
19 !
13 !
e)
155 !
152 !
f)
93!
89 ! 4 !
c)
21!
17 ! 4 !
9!
7 ! 2!
d)
[ 336 , 19535040 , 5985 , 36 , 3652110 , 2919735 ]
Evaluate each of the following expressions, using the definition of the factorial operation to simplify the
work.
a)
10 !
5!
b)
21!
14 !
c)
9!
3 ! 6!
e)
7!
7!

2 ! 5! 4 !3!
f)
15!
15!

9 ! 6! 10!5!
g)
2
5!
3! 2 !
d)
12 !
8 ! 4!
h)
3
11!
7 !4!
[ 30240 , 586051200 , 84 , 495 , 56 , 8008 , 20 , 990 ]
4.
Simplify the following expressions
a)
nn  1!
d)
n! n 2  3n  2


b)
n!n  1
c)
n  1!n 2  n
e)
n!
n  2!
f)
n  2!
n  1!
[ n! , (n+1)! , (n+1)! , (n+2)! , n2-n , (n+2)(n+1)n ]
Page 3 of 45
Prepared by: Scott McEwen
MDM4U
5.
Unit 1 Master
Solve for n, n  W
a)
c)
n  1!  9
n!
3n  1!
 126
n  1!
b)
n!
 20
n  2!
d)
2n!
 84n
n  3!
[8,5,6,8]
6.
In how many ways can the letters of each word be arranged?
a)
d)
MAXIMUM
INTERESTING
b)
e)
CANADA
UNINTERESTING
c)
f)
SASKATCHEWAN
MISSISSAUGA
[ 840 , 120 , 39916800 , 2494800 , 129729600 , 415800 ]
7.
List all five-digit numbers than can be formed by using two 4’s and three 6’s.
8.
How many seven-digit integers, are there which include
a)
two 3’s, three 2’s and two 8’s?
b)
four 3’s and three 4’s?
[ 10 ]
[ 210 , 35 ]
9.
A man bought two vanilla ice cream cones, three chocolate cones, four strawberry cones, and one
pistachio cone for his ten children. In how many ways can he distribute the flavours among the
children?
[ 12600 ]
10.
A coin is tossed nine times. In how many ways could the results be six heads and three tails?
11.
Anya is starting out on her evening run. Her route always takes her eight blocks east and five blocks
north to her grandmother’s apartment building. But she likes to vary the path she follows. How many
different possibilities does she have? Hint: Consider the permutations of 13 letters, 8E’s and 5 N’s
12.
How many numbers greater than 300000 are there using only the digits 1 , 1 , 1 , 2 , 2 , 3 ?
13.
Yurak is shelving books in a display in the school library. He has four different books with three copies
of each. In how many ways can he arrange the books on the shelf for display?
[ 84 ]
[ 1287 ]
[ 10 ]
[ 369600 ]
14.
A developer will build 12 houses on the same side of Costly Court in a new subdivision. If he has room
for two houses modeled on Plan A, four modeled on Plan B, and six modeled on Plan C, in how many
different ways can he arrange the houses on the street?
[ 13860 ]
15.
Emily’s minor soccer team played a total of 14 games in the season. Their record was eight wins, four
losses, and two ties. In how many orders could this have happened?
[ 45045 ]
16.a)
b)
c)
d)
How many permutations are there of the letters of the word BASKETBALL?
How many of the arrangements begin with K?
How many of the arrangements start with a B?
In how many of the arrangements would the two L’s be together?
[ 453600 , 45360 , 90720 , 90720 ]
Page 4 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Problem Solving in Combinations
Homework:
1.
In how many ways can a committee of at least one person be formed from seven club members?
2.
Chang arrives at the giant auction sale late in the afternoon. There are only five items left to be sold.
How many different purchases could he make?
3.
In how many different ways could a team of three students be chosen from Lin’s Finite Mathematics
class of 25 students to complete in the Country Mathematics Contest?
4.a)
How many different sums of money can be made from a $2-bill, a $5-bill, and a $10-bill?
[ 127 ]
[ 31 ]
[ 2300 ]
[7]
b)
How many different sums of money can be made from the bills in (a) as well as one more $10-bill?
[ 11 ]
c)
Why does the situation become much more complicated if another $5-bill is added?
5.
A committee of students and teachers is being formed to study the issue of student parking privileges.
Fifteen staff members and 18 students have expressed an interest in serving on the committee. In how
many different ways could a five person committee be formed if it must include at least one student and
one teacher?
6.
In the binary number system which is used in computer operations, there are only two digits allowed: 0
and 1.
[ 225765 ]
a)
How many different binary numbers can be formed using at most four binary digits (ie: 0110) ?
[ 16 ]
b)
If eight binary digits are used (ie: 11001101), how many different finery numbers can be formed?
[ 256 ]
Page 5 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Sec.4.1 An Introduction to Simulations and Experimental Probability
A simulation is an experiment, model or activity that imitates real or hypothetical conditions. It is used to
estimate quantities that are difficult to calculate and for verifying theoretical calculations.
ie: biological experiments, drug testing, IKEA furniture testing, multiple choice guessing, births
Example # 1: Simulating Birth Sequences
Suppose a family plans to have four children. Use a simulation to estimate the likelihood that the family will
have three girls in a row and then a boy.
Solution # 1a)
Toss a single coin four times to simulate a single trial. Let Heads represent the birth of a girl and let
Tails represent the birth of a boy. Repeat this experiment for 50 trials to get a fair representation of the
likelihood of having three girls followed by a boy.
Solution # 1b)
Toss four different coins to simulate the births of all four children at once. Let Heads represent the birth
of a girl and let Tails represent the birth of a boy. The Penny will represent the 1st birth, the Nickel will
represent the 2nd birth, the Dime will represent the 3rd birth and the Quarter will represent the 4th birth. Repeat
this experiment for 50 trials to get a fair representation of the likelihood of having three girls followed by a boy.
Example # 2: Batting Averages
Suppose that Nicholas has a batting average of 0.320. This means that he will hit the ball 320 times out of 1000
at bats. This reduces to 32 out of 100 (or 8 out of 25). Use a simulation to estimate the likelihood that this
player has no hits in one game (assuming 3 hits @ bat/game).
Solution # 2
Fill a container with 25 slips of paper, 8 of which have the word HIT written on them. All the other
papers will have the word MISS written on them, or you could leave them blank for saving time. Draw a sheet
of paper out of the container and record “hit or miss”, return the paper to the container and draw again. You
will draw 3 papers to represent Nicholas’s likelihood of hitting in a game. Repeat this experiment for 50 trials
to get a fair representation of the likelihood that Nicholas will have no hits in a game.
Page 6 of 45
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MDM4U
Unit 1 Master
Using Graphing Calculators for Simulations
The graphing calculator can be used for simulations. In Example # 1 for the births, you can generate a
list of random numbers to represent the births of girls and boys.
On the TI-83+ select the MATH button, use the right arrow 3 times so that PRB (for probability) is
selected. Either scroll down or select #5 for RANDINT( , you now have to set a lower bound, upper bound and
number of integers. In the case of example # 1, we will let our lower bound be 1 (which will represent the birth
of a girl) and let our upper bound be 2 (which will represent the birth of a boy) and choose 4 because we wish to
represent the births of four children in total.
This function will generate a list of four numbers at random to simulate the birth of the four children.
Repeat this experiment for 50 trials to get a fair representation of the birth sequences.
ie: 2 2 1 2 (would represent a boy, boy, girl, boy for the four births)
In the case of example # 2 we would need a random integer between 1 & 1000 with 3 numbers each
time. The numbers 1 – 320 would represent HITS and the numbers 321 – 1000 would represent MISS. Repeat
this experiment for 50 trials to find the likelihood that Nicholas will have no hits in a game.
ie: 978 279 276 (would represent a miss, hit, hit during one game)
Using Excel Spreadsheets for Simulations
For Example # 1:
Trial
1
2
3
4
5
6
7
8
9
10
Birth 1 Birth 2 Birth 3 Birth 4
2
1
1
2
1
1
2
2
1
1
1
2
1
1
2
1
2
1
1
1
2
2
2
1
1
2
2
1
1
1
1
1
1
1
2
1
2
1
1
1
Under the Birth 1 column you would enter “=int(2*rand( ) + 1). You then use the fill handle on this cell to drag
this same formula into all other columns and rows necessary.
Page 7 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Sec.4.2 Theoretical Probability
Simple Event
 an event that consists of exactly one outcome.
Experimental Probability
 only estimates the likelihood that an event occurs in a given experiment.
Theoretical Probability
 is deduced from analysis of all possible outcomes.
Sample Space
 is represented by “S” and is the collection of all possible outcomes.
Event Space
 is represented by “A” and is the collection of all those outcomes that we want.
We will look at the probability of experiments where the outcomes are easily determined.
P  A 
n A 
# of desired outcomes

