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Chapter 5: Probability: What are the Chances?
Section 5.2
Probability Rules
The Practice of Statistics, 4th edition – For AP*
STARNES, YATES, MOORE
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Chapter 5
Probability: What Are the Chances?
 5.1
Randomness, Probability, and Simulation
 5.2
Probability Rules
 5.3
Conditional Probability and Independence
+ Section 5.2
Probability Rules
Learning Objectives
After this section, you should be able to…

DESCRIBE chance behavior with a probability model

DEFINE and APPLY basic rules of probability

DETERMINE probabilities from two-way tables

CONSTRUCT Venn diagrams and DETERMINE probabilities
Models
Descriptions of chance behavior contain two parts:
Definition:
The sample space S of a chance process is the set of all
possible outcomes.
A probability model is a description of some chance process
that consists of two parts: a sample space S and a probability
for each outcome.
Probability Rules
In Section 5.1, we used simulation to imitate chance behavior.
Fortunately, we don’t have to always rely on simulations to determine
the probability of a particular outcome.
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 Probability
Roll the Dice
Sample
Space
36
Outcomes
Since the dice are fair, each
outcome is equally likely.
Each outcome has
probability 1/36.
Probability Rules
Give a probability model for the chance process of rolling two
fair, six-sided dice – one that’s red and one that’s green.
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 Example:
Flipping Coins
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 Example:
HHH
HHT
HTH
THH
TTT
TTH
THT
HTT
Sample
Space
8
Outcomes
Since the coin is fair, each
outcome is equally likely.
Each outcome has
probability 1/8.
Probability Rules
Give a probability model for the chance process of flipping a fair
coin three times.
Models
Definition:
An event is any collection of outcomes from some chance
process. That is, an event is a subset of the sample space.
Events are usually designated by capital letters, like A, B, C,
and so on.
Probability Rules
Probability models allow us to find the probability of any
collection of outcomes.
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 Probability
If A is any event, we write its probability as P(A).
In the dice-rolling example, suppose we define event A as “sum is 5.”
There are 4 outcomes that result in a sum of 5.
Since each outcome has probability 1/36, P(A) = 4/36.
Suppose event B is defined as “sum is not 5.” What is P(B)? P(B) = 1 – 4/36
= 32/36
Rules of Probability
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 Basic

The probability of any event is a number between 0 and 1.

All possible outcomes together must have probabilities whose sum
is 1.

If all outcomes in the sample space are equally likely, the
probability that event A occurs can be found using the formula
P(A) 


number of outcomes corresponding to event
total number of outcomes in sample space
A
The probability that an event does not occur is 1 minus the
probability that the event does occur.

If two events have no outcomes in common, the probability that
one or the other occurs is the sum of their individual probabilities.
Definition:
Two events are mutually exclusive (disjoint) if they have no
outcomes in common and so can never occur together.
Probability Rules
All probability models must obey the following rules:
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Rules of Probability
Probability Rules

 Basic
• For any event A, 0 ≤ P(A) ≤ 1.
• If S is the sample space in a probability model,
P(S) = 1.
• In the case of equally likely outcomes,
number of outcomes corresponding to event
P(A) 
total number of outcomes in sample space
A
• Complement rule: P(AC) = 1 – P(A)
• Addition rule for mutually exclusive events: If A
and B are mutually exclusive,
P(A or B) = P(A) + P(B).
Distance Learning
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 Example:
Age group (yr):
Probability:
18 to 23
24 to 29
30 to 39
40 or over
0.57
0.17
0.14
0.12
(a) Show that this is a legitimate probability model.
Each probability is between 0 and 1 and
0.57 + 0.17 + 0.14 + 0.12 = 1
(b) Find the probability that the chosen student is not in the
traditional college age group (18 to 23 years).
P(not 18 to 23 years) = 1 – P(18 to 23 years)
= 1 – 0.57 = 0.43
Probability Rules
Distance-learning courses are rapidly gaining popularity among
college students. Randomly select an undergraduate student
who is taking distance-learning courses for credit and record
the student’s age. Here is the probability model:
Randomly select a student who took the 2010 AP Statistics
exam and record the student’s score. Here is the probability
model:
Score:
Probability:
1
2
3
4
5
0.223
0.183
0.235
0.224
0.125
(a) Show that this is a legitimate probability model.
Each probability is between 0 and 1 and
0.223+0.183+0.235+0.224+0.125=1
(b) Find the probability that the chosen student scored 3 or
better.
There are two ways to find this probability:
By the addition rule, P(3 or better) = 0.235 +
0.224 + 0.125 = 0.584
By the complement rule and addition rule,
P(3 or better) = 1 – P(2 or less) = 1 – (0.233 +
0.183) = 0.584.
Probability Rules

