Unit 7 - Probability
Analytical Geometry
Bellringer 8/11/14
• What do you know about Probability?
The Probability Song
(T.D. Version)
Today we’re going to learn about probabilities,
and we’re going to make diagrams that look like trees.
We’re going to flip some coins so we can draw our
trees,
and we’re going to calculate our probabilities.
Probability’s the likelihood of a chance event. Hey!!
Ho!! Hey!!
It’s between 0 and 1 and that’s just common
sense. Hey!! Hey!!
An event is any outcome from an experiment.
If you can figure them all out, then you’re intelligent.
(Repeat)
Today we’re going to learn about probabilities,
On the way to Probability….
What is Probability?
Probability is a measure of the likelihood of an event. It is the ratio of the number of
ways a certain event can occur to the number of possible outcomes.
1. The probability of a given
event is a fraction between
0 and 1.
2. The sum of all probabilities of
every outcome should always
equal 1.
3.
What does it mean to have a probability of zero?
A probability of zero means that an event is impossible.
What does it mean to have a probability of One?
Question: A single 6-sided die is rolled. What is the probability of rolling
a number less than 7?
Answer: Rolling a number less than
7 is a certain event since a single die has 6 sides, numbered 1 through
6.
A probability of one means that an event is certainly going to happen.
Why are probabilities between 0 and 1?
Probabilities are always between 0 and 1 because every event has a chance of
occurrence that lies between 0% (no chance of happening, corresponds to 0) and
100% (certainty of happening, corresponds to 1).
Tree
Diagrams
• Tree diagram: A tree-shaped diagram that illustrates sequentially the
possible outcomes of a given event.
• Here is a tree diagram for the toss of a coin:
Create a tree diagram for the
possible outcomes and
probabilities for the two tosses.
Probabilities
Tree Diagram
The Sum of all Possible outcomes
Bellringer 8/12/14
• Kendra is playing a card game with a standard 52-card deck. She wants her first
draw to be a heart or an ace.
1. How many ways can Kendra draw a heart or an ace?
2. How many ways can Kendra draw a card that is neither a heart nor an ace?
Introduction
Probability is a number from 0 to 1 inclusive or a percent from 0% to 100% inclusive
that indicates how likely an event is to occur. In the study of probability, an
experiment is any process or action that has observable results. The results are called
outcomes. The set of all possible outcomes is called the sample space. An event is
an outcome or set of outcomes of an experiment; therefore, an event is a subset of
the sample space, meaning a set whose elements are in another set, the sample
space. In this lesson, you will learn to describe events as subsets of the sample space.
For example, drawing a card from a deck of cards is an experiment. All the cards in
the deck are possible outcomes, and they comprise the sample space. The event
that a jack of hearts is drawn is a single outcome; it is a set with one element: {jack of
hearts}. An element is an item or a member in a set. The event that a red number
card less than 5 is drawn is a set with eight elements: {ace of hearts, ace of
diamonds, 2 of hearts, 2 of diamonds, 3 of hearts, 3 of diamonds, 4 of hearts, 4 of
diamonds}. Each event is a subset of the sample space.
Note that the words “experiment” and “event” in probability have very specific
meanings that are not always the same as the everyday meanings. In the example
on the previous slide, “drawing a card” describes an experiment, not an event.
“Drawing a jack of hearts” and “drawing a red number card less than 5” describe
events, and those events are sets of outcomes.
The important distinction to remember is that an event is a set of one or more
outcomes, while the action or process is the experiment.
Bellringer 8/13/14
1. What is the Sample Space of rolling 2 dice?
2. Write an event for the experiment mentioned in Question1.
Key Concepts
A set is a list or collection of items. Set A is a subset of set B,
denoted by A ⊂ B, if all the elements of A are also in B. For
example, if B = {1, 2, 3, x}, then some subsets of B are {1, 2, 3}, {2,
x}, {3}, and {1, 2, 3, x}. Note that a set is a subset of itself.
An empty set is a set that has no elements. An empty set is
denoted by . An empty set is also called a null set.
Equal sets are sets with all the same elements. For example,
consider sets A, B, and C as follows:
A is the set of integers between 0 and 6 that are not odd.
B is the set of even positive factors of 4.
C = {2, 4}
A = B = C because they all have the same elements, 2 and 4.
The union of sets A and B, denoted by
, is the set of
elements that are in either A or B or both A and B.
Bellringer 8/14/14
1)What is the probability of rolling an even
number on a 6-sided die?
1)For the previous question write the set for
the Sample space and the subset for the
event.
Common Errors/Misconceptions
1) confusing the meanings of event and experiment
2) confusing union and intersection of sets
3) neglecting order, thereby neglecting to
identify different outcomes such as HT and TH
Key Concepts, continued
An intersection is a set whose elements are each in both of two other sets. The
intersection of sets A and B, denoted by
the set of elements that are in both A and B.
For example, if A = {a, b, c} and B = {m, a, t, h}, then:
, is
A È B = {a, b, c, m, t, h}
A Ç B = {a}
An event is an outcome or set of outcomes, so an event is a subset of the sample
space.
The complement of set A, denoted by , is the set of elements that are in some
universal set, but not in A. The complement of A is the event that does not occur.
Sometimes it is helpful to draw tables or diagrams to visualize outcomes and the
relationships between events.