nS  # of all possible outcomes
Example #1:
A die is rolled once. What is the probability of rolling a four?
Solution # 1:
Let S represent the collection of all possible outcomes. S = { 1 , 2 , 3 , 4 , 5 , 6 }
Let A represent the successful outcomes. A = { 4 }
Page 8 of 45
Prolling a four  
# of four
total #
Prolling a four  
1
6
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Example # 2:
A bag contains 6 red smarties, 2 blue smarties, and 7 green smarties. If you take one, what is the probability
that it is green?
Solution # 2:
Let S represent the collection of all smarties. S = { 15 smarties all together }
Let A represent the green smarties. A = { 7 green }
Pgreen smarties  
# of green
total #
Pgreen smarties  
7
15
Example # 3:
If a single die is rolled what is the probability of;
a) rolling an odd number
b) rolling a number less than 3
c) rolling either a three or a five
Solution # 3:
a) rolling an odd number
Prolling an odd #  
# of odd numbers
total #
Prolling an odd #  
3
6
1
Prolling an odd #  
2
b) rolling a number less than 3
Prolling a #  3 
# of numbers  3
total #
Prolling a #  3 
2
6
1
Prolling a #  3 
3
Page 9 of 45
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c) rolling either a three or a five
Prolling a 3 or 5 
# of 3 or 5
total #
Prolling a 3 or 5 
2
6
1
Prolling a 3 or 5 
3
Example # 4:
Find the probability that a number picked at random between 1 and 10 is a perfect square?
Solution # 4:
n { S } = 10 elements in total
n { A } = 3 elements that are perfect squares between 1 & 10
n A
n S 
3
P# is perfect square  
10
P# is perfect square  
Page 10 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Example # 5:
When drawing a single card from a standard deck of cards, what is the probability that;
a) a queen is drawn
b) a heart is drawn
c) a black card is drawn
Solution # 5:
a) a queen is drawn
n A
n S 
4
Pqueen  
52
1
Pqueen  
13
Pqueen  
b) a heart is drawn
n A
n S 
13
P heart  
52
1
P heart  
4
P heart  
c) a black card is drawn
n A
n S 
26
Pblack card  
52
1
Pblack card  
2
Pblack card  
Page 11 of 45
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MDM4U
Unit 1 Master
One immediate consequence of the definition of probabilities is the relationship between an event and it’s
compliment, that is the set of outcomes in which the event does not occur.
P A '  1  P A
In General:
Example # 6:
What is the probability of not selecting a prime number in a random selection of a number from 1 to 20?
Solution # 6:
Let A represent the prime numbers between 1 and 20.
A = { 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 }
n(A)=8
&
n ( S ) = 20
n A
n S 
8
P selecting a prime #  
20
2
P selecting a prime #  
5
P selecting a prime #  
Therefore,
Pnot selecting a prime #  1  P A
Pnot selecting a prime #  1 
Pnot selecting a prime # 
Page 12 of 45
2
5
3
5
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MDM4U
Unit 1 Master
Example # 7:
My music collection at home consists of eight rock CD’s, twelve blues CD’s and 4 classic CD’s. If a CD is
chosen at random, find the probability that the one chosen is;
a) rock
b) not classical
c) neither blues nor classical
Solution # 7:
a) rock
n A
n S 
8
P rock  
24
1
P rock  
3
P rock  
b) not classical
Pnot classical   1  Pclassical 
Pnot classical   1 
4
24
Pnot classical  
20
24
5
Pnot classical  
6
c) neither blues nor classical
Pneither blues nor classical   Prock 
Pneither blues nor classical  
8
24
1
Pneither blues nor classical  
3
Page 13 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Sec.4.3a Finding Probability Using Sets
A Set