AP Statistics Scores
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 Example:
Tables and Probability
Consider the example on page 303. Suppose we choose a student at
random. Find the probability that the student
(a) has pierced ears.
(b) is a male with pierced ears.
(c) is a male or has pierced ears.
Define events A: is male and B: has pierced ears.
(a)
(b) Each
(c)
We want
student
to find
is equally
P(male likely
or
and
pierced
pierced
to beears),
chosen.
ears),
that
that
103
is,is,
students
P(A
P(A
orand
B).have
There
B).
pierced
Look90atmales
are
ears.
the intersection
in
So,
theP(pierced
classofand
the
ears)
103
“Male”
=
individuals
P(B)
row =and
103/178.
with
“Yes”pierced
column.
ears.
There
are 19 males
However,
19 males
with pierced
have pierced
ears. So,
ears
P(A
– don’t
and B)
count
= 19/178.
them twice!
P(A or B) = (19 + 71 + 84)/178. So, P(A or B) = 174/178
Probability Rules
When finding probabilities involving two events, a two-way table can display
the sample space in a way that makes probability calculations easier.
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 Two-Way
Tables and Probability
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 Two-Way
Probability Rules
What is the relationship between educational achievement and home
ownership? A random sample of 500 people who participated in the 2000
census was chosen. Each member of the sample was identified as a high
school graduate (or not) and as a home owner (or not). The two-way table
displays the data. Suppose we choose a member of the sample at
random. Find the probability that the member
High School Graduate?
(a) Is a high school graduate.
Yes
No
Total
Homeowner
221
119
340
(b) Is a high school graduate and
owns a home.
Not a Homeowner
Total
89
310
71
190
160
500
(c) Is a high school graduate or
owns a home.
Define events A: a high school graduate and B: a homeowner.
(c) Since there are 221 + 89 + 119 = 429 people who graduated from high
school or own a home, P(A or B) is 429/500. Note that is inappropriate to
(a) Since
(b)
310
of the
500 this
members
of thesince
sample
fromB high
compute
P(A)221
+ P(B)
to find
probability
the graduated
events A and
are not
school,
school
and
P(A)own
= 310/500.
a home,
B) = who
221/500.
mutually
exclusive—there
areP(A
221and
people
are both high school graduates
and own a home. If you did add these probabilities, the result would be
650/500, which is clearly wrong since the probability is greater than 1.
Tables and Probability
The Venn diagram below illustrates why.
Probability Rules
Note, the previous example illustrates the fact that we can’t use
the addition rule for mutually exclusive events unless the
events have no outcomes in common.
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 Two-Way
General Addition Rule for Two Events
If A and B are any two events resulting from some chance process, then
P(A or B) = P(A) + P(B) – P(A and B)
Diagrams and Probability
The complement AC contains exactly the outcomes that are not in A.
The events A and B are mutually exclusive (disjoint) because they do not
overlap. That is, they have no outcomes in common.
Probability Rules
Because Venn diagrams have uses in other branches of
mathematics, some standard vocabulary and notation have
been developed.
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 Venn
Diagrams and Probability
Probability Rules
The intersection of events A and B (A ∩ B) is the set of all outcomes
in both events A and B.
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 Venn
The union of events A and B (A ∪ B) is the set of all outcomes in either
event A or B.
Hint: To keep the symbols straight, remember ∪ for union and ∩ for intersection.
Diagrams and Probability
Define events A: is male and B: has pierced ears.
Probability Rules
Recall the example on gender and pierced ears. We can use a Venn
diagram to display the information and determine probabilities.
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 Venn
Diagrams and Probability
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 Venn
A
B
High School Graduate?
Yes
No
Total
Homeowner
221
119
340
Not a Homeowner
89
71
160
Total
310
190
500
89
221
119
71
Probability Rules
Here is the two-way table summarizing the relationship between educational
status and home ownership from the previous example:
Define events A: a high school graduate and B: a homeowner.
Region in Venn Diagram
In Words
In Symbols
Count
In the intersection of 2 circles
HS grad and owns home
A∩B
221
Inside circle A, outside circle B
HS grad and doesn’t own home
A ∩ BC
89
Inside circle B, outside circle A
Not HS grad but owns home
AC ∩ B
119
Outside both circles
Not HS grad and doesn’t own home.
AC ∩ B C
71
Example: Phone Usage
According to the National Center for Health Statistics,
(http://www.cdc.gov/nchs/data/nhis/earlyrelease/wireless200905_tables.htm#T
1), in December 2008, 78% of US households had a traditional landline
telephone, 80% of households had cell phones, and 60% had both. Suppose
we randomly selected a household in December 2008.
(a) Make a two-way table that displays the sample space of this chance process.
(b) Construct a Venn diagram to represent the outcomes of this chance process.
(c) Find the probability that the household has at least one of the two types of
phones.
(d) Find the probability the household has a cell phone
only.
(c) To find the probability
(d) that
P(cell
thephone
household
only) has
= P(A
atc least
B) = 0.20
one of the two types of
phones, we need to find the probability that the household has a landline, a cell
Cell Phone
No Cell Phone
Total
phone, or both. Landline
P(A B) = P(A) + P(B)
–
P(A
B)
=
0.78
+
0.80
–
0.60
0.60
0.18
0.78 = 0.98.
There is a 98% No
chance
that the household
has at0.02
least one of the
two types of
Landline
0.20
0.22
phones.
Total
0.80
0.20
1.00
Probability Rules