A Venn diagram is a diagram that shows how two or more sets in a universal set are
related. In this diagram, members in event A also fit the criteria for members in event
B, so the circles overlap:
Bellringer 8/18/14
1) What is the Probability of your phone
landing screen side down when you
drop it?
2) How do you express Probability as a
ratio?
**Turn in your Take home test to the appropriate current work folder for your class**
1st – Green, 2nd – Gray, 7th - Blue
Probability words to know!
Probability: a measure of the
likelihood of an event. It is the ratio
of the number of ways a certain
event can occur to the number of
possible outcomes. The probability
of a given event is a fraction
between 0 and 1.
The sum of all
probabilities of every outcome
should always equal 1.
• Experiment: a process or ACTION
with an observable result.
• Outcomes: the observable results
of an experiment.
• Sample Space: the set of ALL
possible outcomes of an
experiment.
• Event: an outcome or set of
outcomes of an experiment.
• Subset: a set whose elements are
all in another set.
Need to know about Sets!
• Set: a list or collection of items
• Subset – denoted by ⊂ , as in A ⊂
B where A is a subset of B, where
all of the elements of A are also in
B
• Empty set: also known as a null
set, is a set with no elements
denoted by ∅
• Equal sets: sets with the exact
same elements
• Union of Sets: joining of all of the
elements of two sets denoted by
𝐴 ∪ 𝐵, where A and B are being
combined
• Intersection of sets: the elements
that are alike or shared by two
sets denoted by 𝐴 ∩ 𝐵, the
intersection of A and B contains
the elements that are in both A
and B
• Complement of a set: the set of
elements that are in some
universal set, but not in set A,
denoted by Ā or A’
Common Errors/Misconceptions
• confusing the meanings of event and experiment
• confusing union and intersection of sets
• neglecting order, thereby neglecting to identify
different outcomes such as HT and TH
Guided Practice
• Example 1
• Hector has entered the following names in the contact list of his new cell phone:
Alicia, Brisa, Steve, Don, and Ellis. He chooses one of the names at random to call.
Consider the following events.
• B: The name begins with a vowel.
• E: The name ends with a vowel.
• Draw a Venn diagram to show the sample space and the events B and E. Then
describe each of the following events by listing outcomes.
B
E
𝐵∪𝐸
𝐵∩𝐸
𝐵
𝐵∪𝐸
Guided Practice: Example 1, continued
1. Draw a Venn diagram. Use a rectangle for
the sample space. Use circles or elliptical
shapes for the events B and E.
Write the students’ names in the appropriate
sections to show what events they are in.
23
7.1.1: Describing Events
Guided Practice: Example 1, continued
2. List the outcomes of B.
B = {Ellis, Alicia}
24
7.1.1: Describing Events
Guided Practice: Example 1, continued
3. List the outcomes of E.
E = {Alicia, Brisa, Steve}
25
7.1.1: Describing Events
Guided Practice: Example 1, continued
4. List the outcomes of
B = {Ellis, Alicia}
E = {Alicia, Brisa, Steve}
B Ç E is the intersection of events B and E. Identify
the outcome(s) common to both events.
B Ç E = {Alicia}
26
7.1.1: Describing Events
Guided Practice: Example 1, continued
5. List the outcomes of
B = {Ellis, Alicia}
E = {Alicia, Brisa, Steve}
B È E is the union of events B and E. Identify the
outcomes that appear in either event or both events.
B È E = {Ellis, Alicia, Brisa, Steve}
27
7.1.1: Describing Events
Guided Practice: Example 1, continued
6. List the outcomes of B .
B = {Ellis, Alicia}
Sample space = {Alicia, Brisa, Steve, Don, Ellis}
B is the set of all outcomes that are in the sample
space, but not in B.
B = {Brisa, Steve, Don}
28
7.1.1: Describing Events
Guided Practice: Example 1, continued
7. List the outcomes of
.
B = {Ellis, Alicia}
E = {Alicia, Brisa, Steve}
B È E = {Ellis, Alicia, Brisa, Steve}
Sample space = {Alicia, Brisa, Steve, Don, Ellis}
B È E is the set of all outcomes that are in the sample
space, but not in B È E .
B È E = {Don}
✔
29
7.1.1: Describing Events
Guided Practice
• Example 2
• An experiment consists of rolling a pair of dice. How
many ways can you roll the dice so that the product
of the two numbers rolled is less than their sum?
Guided Practice: Example 2, continued
1. Begin by showing the sample space.
This diagram of ordered pairs shows the sample space.
(1, 1)
(2, 1)
(3, 1)
(4, 1)
(5, 1)
(6, 1)
(1, 2)
(2, 2)
(3, 2)
(4, 2)
(5, 2)
(6, 2)
(1, 3)
(2, 3)
(3, 3)
(4, 3)
(5, 3)
(6, 3)
(1, 4)
(2, 4)
(3, 4)
(4, 4)
(5, 4)
(6, 4)
(1, 5)
(2, 5)
(3, 5)
(4, 5)
(5, 5)
(6, 5)
(1, 6)
(2, 6)
(3, 6)
(4, 6)
(5, 6)
(6, 6)
Key: (2, 3) means 2 on the first die and 3 on the second
die.
7.1.1: Describing Events
31
Guided Practice: Example 2, continued
2. Identify all the outcomes in the event “the
product is less than the sum.”