ie:
is a collection or group of objects witch have a specific relationship.
a set of dishes (all have the same pattern)
the set of natural numbers (no fractions or decimals, 1 and up)
An Element

ie:
is all the members of a set and are usually listed in curly brackets { } and each entry is separated by
comma’s.
D = { tea cup, saucer, bowl } this is called a finite set. n ( D ) = 3
N = { 1 , 2 , 3 , 4 , 5 , ………. } this is an infinite set
n ( N ) = undefined
Null Set

is the set with no members in it. It is denoted by  , or empty { } but not by{ 0 }.
Disjoint Sets

two sets that have no elements in common
ie:
C = { 4 , 6 , 7 , 9 , 10 }
A = { 1 , 5 , 11 , 13 }
C & A are disjoint sets.
Subset

ie:
if all elements of one set are contained (also found in) a second set, then the first is a subset of the
second.
C={1,3,5,7}
G={1,3,5,7,9}
C is called a subset of G
 CG
Compliment

C ` is the compliment of C and contains all elements not in C but in the sample space S.
Let our sample space be the odd #’s between 1 & 10
S  1 , 3 , 5 , 7 , 9 
If C is defined as C = { 1 , 3 , 5 , 7 } the same as above, then C ` = { 9 }
Page 14 of 45
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MDM4U
Unit 1 Master
Example #1:
Describe the relationship between the sets described below using the terms equal, disjoint and subset.
a)
N = { the natural numbers } & W = { the whole numbers }
b)
R = { colours in a rainbow } & B = { grey, black, brown }
c)
D = { prime divisors of 34 }
& E = { 2 , 17 }
Solution # 1:
a) N = { the natural numbers } & W = { the whole numbers }
 N  W  the natural numbers are contained within the whole numbers
b) R = { colours in a rainbow } & B = { grey, black, brown }
 R & B  are disjoint, no element is the same in the two sets.
c) D = { prime divisors of 34 } & E = { 2 , 17 }
 D & E  are equal because all the elements in D are also in E.
All sets are subsets of a universal set for that particular situation
ie:
Page 15 of 45
the vowels
V = { A , E , I , O , U } are a subset of the universal set
called the alphabet
V  U
where U = { A , B , C , D , E , F , G , …………, X , Y , Z }
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MDM4U
Unit 1 Master
Let's say that our universe contains the numbers 1, 2, 3, and 4.
Let A be the set containing the numbers 1 and 2. That is, A = {1, 2}.
Let B be the set containing the numbers 2 and 3; that is, B = {2, 3}.
Then we have the following relationships:
The Union

contains all the elements from both sets without repeats
ie:
A  B  means everything in A or B
The Intersection