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 Alternate
A: Landline
B: Cell phone
0.18
0.60
0.20
0.02
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+ Section 5.2
Probability Rules
Summary
In this section, we learned that…

A probability model describes chance behavior by listing the possible
outcomes in the sample space S and giving the probability that each
outcome occurs.

An event is a subset of the possible outcomes in a chance process.

For any event A, 0 ≤ P(A) ≤ 1

P(S) = 1, where S = the sample space

If all outcomes in S are equally likely,

P(AC) = 1 – P(A), where AC is the complement of event A; that is, the
event that A does not happen. 
P(A) 
number of outcomes corresponding to event
total number of outcomes in sample space
A
+ Section 5.2
Probability Rules
Summary
In this section, we learned that…

Events A and B are mutually exclusive (disjoint) if they have no outcomes
in common. If A and B are disjoint, P(A or B) = P(A) + P(B).

A two-way table or a Venn diagram can be used to display the sample
space for a chance process.

The intersection (A ∩ B) of events A and B consists of outcomes in both A
and B.

The union (A ∪ B) of events A and B consists of all outcomes in event A,
event B, or both.

The general addition rule can be used to find P(A or B):
P(A or B) = P(A) + P(B) – P(A and B)
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Looking Ahead…
In the next Section…
We’ll learn how to calculate conditional probabilities as well
as probabilities of independent events.
We’ll learn about
 Conditional Probability
 Independence
 Tree diagrams and the general multiplication rule
 Special multiplication rule for independent events
 Calculating conditional probabilities