For (1, 1), the product is 1 × 1 = 1 and the sum is
1 + 1 = 2, so the product is less than the sum.
For (1, 2), the product is 1 × 2 = 2 and the sum is
1 + 2 = 3, so the product is less than the sum.
The tables on the next two slides show all the
possible products and sums.
32
7.1.1: Describing Events
Guided Practice: Example 2, continued
Outcome Product Sum
(1, 1)
(1, 2)
(1, 3)
(1, 4)
(1, 5)
(1, 6)
(2, 1)
(2, 2)
(2, 3)
7.1.1: Describing Events
1
2
3
4
5
6
2
4
6
2
3
4
5
6
7
3
4
5
Product
Product
Outcome Product Sum
< sum
< sum
(2, 4)
Yes
8
6
No
(2, 5)
Yes
10
7
No
(2, 6)
Yes
12
8
No
(3, 1)
Yes
3
4
Yes
(3, 2)
Yes
6
5
No
(3, 3)
Yes
9
6
No
(3, 4)
Yes
12
7
No
(3, 5)
No
15
8
No
(3, 6)
No
18
9
No
(continued)
33
Guided Practice: Example 2, continued
Product
Product
Outcome Product Sum
< sum
< sum
5
Yes
(5, 4)
20
9
No
6
No
(5, 5)
25
10
No
7
No
(5, 6)
30
11
No
8
No
(6, 1)
6
7
Yes
9
No
(6, 2)
12
8
No
10
No
(6, 3)
18
9
No
6
Yes
(6, 4)
24
10
No
7
No
(6, 5)
30
11
No
8
No
(6, 6)
36
12
No
Outcome Product Sum
(4, 1)
(4, 2)
(4, 3)
(4, 4)
(4, 5)
(4, 6)
(5, 1)
(5, 2)
(5, 3)
4
8
12
16
20
24
5
10
15
34
7.1.1: Describing Events
Guided Practice: Example 2, continued
By checking all the outcomes in the sample space,
you can verify that the product is less than the sum
for only these outcomes:
(1, 1)
(2, 1)
(1, 2)
(3, 1)
(1, 3)
(4, 1)
(1, 4)
(5, 1)
(1, 5)
(6, 1)
(1, 6)
35
7.1.1: Describing Events
Guided Practice: Example 2, continued
3. Count the outcomes that meet the event
criteria.
There are 11 ways to roll two dice so that the product
is less than the sum.
✔
36
7.1.1: Describing Events
Bellringer 8/19/14
1. Create an experiment with 2
events.
2. Draw a Venn Diagram to illustrate
the experiment with the two events.
3. Put away your notebooks and other
items. I should only see a pencil
and eraser on your desk!
• Key Concepts
•
The conditional probability of B given A is the probability that event B occurs, given that
event A has already occurred.
•
If A and B are two events from a sample space with P(A) ≠ 0, then the conditional
probability of B given A, denoted P ( B A) , has two equivalent expressions:
( )
P BA =
P ( A and B )
P ( A)
=
number of outcomes in ( A and B )
number of outcomes in A
(
)
•
The second formula can be rewritten as P B A
,is read “the probability of B given A.”
•
Using set notation, conditional probability is written like this:
•
The “conditional probability of B given A” only has meaning if event A has occurred.
That is why the formula for P B A has the requirement that P(A)≠0.
•
The conditional probability formula can be solved to obtain a formula for P(A and B), as
shown on the next slide.
( )
P BA =
P ( A Ç B)
P ( A)
( )
• Key Concepts, continued
(
)
P BA =
P ( A and B )
P ( A)
P ( A and B )
P ( A) • P B A =
• P ( A)
P ( A)
( )
Multiply both sides
by P(A).
( )
Simplify.
P ( A) • P B A = P ( A and B )
( )
P ( A and B ) = P ( A) • P B A
•
Write the
conditional
probability formula.
Reverse the left
and right sides.
Remember that independent events are two events such that the probability of both events
occurring is equal to the product of the individual probabilities. Two events A and B are
independent if and only if
• P(A and B) = P(A) • P(B). Using set notation, P A Ç B = P A • P B
.
• The occurrence or non-occurrence of one event has no effect on the probability of the other
event.
(
•
)
( ) ( )
If A and B are independent, then the formula for P(A and B) is the equation used in the
definition of independent events, as shown below.
( )
P ( A and B ) = P ( A) • P B A
P ( A and B ) = P ( A) • P ( B )
formula for P(A and B)
formula for P(A and B) if A and B are independent
• Key Concepts, continued
•
•
•
The following statements are equivalent. In other words, if any one of them is
true, then the others are all true.
•
Events A and B are independent.
•
The occurrence of A has no effect on the probability of B; that is, P B A = P B .
•
The occurrence of B has no effect on the probability of A; that is, P A B = P A .
•
P(A and B) = P(A) • P(B).
( )
( )
( )
( )
Note: For real-world data, these modified tests for independence are
sometimes used:
P BA »P B .
( )
( )
•
Events A and B are independent if the occurrence of A has no
significant effect on the probability of B; that is,
•
Events A and B are independent if the occurrence of B has no
significant effect on the probability of A; that is,
( )
P A B » P ( A) .
When using these modified tests, good judgment must be used when
deciding whether the probabilities are close enough to conclude that the
events are independent.