contains all the elements that are common to both sets
ie:
A  B  means everything in A and B
Recall the Compliment
ie: A ` means everything in the universe not in A
Try This?
a) Find  A  B `
b) Find  A  B `
 A  B`
 A  B`
Page 16 of 45
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MDM4U
Unit 1 Master
Example # 2:
Determine the elements in the union & intersection of M & N.
M = { 1 , 3 , 6 , 9 , 12 }
&
N = { 2 , 4 , 6 , 8 , 10 , 12 }
Solution # 2:
Union of M & N is: M  N  { 1 , 2 , 3 , 4 , 6 , 8 , 9 , 10 , 12 }
Intersection of M & N is :
M  N  { 6 , 12 }
Example # 3:
Determine the elements in the following, given;
S={K,E,Y,B,O,A,R,D}
A={E,O,A}
B={B,O,A,R,D}
Find;
a) A `
b) B `
c)
 A  B
d)
 A  B
e)
 A  B`
Solution # 3:
a) A `= { K , Y , B , R , D }
b) B ` = { K , E , Y }
c) A  B  { O , A }
d) A  B  { E , B , O , A , R , D }
e)
A  B'  { K , E ,Y , B , R , D }
Page 17 of 45
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Unit 1 Master
Sec.4.3b Finding Probability Using Venn Diagrams
Try This?
Example #1:
Find the probability of rolling an even number or a number greater than 3 when rolling a die. These two events
are dependent on each other. ie: there is overlap between the two events.
Solution # 1:
S={1,2,3,4,5,6}
n(S) = 6
Let A be the event of rolling an even number.
A={2,4,6}
n(A) = 3
Let B be the event of rolling a number greater than 3
B={4,5,6}
n(B) = 3
(A  B) = { 4 , 6 }
n(A  B) = 2
 P  A  B   P  A  P B   P  A  B 
3 3 2
 
6 6 6
2

3

*** you must take away the intersection ***
Now let’s look at a situation where the two events are not dependent on each other. They are independent or
disjointed events, sometimes also referred to as mutually exclusive.
Page 18 of 45
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MDM4U
Unit 1 Master
Example # 2:
If two dice are rolled, one red and one green (so that you can tell the difference between them), find the
probability that a sum of;
a)
3 or 12 will occur
Solution # 2a
Let A be the event of rolling a sum of 3
Let B be the event of rolling a sum of 12
** These two events are mutually exclusive, independent **
P  A 
P B  
2
36
Dice # 1
Sums
of two
dice
1
2
3
4
5
6
b)
P  A  B   P  A  P  B 
1
36
1
2
Dice # 2
3
4
5
2
3
4
5
6
7
3
4
5
6
7
8
4
5
6
7
8
9
5
6
7
8
9
10
6
7
8
9
10
11
2
1

36 36
3

36
1

12

6
7
8
9
10
11
12
6 or a pair will occur
Solution # 2b
Let A be the event of rolling a sum of 6
Let B be the event of rolling a pair
** These two events are not mutually exclusive, they are dependent because there is 6 in both sets! **
P  A 
5
36
P B  
6
36
P A  B  
1
36
** account for 6 in both **
 P  A  B   P  A  P B   P  A  B 
5
6
1


36 36 36
10

36
5

18

Page 19 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Mutually exclusive events occur when 2 events can not happen at the same time.
B
A
** there is nothing to take away when the two sets are disjoint **
 P A  B  P A  PB
Example # 3:
Jody attends a fundraiser at which T-shirts are given away as door prizes. Door prize winners are randomly
given a shirt from a stock of 2 black shirts, 4 blue shirts and 9 white shirts. Assuming that Jody wins the first
door prize what is the probability that she will get a black or white shirt?
Solution # 3:
Let A be the event of drawing a black shirt
Let B be the event of drawing a white shirt
** these are mutually exclusive, independent events **
P  A 
2
15
P B  
9
15
P  A  B   P  A  P B 
2
9

15 15
11

15

Page 20 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Example # 4:
A card is randomly selected from a standard deck of cards. What is the probability that either a heart or a face
card (JQK) is selected.
Solution # 4:
Let A be the event of selecting a heart
Let B be the event of selecting a face card
** these are not mutually exclusive, they are dependent events **
P  A 
P B  
13
52
12
52
P A  B  
3
52
** J of heart, Q or heart, K of hearts **
 P  A  B   P  A  P B   P  A  B 
13 12 3


52 52 52
22

52
11

26

Example # 5:
As a result of a recent survey it was estimated that 85% of the grade twelve population at a local high school
enjoys an alcoholic beverage once a week, 35% smoke at least one cigarette a day and 25% indulge in both
habits. What is the probability that an individual chosen at random from the targeted population either smokes
or drinks.
Solution # 5:
Let A be the event that the individual drinks
Let B be the event that the individual smokes
** these are not mutually exclusive, they are dependent events **
P(A) = 0.85
P(B) = 0.35
P(A  B) = 0.25
 P  A  B   P  A  P  B   P  A  B 
 0.85  0.35  0.25
 0.95
Therefore the probability that an individual either smokes or drinks when chosen at random from the targeted
population is 95 %.
Page 21 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Sec. 4.4 Conditional Probability
When calculating probabilities, additional information may become available that will affect the calculation of
the outcome.
ie: If you are interested in the probability of your passing the next test, the probability will depend on whether
you studied, if you worked late the night before, if you had a good breakfast, have you been doing your
homework regularly, etc…
When additional information is known that will affect the probability of an outcome, we calculate what is called
the conditional probability represented by P(A | B), read as “the probability of event A, given that the event B
has occurred”, or, “the probability of A given B”.
P A | B  
P A  B 
P B 
PB | A 
P  B  A
P  A
**N.B.** The P A  B is identical to the PB  A
Example #1:
What is the probability of rolling a sum greater than 7 with two dice if it is known that the first die rolled is a 3.
Solution # 1:
Let A be the event of a sum greater than 7.
Let B be the event of the first roll being a 3.
P A  B 
P B 
2
6