Bellringer 8/20/14
1.Define Probability
2.Define Experiment
3.Define Event
4.Define Outcome
• Guided Practice
• Example 1
• Alexis rolls a pair of number cubes. What is the probability
that both numbers are odd if their sum is 6? Interpret your
answer in terms of a uniform probability model.
Guided Practice: Example 1, continued
1. Assign variable names to the events and
state what you need to find, using
conditional probability.
Let A be the event “Both numbers are odd.”
Let B be the event “The sum of the numbers is 6.”
You need to find the probability of A given B.
(
)
That is, you need to find P A B .
43
7.2.1: Introducing Conditional Probability
Guided Practice: Example 1, continued
2. Show the sample space.
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
Key: (2, 3) means 2 on the first cube and 3 on the second
cube.
44
7.2.1: Introducing Conditional Probability
Guided Practice: Example 1, continued
3. Identify the outcomes in the events.
The outcomes for A are bold and purple.
A: Both numbers are odd.
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
45
7.2.1: Introducing Conditional Probability
Guided Practice: Example 1, continued
The outcomes for B are bold and purple.
B: The sum of the numbers is 6.
(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)
(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)
(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)
(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)
(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)
(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)
46
7.2.1: Introducing Conditional Probability
Guided Practice: Example 1, continued
4. Identify the outcomes in the events
and B.
Use the conditional probability formula:
( )
P AB =
P ( A Ç B)
P (B)
.
A Ç B = the outcomes that are in A and also in B.
A Ç B = {(1, 5 ) , ( 3, 3 ) , ( 5, 1)}
B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
47
7.2.1: Introducing Conditional Probability
Guided Practice: Example 1, continued
5. Find
and
A Ç B has 3 outcomes; the sample space has 36
outcomes.
P ( A Ç B) =
number of outcomes in A and B
number of outcomes in sample space
=
3
36
B has 5 outcomes; the sample space has 36 outcomes.
P (B) =
number of outcomes in B
number of outcomes in sample space
=
5
36
48
7.2.1: Introducing Conditional Probability
Guided Practice: Example 1, continued
6. Find
( )
P AB =
P ( A Ç B)
P (B)
Write the conditional
probability formula.
3
(
)
P A B = 36
5
Substitute the probabilities
found in step 5.
36
(
)
3 36
P AB =
•
36 5
To divide by a fraction,
multiply by its reciprocal.
49
7.2.1: Introducing Conditional Probability
Guided Practice: Example 1, continued
(
)
3
P AB =
1
(
)
P AB =
36
•
36
5
1
Simplify.
3
5
50
7.2.1: Introducing Conditional Probability
Guided Practice: Example 1, continued
7. Verify your answer.
Use this alternate conditional probability formula:
number of outcomes in A Ç B
P AB =
.
number of outcomes in B
( )
AÇB
has 3 outcomes: {(1, 5), (3, 3), (5, 1)}.
B has 5 outcomes: {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}.
( )
P AB =
number of outcomes in A Ç B
number of outcomes in B
(
)
P AB =
3
5
51
7.2.1: Introducing Conditional Probability
Guided Practice: Example 1, continued
8. Interpret your answer in terms of a uniform
probability model.
The probabilities used in solving the problem are found by
using these two ratios:
number of outcomes in an event
number of outcomes in the sample space
number of outcomes in an event
number of outcomes in a subset of the sample space
52
7.2.1: Introducing Conditional Probability
Guided Practice: Example 1, continued
These ratios are uniform probability models if all outcomes in the sample
space are equally likely. It is reasonable to assume that Alexis rolls fair
number cubes, so all outcomes in the sample space are equally likely.
Therefore, the answer is valid and can serve as a reasonable predictor.
53
7.2.1: Introducing Conditional Probability
Guided Practice: Example 1, continued
You can predict the following: If you roll a pair of number cubes a large
number of times and consider all the outcomes that have a sum of 6, then
3
about of those outcomes will have both odd numbers.
5
✔
54
7.2.1: Introducing Conditional Probability
Bellringer 8/21/14
1.Define Set
2.Define Subset
3.Define Empty Set/Null Set
4.Define Equal Set
• Guided Practice
• Example 2
• A vacation resort offers bicycles and personal watercrafts for rent.
The resort’s manager made the following notes about rentals:
• 200 customers rented items in all—100 rented bicycles and 100
rented personal watercrafts.
• Of the personal watercraft customers, 75 customers were young
(30 years old or younger) and 25 customers were older (31 years
old or older).
• 125 of the 200 customers were age 30 or younger. 50 of these
customers rented bicycles, and 75 of them rented personal
watercrafts.
• Consider the following events that apply to a random customer.
• Y: The customer is young (30 years old or younger).
• W: The customer rents a personal watercraft.
(
)
(
• Are Y and W independent? Compare P Y W and P W Y
interpret the results.
)
, and
Guided Practice: Example 2, continued
1. Determine if Y and W are independent.
First, determine the probabilities of each event.
125
P (Y ) =
P (W ) =
200
100
200
(
)
(
)
P YW =
P WY =
= 0.625
Of 200 customers, 125
were young.
= 0.5
Of 200 customers, 100
rented a personal
watercraft.
75
100
75
125
= 0.75
Of 100 personal watercraft
customers, 75 were young.
= 0.6
Of 125 young customers,
75 rented a personal
watercraft.