36 36
2

6
1

3
P A | B  
1
The probability of rolling a sum greater than 7, given that the first roll was a 3 is 3 , 0.33333 or 33.3%
We can rearrange the conditional probability formula to find the intersection of two events if the conditional
probability is known.
P A | B  
P A  B 
P B 
 P  A  B   PB   P  A | B 
Page 22 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Example # 2:
When drawing two cards from a standard deck of 52 cards (without replacement) what is the probability that
they are both aces?
Solution # 2
Let A be the event that the first card drawn is an ace.
Let B be the event that the second card drawn is an ace.
P B  A  P  A  B   P  A  P B | A
4 3

52 51
1 1
 
13 17
1

221

1
Therefore there is a 221 or approx. 0.45% chance that both aces will be drawn.
Example # 3:
A professional hockey team has 8 wingers. Three of these wingers are 30-goal scorers or “snipers”. Every fall
the team plays a match with the club’s farm team. The coaches agree to select two wingers at random from the
pro-team to play for the farm team. What is the probability that two snipers will play for the farm team.
Solution # 3:
Let A be the event the first winger chosen is a sniper. Let B represent the second winger chosen is a sniper.
P B  A  P  A  B   P  A  P B | A
3 2
 
8 7
3 1
 
4 7
3

28
3
Therefore there is a 28 or approx. 10.71% chance of selecting two wingers that are both snipers.
Page 23 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Example # 4:
2
1
The probability that Ann will go to Queens is 5 . The probability that she will go to another university is 2 .
3
If Ann goes to Queens, the probability that her boyfriend, James, will follow her and go to Queens is 4 . What
is the probability that both Ann and James will attend Queens?
Solution # 4:
Let A be the event that Ann goes to Queens.
Let B be the event that James goes to Queens.
P B  A  P  A  B   P  A  P B | A
2 3

5 4
1 3
 
5 2
3

10

3
Therefore there is a 10 or approx. 30% chance that both will attend Queens.
Example # 5:
A brown bag contains 2 white and 2 red balls. A ball is drawn at random. If the ball chosen is white, then it is
returned to the bag along with an extra white ball. If the ball chosen on the first draw is red, then it is returned
to the bag with 2 extra red balls. Then a second ball is drawn from the bag. Find the probability that the second
ball drawn is red?
Solution # 5:
Let R represent choosing a red ball.
Let W represent choosing a white ball.
**drawing a tree diagram will help sort out the different probabilities**
Pred 2nd pick   Pred & red or Pwhite & red 
 1  4   1  2 
       
 2  6   2  5 
1 1
 
3 5
8

15
8
Therefore the probability of getting a red ball on the second drawn is 15 or approx. 53.3%
Page 24 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Sec.4.5 Tree Diagrams & Dependent vs. Independent Events
Independent Events
 the probability of one event happening is not affected by the outcome of the other events.
Example #1:
Consider an experiment in which you roll a 6 sided dice and then toss a coin. What is the probability of tossing
tails and rolling and even number.
Solution # 1:
Let E be the event of rolling an even number.
Let T be the event of tossing tails.
We can create an outcomes table to show all of the possibilities for these particular events.
Roll
Toss
1
H
1
T
2
H
2
T
3
H
3
T
4
H
4
T
5
H
5
T
6
H
6
T
Page 25 of 45
neven & tail 
n(total )
3

12
1

4
Peven & tail  
Peven & tail   Peven  Ptail 
3 1
 
6 2
1

4
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Example # 2: (Review of Conditional Probability)
A card is drawn from a deck of cards. What is the probability that it is a jack given that it is a face card?
Solution # 2
Let A be the even of drawing a Jack.
Let B be the event of drawing a face card.
P A  B 
P B 
4
 52
12
52
4 52
 
52 12
1

3
P A | B  
Therefore there is a 33.3% chance of drawing a jack knowing that the card is a face card.
Recall, when 2 events have no effect on one another they are independent events.
P  A  B   P  A  P  B 
Example # 3:
Justin estimates that his probability of passing English is 0.9 and his probability of passing Math is 0.8. Find
the probability that Justin will;
a) pass both Math and English
b) pass Math but fail English
c) not pass Math or English
Solution # 3:
Let A be the event of passing Math.
a)
Let B be the event of passing English.
P  A  B   P  A  P B 
 0 .8  0 .9
 0.72
Therefore there is a 72% chance of passing both Math & English.
Page 26 of 45
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MDM4U
b)
Unit 1 Master
P  A  B '  P  A   P  B '
 0 .8  0 .1
 0.08
Therefore there is an 8% chance of passing Math but failing English.
c)
P A ' B '  P A ' PB '
 0 .2  0 .1
 0.02
Therefore there is a 2% chance of not passing Math or English.
Example # 4:
4
George estimates that the probability of getting the next guess right if the previous one was right on a test is 5 .
2
But the probability of getting it right if the previous one was wrong is only 5 . If the probability of getting her
3
first guess right is 4 , what is the probability of getting the second guess right?
Solution # 4:
Let A be the event that the first is correct.
Let B be the event that the second is correct.
**Draw a tree diagram to organize your probabilities**
P2nd is correct   P A  B   P A ' B 
12 2

20 20
14

20
7

10

Therefore there is a 70% chance of getting the second guess correct.
Page 27 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Sec. 4.6a Factorials & Permutations
Example #1:
In how many ways can you arrange 7 smarties that are all different colors.
Solution # 1:
7 6 5 4 3 2 1 = 7 ! = 5040
Factorials
n! nn  1n  2n  3........  3  2 1
Example # 2:
Evaluate each of the following factorial expressions.
a)
9!
d)
8!
6!
b)
0!
e)
75 !
71 !
c)
1!
f)
17 !
15 ! 2 !
Solution # 2
a) 9 ! = 362 880
d)
8  7  6  5  4  3  2 1
6  5  4  3  2 1
 56

Page 28 of 45
b) 0 ! = 1
e)
75  74  73  72  71  70  .........
71  70  .........
 29,170,800

c) 1 ! = 1
f)

17  16  15!
15! 2!
 132
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Example # 3:
a) Albert, Jo, Kim and Jane are to make speeches in English class. How many different arrangements are there
for who will go 1st, 2nd, 3rd and 4th ?
b) If the English class has 10 students to present but only has time for 4 students how many possible
arrangements are there?
Solution # 3:
a) 4 3 2 1 = 4 ! = 24
b) 10 9 8 7 = 10P4 =
10!
10!

 5040
10  4! 6!
A permutation is the number of different arrangements that can be made from a selection of objects in a definite
order.
In general, if we have “n” distinct objects and we want to pick “r” of them to arrange, the number of possible
n!
different arrangements is n  r  ! , denoted by P(n,r) or nPr.
Example # 4:
An investment club of 5 members wants to select a president and vice-president. In how many ways can this be
done?
Solution # 4:
Choosing 2 people from 5 where order matters. Therefore, 5P2 or P(5,2)
5!
5!

 5  4  20
5  2 ! 3 !
Page 29 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Example # 5:
a) Find the number of permutations of the letters of the word “DIPLOMA” for which the letter “L” must be in
the middle?
b) How many ways are there of arranging the letters of the word “DIPLOMA” so that the “O” and “I” are
together?
Solution # 5:
a) 6 5 4
L 3 2 1 = 6 ! = 720
b) treat “OI” as a boxed item, there are 6 ! ways of arranging that box and then 2 ! ways of arranging the “OI”
inside the box.
Therefore, 6 ! 2 ! = 1440
Example # 6:
At a used car lot, six different model cars are lined up at the front of the lot. In how many ways can this be
done if:
a) the only car with a sunroof must be at the right end of the line?
b) the three black cars must be together?
Solution # 6:
a) 5 4
3 2 1 Sunroof = 5 ! = 120
b) “3 black cars” 3 2 1 = 4 ! 3 ! = 144
Page 30 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Arrangement of objects when some are alike
Example # 7:
Consider taking a photograph of three friends, Joel, Josh & Lisa.
a) In how many ways can the friends be lined up?
b) Now consider that Joel & Josh are identical twins. How many different ways would the photo look to
people who do not know the twins?
Solution # 7:
a) 3 ! = 6
b)
Joel
Josh
Lisa
JJL
Josh
Joel
Lisa
JJL
Joel
Lisa
Josh
JLJ
Josh
Lisa
Joel
JLJ
Lisa
Joel
Josh
LJJ
Lisa
Josh
Joel
LJJ
There are only 3 distinct ways to view the photo.
3!
3
2!
In general, “n” objects, of which some are alike, can be arranged in:
n!
a !b!c !
Example # 8:
In how many ways can the letters of the word “MINIMUM” be arranged?
Solution # 8:
7!
 420
3
!
2
!
a)
Page 31 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Example # 9:
Anya is starting out on her evening run. Her route always takes her 8 blocks east and five blocks north to her
grandmothers house. She likes to vary her path every night. How many different possibilities are there?
Solution # 9:
13 !
 1287
a) 8 ! 5 !
Example # 10:
Determine the number of arrangements possible using all the letters of the word “PINEAPPLE” if;
a) there are no restrictions
b) the “N” must go first
c) the “A” must be first and the “N” must be last
Solution # 10:
a)
9!
 30240
3!2!
b) Place the “N” and arrange the other 8 letters. Therefore,
1 8 !
 3360
3!2!
c) Place the “A” & “N” and arrange the other 7 letters. Therefore,
1 7 !
 420
3!2!
Page 32 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Sec. 4.6b Probabilities and Permutations
Since,
P event  

nevent set 
noutcome set 
nwith restrictio ns 
nwithout restrictio ns 
Example #1:
When the kindergarden class of 18 students is lining up, what is the probability that Beth is first?
Solution #1:
n(Beth first) = 17 !
PBeth 1st  
n(no restrictions) = 18 !
17 ! 1

18 ! 18
Example #2:
In making a 7 digit number by arranging the digits 1 , 2 , 2 , 3 , 5 , 5 , 6 what is the probability of getting a
number less than 3 000 000?
Solution #2:
3 6 5 4 3 2 1
nrestrictions  
3 6 !
2!2!
nwithout restrictio ns  
7!
2!2!
3 6 !
2!2! 3
P 3,000,000  