7.2.1: Introducing Conditional Probability
57
Guided Practice: Example 2, continued
(
)
Y and W are dependent because P Y W ¹ P (Y )
and P W Y ¹ P (W ) .
(
)
58
7.2.1: Introducing Conditional Probability
Guided Practice: Example 2, continued
2. Compare
( )
P (W Y ) = 0.6
P Y W = 0.75
(
)
(
)
P YW >P WY .
0.75 > 0.6; therefore,
59
7.2.1: Introducing Conditional Probability
Guided Practice: Example 2, continued
3. Interpret the results.
(
P YW
) represents the probability that a customer is
young given that the customer rents a personal
watercraft.
P WY
represents the probability that a customer
rents a personal watercraft given that the customer is
young.
(
)
60
7.2.1: Introducing Conditional Probability
Guided Practice: Example 2, continued
The dependence of the events Y and W means that a
customer’s age affects the probability that the
customer rents a personal watercraft; in this case,
being young increases that probability because
P W Y > P (W ). The dependence of the events Y
and W also means that a customer renting a personal
watercraft affects the probability that the customer is
young; in this case, renting a personal watercraft
increases that probability because P Y W > P (Y ) .
(
)
(
)
61
7.2.1: Introducing Conditional Probability
Guided Practice: Example 2, continued
(
)
(
)
P Y W > P W Y means that it is more likely that a
customer is young given that he or she rents a
personal watercraft than it is that a customer rents a
personal watercraft given that he or she is young.
✔
62
7.2.1: Introducing Conditional Probability
Bellringer 8/22/14
1.Define Union
2.Define Intersection
3.Define Complement
4.Define Element
Bellringer 8/25/14
1. Draw a Venn Diagram where inside of the
universal set you have two circles that overlap
representing two sets.
a) Label Set A, Set B, and the Universal Set
b) Put dots in the intersection of set A and set B
c) Put zig zags over the union of set A and set B
d) Lightly shade in the Complement of A union B
2. Write the intersection, union, and complement
of A and B using the correct notation.
Bellringer 8/26/14
A = {1, 3, 5}
B = {2, 4, 6}
C= {2, 4}
1.How would you describe the events A and
B? B and C?
2.How would the Venn Diagram of sets A
and B look compared to how we usually
draw them? B and C?
• REVIEW
• Remember that probability is a number from 0 to 1 inclusive or a percent from 0% to
100% inclusive that indicates how likely an event is to occur. A probability of 0 indicates
that the event cannot occur. A probability of 1 indicates that the event is certain to
occur. The following diagram shows some sample probabilities.
• When all the outcomes of an experiment are equally likely, the probability of an event
E, denoted P(E), is
given by P ( E ) =
number of outcomes in E
number of outcomes in the sample space
.
• For situations involving more than one event, the Addition Rule sometimes applies. You
will learn about this rule.
• Key Concepts
• Probability is the likelihood of an event.
• A probability model is a mathematical model for observable facts or
occurrences that are assumed to be random. The facts or occurrences
are called outcomes. Actions and processes that produce outcomes are
called experiments.
• When all the outcomes of an experiment are assumed to be equally
likely, then the probability model is a uniform probability model.
• The frequency of an event is the number of times it occurs. The relative
frequency of an event is the number of times it occurs divided by the
number of times the experiment is performed.
• Key Concepts, continued
•
•
According to the Addition Rule, if A and B are any two events, then the
probability of A or B, denoted
• P(A or B), is given by the formula
• P(A or B) = P(A) + P(B) – P(A and B).
Using set notation, the rule is
P ( A È B ) = P ( A) + P ( B ) - P ( A Ç B ) .
•
The Addition Rule contains four probabilities:
• P(A or B), P(A), P(B), and P(A and B); if three of them are known, then the
equation can be solved to find the fourth.
•
If events A and B have no outcomes in common, then they are mutually
exclusive events, also known as disjoint events. If A and B are mutually
exclusive events, then they cannot both occur, so P(A and B) = 0.
•
This leads to the following special case of the Addition Rule for mutually
exclusive events:
P ( A or B ) = P ( A) + P ( B ) , or P ( A È B ) = P ( A) + P ( B )
Bellringer 8/27/14
Favorite Book Genre
Number that chose the
genre
Non-Fiction
10
Sci-Fi/Fantasy
15
Mystery/Suspense
12
Other
3
Relative frequency
Totals
• The above table represents the experiment
where I asked teachers to choose a favorite
book genre.
• Complete the table by finding the relative
frequencies for each answer
Pop Quiz!!
A
B
1. Find 𝐴 ∪ 𝐵.
2. Find 𝐴 ∪ 𝐵 .
3. What elements are in the Universal Set (also known as
Sample Space)?
4. What is the complement of the Universal set?
U
BellRinger 8/29/14
1. How do you think the day with the sub went?
2. Did you use your notes when you got stuck on your assignment?
3. Did you get off task often?
4. Did the sub show you the answer key?
5. Did you do everything I asked you to do while the sub was
there?
6. What do you think the note from the sub said?
7. How do you think I felt when I saw the note from the sub and the
work that you turned in?
8. What could have been done to make yesterday go better?
Bellringer 9/2/14
1. Draw a Venn Diagram of set A={1, 2, 5, 7, 8} and set B={3,
7, 9, 13} where the Universal set contains numbers 1-15
2. Do they intersect? How do you know?
3. Where is the complement of the union of A and B?
Key Concepts
•
•
According to the Addition Rule, if A and B are any two events, then the probability of
A or B, denoted
• P(A or B), is given by the formula
• P(A or B) = P(A) + P(B) – P(A and B).
Using set notation, the rule is
P ( A È B ) = P ( A) + P ( B ) - P ( A Ç B ) .
•
The Addition Rule contains four probabilities:
• P(A or B), P(A), P(B), and P(A and B); if three of them are known, then the equation
can be solved to find the fourth.
•
If events A and B have no outcomes in common, then they are mutually exclusive
events, also known as disjoint events. If A and B are mutually exclusive events, then
they cannot both occur, so P(A and B) = 0.
•
This leads to the following special case of the Addition Rule for mutually exclusive
events:
P ( A or B ) = P ( A) + P ( B ) , or P ( A È B ) = P ( A) + P ( B )
Bellringer 9/3/14
1. What are Mutually Exclusive Events?
2. What are 2 examples of Disjoint Events?
3. What is the P(A and B) when A and B are
Mutually Exclusive?
Example 1
• Donte is playing a dice game with a 20-sided die. He’s hoping for an even
number or number greater then 15 on his first roll. What is the probability
that he rolls an even number or a number greater then 15 on his first roll?
Guided Practice: Example 1, continued
1. Identify the sample space and count the
outcomes.
The sample space is the set of all cards in the deck,
so there are 52 outcomes.
76
7.1.2: The Addition Rule
Guided Practice: Example 1, continued
2. Identify the outcomes in the event and
count the outcomes.
You can use a table to show the sample space. Then
identify and count the cards that are either a club or a
face card or both a club and a face card.
Suit
2
3
4
5
6
7
8
9
1
0
Q K
A
✔ ✔ ✔
Spade
Club
J
✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔ ✔
✔ ✔ ✔ ✔
Diamon
✔ ✔ ✔
The
d event of “club or face card” has 22 outcomes.
Heart
7.1.2: The Addition Rule
✔ ✔ ✔
77
Guided Practice: Example 1, continued
3. Apply the formula for the probability of an
event.
P (E ) =
number of outcomes in E
number of outcomes in the sample space
22 11
P ( club or face card) =
=
» 0.42
52 26
78
7.1.2: The Addition Rule
Guided Practice: Example 1, continued
4. Apply the Addition Rule to verify your
answer.
Let A be the event “club.” There are 13 clubs, so A
has 13 outcomes.
Let B be the event “face card.” There are 12 face
cards, so B has 12 outcomes.
The event “A and B” is the event “club and face card,”
which has 3 outcomes: jack of clubs, queen of clubs,
and king of clubs.
79
7.1.2: The Addition Rule
Guided Practice: Example 1, continued
Apply the Addition Rule.
P ( A or B ) = P ( A) + P ( B ) - P ( A and B )
P ( club or face card) = P ( club ) + P ( face card) P ( club and face card)
13 12 3 22 11
P ( club or face card) =
+
=
=
» 0.42
52 52 52 52 26
80
7.1.2: The Addition Rule
Guided Practice: Example 1, continued
The Addition Rule answer checks out with the
probability found in step 3, so the probability of the
event is approximately 0.42.
✔
81
7.1.2: The Addition Rule
Example 2
• Students at Rolling Hills High School receive an achievement award for
either performing community service or making the honor roll. The school
has 500 students and 180 of them received the award. There were 125
students who performed community service and 75 students who made
the honor roll. What is the probability that a randomly chosen student at
Rolling Hills High School performed community service and made the
honor roll?
Guided Practice: Example 2, continued
1. Define the sample space and state its
number of outcomes.
The sample space is the set of all students at the
school; it has 500 outcomes.
83
7.1.2: The Addition Rule
Guided Practice: Example 2, continued
2. Define events that are associated with the
numbers in the problem and state their
probabilities.
Let A be the event “performed community service.”
125
Then P ( A) =
.
500
Let B be the event “made the honor roll.” Then
75
P (B) =
.
500
84
7.1.2: The Addition Rule
Guided Practice: Example 2, continued
The event “A or B” is the event “performed community
service or made the honor roll,” and can also be
written A È B.
P ( A È B) =
180
because 180 students
500
received the award for either community service or
making the honor roll.
85
7.1.2: The Addition Rule
Guided Practice: Example 2, continued
3. Write the Addition Rule and solve it for
P(A and B), which is the probability of the
event “performed community service and
made the honor roll.” P(A and B) can also
be written
86
7.1.2: The Addition Rule
Guided Practice: Example 2, continued
P ( A È B ) = P ( A) + P ( B ) - P ( A Ç B )
180
500
180
=
125
=
200
+
75
500 500
- P ( A Ç B)
- P ( A Ç B)
500 500
20
= -P ( A Ç B )
500
20
= P ( A Ç B)
500
Substitute the known
probabilities.
Simplify.
Subtract
200
500
from both sides.
Multiply both sides by –1.
87
7.1.2: The Addition Rule
Guided Practice: Example 2, continued
The probability that a randomly chosen student at Rolling Hills High
School has performed community service and is on the honor roll is
20
1
= , or 4%.
500 25
✔
88
7.1.2: The Addition Rule
Introduction / Key Concepts
• In probability, events are either dependent or independent.