7!
7
2!2!
Page 33 of 45
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MDM4U
Unit 1 Master
Example #3:
When the 5 starting skaters on the hockey team line up at the blue line, what is the probability that the two
defence men are not beside each other?
Solution #3:
n(2 defence men are together) = 4 3 2 1 x 2 !
Ptogether  
nwithout restrictions   5 !
2 ! 4 ! 2

5!
5
 Pnot together  1  Ptogether
 1

2
5
3
5
Example #4:
In how many ways can a President, Vice-President and a Secretary be arranged around a circular table?
Solution #4:
For every 3 different arrangements in a line, there is only 1 corresponding arrangement in a circle.
Therefore, you have to fix one person around a circular table to get a new arrangement.
3!
2
3
In General,
# of ways 
Page 34 of 45
n!
 n  1!
n
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MDM4U
Unit 1 Master
Sec. 4.7a Combinations
A selection from a group of items without regard to order is called a combination.
Example #1:
List all the two letter permutations that can be made using the letters “A , B , C” without repetition.
Solution # 1:
AB
AC
BC
BA
CA
CB
Therefore, the # of different permutations is 3 ! = 6
Example # 2:
Now, a construction crew of two people is to be chosen from Al, Ben and Charlie. List all possible groupings.
Solution # 2
AB
AC
BC
Therefore, the # of different groups is;
3
3!
2!
Since arranging the two people once selected does not matter.
The combinations of “n” distinct objects taken “r” at a time is a selection of “r” of these objects without regard
to order.
 n
n!
C n, r   n C r    
 r  n  r ! r !
Notation:
Page 35 of 45
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MDM4U
Unit 1 Master
Example # 3:
Evaluate the following combinations;
a)
8
  
 3
a)
8
   56
 3
b)
8
  
 5
b)
8
   56
 5
Solution # 3:
Why are they the same (using the definition of combinations and factorials)? Connection to Pascal’s Triangle?
Example # 4:
From 20 members of the parent council;
a) how many ways can you choose a president, vice-president and a secretary?
b) in how many ways can you choose three people to bring cookies?
Solution # 4:
20
P3 
a)
20 !
 20  19  18  6840
20  3!
20
b)
C3 
20 !
20  19  18

 1140
20  3! 3 !
3 2
Example # 5:
To play lotto 649, you must select 6 different numbers from 1 to 49 (the order of the numbers does not matter).
How many ways can this be done?
Solution # 5:
49
C6 
49 !
 13,983,816
49  6! 6 !
Page 36 of 45
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MDM4U
Unit 1 Master
Example # 6:
Joan runs a small landscaping business. She has on hand 12 kinds of rose bushes, 16 kinds of small shrubs, 11
kinds of evergreen seedlings and 18 kinds of flowering lilies. In how many ways can Joan fill an order if a
customer wants;
a) 15 different varieties consisting of 4 roses, 3 shrubs, 2 evergreens and 6 lilies?
b) Either 4 different roses or 6 different lilies?
Solution # 6a)
Roses =
12
C4 and
Therefore,
12
Shrubs =
16
C3 and
Evergreens =
11
C2
and
Lilies =
18
C6
C 4 16 C3 11 C 2 18 C 6
 2.83  1011
Solution # 6b)
12
Roses = 12 C4 or Lilies =
18 C 6
Therefore,
C 4  18 C 6
 495  18564
 19059
Example # 7:
A delegation of three people is to be chosen from a group of community volunteers consisting of four lawyers, a
minister, and three retail merchants. In how many ways can this group be formed if at least one retailer must be
a member of the delegation?
Solution # 7a
Direct Reasoning (consider 3 cases)
i) 1 retailer, or
ii) 2 retailers, or
iii) 3 retailers
 3  5 
    3  10  30
1  2 
 3  5 
    3  5  15
 2 1 
 3  5 
    1 1  1
 3  0 
Therefore, # of ways is 30 + 15 + 1 = 46
Solution # 7b
Indirect Reasoning
(find the total # of 3 person groups without restrictions, then subtract the # with no retailers)
8  5
      46
 3  3 
Page 37 of 45
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MDM4U
Unit 1 Master
Example # 8:
There are 8 boys and 12 girls in a drama class. How many ways can a committee of 5 be selected in each case;
a) There must be exactly 2 boys and 3 girls?
b) There must be at least 2 boys?
Solution # 8:
 8 12 
    28  220  6160
a)  2  3 
b)
Direct Reasoning;
2 boys + 3 boys + 4 boys +5 boys
Indirect Reasoning
All – 1 boy – no boys
 20 
no restrictions     15504
5 
 8 12 
1 boy      3960
1  4 
12 
no boy     792
5 
Therefore, 15504 – 3960 – 792 = 10752
Page 38 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Sec. 4.7b Probability & Odds
Try This?
Example #1:
When drawing two cards from a regular deck of 52 cards without replacement, what is the probability that they
are both aces?
Solution # 1a:

4
3
Logically the chance for the first ace is 52 and the chance for the second ace is 51 . As seen in our
previous examples from the other day, this is approx. 0.0045
Solution # 1b:

Using combinations
Ptwo aces 

# of combinations with restrictio ns
# of combinations without restrictio ns
 4
 
2
    0.0045
 52 
 
2 
Example #2:
From a class of 12 boys and 11 girls, what is the probability that a group of 5 people chosen at random contains:
a) exactly 3 girls
Solution # 2a
a)
1112 
  