• Two events are independent if the occurrence or non-occurrence of one
event has no effect on the probability of the other event. If two events are
independent, then you can simply multiply their individual probabilities to
find the probability that both events will occur.
• If events are dependent, then the outcome of one event affects the
outcome of another event. So it is important to know whether or not two
events are independent.
Introduction / Key Concepts Continued
• Two events A and B are independent if and only if
–P(A and B) = P(A) • P(B).
–But sometimes you only know two of these three
probabilities and you want to find the third. In such
cases, you can’t test for independence, so you
might assess the situation or nature of the experiment
and then make an assumption about whether or not
the events are independent. Then, based on your
assumption, you can find the third probability in the
equation.
Guided Practice
• Trevor tosses a coin 3 times. Consider the following events.
A: The first toss is heads.
B: The second toss is heads.
C: There are exactly 2 consecutive heads.
• For each of the following pairs of events, determine if the events are
independent.
A and B (This is A Ç B in set notation.)
A and C (This is A Ç C in set notation.)
B and C (This is B Ç C in set notation.)
Guided Practice: Example 1, continued
1. List the sample space.
Sample space = {HHH, HHT, HTH, HTT, THH, THT,
TTH, TTT}
92
7.1.3: Understanding Independent Events
Guided Practice: Example 1, continued
2. Use the sample space to determine the
relevant probabilities.
P ( A) =
P (B) =
P (C ) =
4
8
4
8
2
8
=
=
=
1
2
1
2
1
4
There are 4 outcomes with
heads first.
There are 4 outcomes with
heads second.
There are 2 outcomes with
exactly 2 consecutive heads.
93
7.1.3: Understanding Independent Events
Guided Practice: Example 1, continued
P ( A Ç B) =
P (A Ç C) =
P (B Ç C ) =
2
8
=
1
4
1
8
2
8
=
1
4
There are 2 outcomes with
heads first and heads
second.
There is 1 outcome with
heads first and exactly 2
consecutive heads.
There are 2 outcomes with
heads second and exactly 2
consecutive heads.
94
7.1.3: Understanding Independent Events
Guided Practice: Example 1, continued
3. Use the definition of independence to
determine if the events are independent in
each specified pair.
P ( A Ç B ) = P ( A) · P ( B )
1 1 1
= ·
4 2 2
A and B are independent.
95
7.1.3: Understanding Independent Events
Guided Practice: Example 1, continued
P ( A Ç C ) = P ( A ) · P (C )
1 1 1
= ·
8 2 4
A and C are independent.
P ( B Ç C ) = P ( B ) · P (C )
1 1 1
¹ ·
4 2 4
B and C are dependent.
✔
96
7.1.3: Understanding Independent Events
Bellringer 9/8/14
• Set A = {1,3,5}
• Set B = {2,4,6}
Find:
1.
A∪B
2. 𝐴 ∩ 𝐵
3. 𝐴
4. 𝐴 ∩ 𝐵
Objectives
Construct and interpret two-way frequency
tables of data when two categories are
associated with each object being classified.
Vocabulary
joint relative frequency
marginal relative frequency
conditional relative frequency
A two-way table is a useful way to organize data
that can be categorized by two variables. Suppose
you asked 20 children and adults whether they
liked broccoli. The table shows one way to arrange
the data.
The joint relative frequencies
are the values in each category
divided by the total number of
values, shown by the shaded
cells in the table. Each value is
divided by 20, the total number
of individuals.
The marginal relative frequencies are found by
adding the joint relative frequencies in each row
and column.
To find a conditional relative frequency , divide
the joint relative frequency by the marginal relative
frequency. Conditional relative frequencies can be
used to find conditional probabilities.
Example 1: Finding Joint and Marginal Relative
Frequencies
The table shows the results of randomly selected
car insurance quotes for 125 cars made by an
insurance company in one week. Make a table of
the joint and marginal relative frequencies.
Example 1: Continued
Divide each value by the total of 125 to find the
joint relative frequencies, and add each row and
column to find the marginal relative frequencies.
0 acc.
1 acc.
2 + acc.
Total
Teen
0.12
0.032
0.072
0.224
Adult
0.424
0.256
0.096
0.776
Total
0.544
0.288
0.168
1
Bell Ringer 9/9/14
A bag contains 4 red and 2 yellow marbles. A
marble is selected, kept out of the bag, and
another marble is selected. Find each
conditional probability of selecting the second
marble.
1. P(red | red) 0.6
2. P(red | yellow) 0.8
3. P(yellow | yellow) 0.2
4. P(yellow | red)
0.4
Check It Out! Example 1
The table shows the number of books sold at a
library sale. Make a table of the joint and
marginal relative frequencies.
Check It Out! Example 1 Continued
Divide each value by the total of 210 to find the joint
relative frequencies, and add each row and column
to find the marginal relative frequencies.
Hardcover
Paperback
Total
Fiction
0.133
0.448
0.581
Nonfiction
0.248
0.171
0.419
Total
0.381
0.619
1
Example 2: Using Conditional Relative Frequency to
Find Probability
A reporter asked 150 voters if they plan to vote
in favor of a new library and a new arena. The
table shows the results.
Example 2A Continued
A. Make a table of the joint and marginal relative
frequencies.