3 2
   
 23 
 
5 
 0.3236
b) at least one boy
Solution # 2b
Pat least1
b)
 1  Pno boys 
1112 
  
5 0
 1    
 23 
 
5 
 1  0.01373
 0.98627
Page 39 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
ODDS
1
We learned that the probability of rolling a 6 in a single roll of a die is 6 . The odds of rolling a six is 1:5 . The
odds against rolling a six are 5:1.
The odds for (in favour) of an event A can be expressed as the ratio P(A):P(A’), remember that P(A’) = 1 –
P(A).
P  A '
P  A
odds in favour of A 
odds against A 
P  A '
P  A
Example #3:
A messy drawer contains three red socks, five white socks and four black socks. What are the odds in favour of
randomly drawing a red sock?
Solution # 3:
Let A be the event of drawing a red sock.
P  A 
P A'  1 
3 1

12 4
1 3

4 4
P  A
P A'
1
 4
3
4
1 4 1
  
4 3 3
odds in favour 
Therefore the odds in favour of drawing a red sock are 1:3
Example #4:
If the chance of a snowstorm in Windsor, in January is estimated at 0.4, what are the odds against Windsor
having a snowstorm next January?
Solution # 4:
Let A be the event of a snowstorm in January.
P(A) = 0.4
P(A’) = 0.6
odds against 
0.6 3

0.4 2
Odds against a snow storm are 3:2. Since this is greater than a 1:1 ratio, snow is less likely to occur.
Page 40 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Example # 5:
A university professor, in an effort to promote good attendance habits, states that the odds of passing her course
are 8:1 when a student misses fewer than five classes. What is the probability that a student with good
attendance will pass?
Solution # 5:
Let A be the event that a student with good attendance passes.
odds in favour of A 
P  A
P A'
8
P  A

1 1  P  A
8  8 P  A  P  A
P  A 
8
9
Therefore the odds in favour of a student with good attendance passing is 8:9
Example # 6:
In choosing two cards from a deck of 52, what are the odds in favour of both cards being black?
Solution #6:
Pboth black 
 26 
 
2
 
 52 
 
2 

25
102
25
odds in favour  102
77
102
25

77
Therefore the odds in favour of both cards chosen being black is 25:77
Page 41 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Example #7:
If drawing 5 cards from a deck of 52, what are the odds in favour of a full house?
Solution # 7:
P full house 
 4  4
13 12 
3
2
    
 52 
 
5 
3744
2,598,960
6

4165

6
odds in favour  4165
4159
4165
6

4159
Therefore there is a 6:4159 odds in favour of getting a full house.
Page 42 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
Sec. 4.7c Problem Solving with Combinations
Example #1:
Twenty people are to travel in a bus from the airport to the hotel at a resort. The bus is designed for use in
tropical climate, it can carry 12 passengers outside and eight inside. If four of the passengers refuse to travel
outside and five will not travel inside, in how many ways can the passengers be seated if the demands of the
passengers are all met?
Solution # 1:
Seat all the picky people first.
Inside – 4 must be, 4 left
Outside – 5 must be, 7 left
Therefore, 9 already seated, 11 left to sit!
11 4 
    330
 7  4 
outside
inside
Example # 2:
An artist has an apple, an orange and a pear in his refrigerator. In how many ways can the artist choose one or
more piece of fruit for a still life painting?
Solution # 2
Artist has 2 choices for each piece of fruit.
Include or Not!
# of choices = 2 x 2 x 2 = 8
But not choosing each will leave nothing to paint.
We need to have at least one piece of fruit.
Therefore, 23 – 1 = 7
Page 43 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
The total number of combinations containing at least one item chosen from a group of “n” distinct items
is 2n – 1. If we want to include the null set we need to use just 2n.
Example # 3:
In how many ways can you form a committee if you have a delegate of six members?
Solution # 3:
 6  6  6  6  6  6  6
                     2 6
 0  1   2   3   4   5   6 
However, you can’t have a committee with 0 people or one person therefore the first two combinations don’t
count.
26 – 1 – 6 = 64 – 1 – 6 = 57
Therefore, there is 57 ways to choose at least one person from a 6 member delegate.
All possible combinations with some identical items
Example # 4:
Katie is responsible for stocking the coffee room at her office. She can purchase up to 3 cases of cookies, four
cases of soft drinks and two cases of coffee packets without having to send the order through accounting. How
many different direct purchases can Katie make?
Solution # 4:
Cookies
0
1
2
3
3
4
Soft drinks
0
1
2
Coffee
0
1
2
= (Cookies x Soft Drinks x Coffee) – 1
= (4 x 5 x 3) – 1
= 59
Page 44 of 45
Prepared by: Scott McEwen
MDM4U
Unit 1 Master
If at least one item is chosen, the total number of selections that can be made from “p” items of one kind,
“q” items of another kind, etc……. is (p+1)(q+1)……- 1
Example # 5:
Patti wants to throw coins in the fountain at the mall and make a wish.
a) She has a penny, a nickel, a dime and a quarter in her pocket. How many different sums of money can she
throw into the fountain?
b) How many different sums of money could she throw if she had three more pennies?
Solution # 5:
a)
P
(2
b)
x
P
(5
Page 45 of 45
N
2
D
x
N
x
2
2
Q
x
D
x
2
2)
-1=
15
-1=
39
Q
x
2)
Prepared by: Scott McEwen