Library
Arena
Yes
No
Total
Yes
0.14
0.38
0.52
No
0.2
0.28
0.48
Total
0.34
0.66
1
Example 2B Continued
B. If you are given that a voter plans to vote no to
the new library, what is the probability the voter
also plans to say no to the new arena?
0.28 ≈ 0.58
0.48
Check It Out! Example 2
The classes at a dance academy include ballet
and tap dancing. Enrollment in these classes is
shown in the table.
2a. Copy and complete the table of the joint relative
frequencies and marginal relative frequencies.
Check It Out! Example 2 continued
Ballet
Tap
Yes
No
Total
Yes
0.19
0.43
0.62
No
0.26
0.12
0.38
Total
0.45
0.55
1
2b. If you are given that a student is taking ballet,
what is the probability that the student is not taking
tap?
0.43 ≈ 0.69 or 69%
0.62
Bellringer 9/10/14
Kira likes number puzzles and is tackling the puzzle
below. The puzzle is incomplete. When it is complete,
each total at the end of a row will be the sum of the
numbers in that row, and each total at the end of a
column will be the sum of the numbers in that column.
Find all the missing numbers to solve the puzzle.
Bellringer 9/10/14
Kira likes number puzzles and is tackling the puzzle
below. The puzzle is incomplete. When it is complete,
each total at the end of a row will be the sum of the
numbers in that row, and each total at the end of a
column will be the sum of the numbers in that column.
Find all the missing numbers to solve the puzzle.
This sum is 25.
This sum is 30.
Totals
10
12
8
10
Totals
25
30
34
40
100
Completed puzzle:
Totals
Totals
10
14
12
36
5
17
8
30
10
9
15
34
25
40
35
100
115
7.2.2: Using Two-Way Frequency Tables
Explanation:
• For the first column, the sum given is 25 and
the numbers provided add to 20. Therefore,
the missing cell is 5.
• The second row has a sum of 30. Using the
information completed from the first column,
we know that the first cell in the first row is 5.
We are given 8 and the total is 30. Therefore,
the missing cell is 30 – (8 + 5) = 17.
• The total row at the bottom sums to 100 and
the first two cells are given as 25 and 40. Use
100 – (25 + 40) = 35 to find the missing number.
116
7.2.2: Using Two-Way Frequency Tables
• The third column we now know sums to 35,
and 12 and 8 were given as two entries in that
column. Use 35 – (12 + 8) = 15 to find the
missing cell.
• The third row now has two entries, 10 and 15.
The row must sum to 34. Use 34 – (10 + 15) = 9
to find the missing cell.
• The second column sums to 40 and we now
have two entries in the column, 17 and 9. Use
40 – (17 + 9) = 14 to find the missing cell.
• The sum is missing from the first row. All the
cells have been filled in for the first row. Sum
the cells to find the missing entry: 10 + 14 + 12
= 36.
117
7.2.2: Using Two-Way Frequency Tables
Example 3: Comparing Conditional Probabilities
A company sells items in a store, online, and
through a catalog. A manager recorded whether or
not the 50 sales made one day were paid for with
a gift card.
Use conditional probabilities to determine for
which method a customer is most likely to pay
with a gift card.
Example 3 Continued
Gift Card
Store
Online
Catalog
TOTAL
0.12
0.18
0.10
0.40
Another
Method
0.18
0.26
0.16
0.60
TOTAL
0.30
0.44
0.26
1
P(gift card if in store) = 0.4
P(gift card if online) ≈ 0.41
P(gift card if by catalog) ≈ 0.38
so most likely if buying online.
A customer is most likely to pay with a gift card if
buying online.
Check It Out! Example 3
Francine is evaluating three driving schools. She
asked 50 people who attended the schools
whether they passed their driving tests on the
first try. Use conditional probabilities to
determine which is the best school.
Use conditional probabilities to determine
which is the best school.
Check It Out! Example 3 Continued
Pass
Fail
TOTAL
Al’s Driving
0.28
0.16
0.44
Drive Time
0.22
0.14
0.36
Crash
Course
0.10
0.10
0.20
TOTAL
0.60
0.40
1
Al’s Driving has the best pass rate, about 64%,
versus 61% for Drive Time and 50% for Crash
Course.
Lesson Quiz: Part I
1. At a juice-bottling factory, quality-control
technicians randomly select bottles and mark
them pass or fail. The manager randomly selects
the results of 50 tests and organizes the data by
shift and result. The table below shows these
results.
Lesson Quiz: Part I continued
1. Make a table of the joint and marginal
relative frequencies.
Lesson Quiz: Part 2
2. Find the probability that a bottle was
inspected in the afternoon given that it
failed the inspection.
Lesson Quiz: Part 3
3. Use conditional probabilities to determine on
which shift a bottle is most likely to pass
inspection.
Bellringer 9/11/14
Define and Give an Example:
1. Set
2. Union
3. Intersection
4. Complement
Bellringer 8/12/14
–You only have 5 minutes from the time the bell rings!
1. What is the probability of rolling a 3, and then a 7 on a 6 sided die?
2. What is the Probability of spinning a spinner with the numbers 1-10 on it
and it landing on a 2 or an 8?
3. What is the Probability of spinning a spinner with the numbers 1-10 on it
and it landing on a 2 and then an 8?
– Turn in your bellringer and pick up a whiteboard, marker, and eraser when you
are finished
• No, 2 and 3 are not the same question or answer
• Yes we will go over it