Download 5.2 full notes

+ Chapter 5
Probability: What Are the Chances?



5.1
5.2
5.3
Randomness, Probability, and Simulation
Probability Rules
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
Sample Space - the collection of all possible
outcomes of a chance experiment
Roll a die
S = {1,2,3,4,5,6}
Event - any collection of outcomes from the
sample space
-
Complement
Consists of all outcomes that
are not in the event
- Not rolling a even # = 1- P(rolling an even #)
Union - the event A or B happening consists of
all outcomes that are in at least one of the two
events E  A U B
- Rolling a prime # or even number
E={2,3,4,5,6}
Intersection - the event A and B happening
consists of all outcomes that are in both events
E  AB
- Drawing a red card and a “2” E = {2 hearts, 2 diamonds}
Mutually Exclusive (disjoint) -two events that
have no outcomes in common
- Roll a “2” or a “5”
Addition rule:
If A and B are disjoint events, then
P(A or B ) = P(A) + P(B)
If A and B are NOT disjoint events, then
P(A or B ) = P(A) + P(B) – P (A ∩ B )
Multiplication rule: If there are n ways to do a first
event & m ways to do a second event.
Then the number of all possible outcomes = n∙m
At least one: The probability that at least one
outcome happens is 1 minus the probability that no
outcomes happen.
P(at least 1) = 1 – P(none)
Disjoint example: A large auto center sells cars
made by many different manufacturers. Three of
these are Honda, Nissan, and Toyota.
Suppose that P(H) = .25, P(N) = .18, P(T) = .14.
Are these disjoint events?
P(H or N or T) =
yes
.25 + .18+ .14 = .57
P(not (H or N or T) = 1 - .57 = .43
NonDisjoint example: Musical styles other than
rock and pop are becoming more popular. A
survey of college students finds that the
probability they like country music is .40. The
probability that they liked jazz is .30 and that
they liked both is .10.
Are these disjoint events?
No
What is the probability that they like country or jazz?
P(C or J) = .4 + .3 -.1 = .6
For a sales promotion the manufacturer places
winning symbols under the caps of 10% of all Dr.
Pepper bottles. You buy a six-pack. What is the
probability that you win something?
P(at least one winning symbol) =
1 – P(no winning symbols)
1-
6
.9
1 - .531441= .4686
Rules of probability SUMMARY
• What is the range of values for any given probability? 0
• What is the sum of the probability of all events?
≤ P(A) ≤ 1
Sum = 1
• Compliment rule: For any event A, P(Ac) = 1 – P(A)
• Addition rule: If A and B are disjoint events, then
P(A or B ) = P(A) + P(B)
If A and B are NOT disjoint events, then
P(A or B ) = P(A) + P(B) – P (A ∩ B )

Multiplication rule: If there are n ways to do a first event & m ways
to do a second event. Then the number of
all possible outcomes = n∙m
•
At least one: P(at least 1) = 1 – P(none)
Simple Probability – Example # 1
• License plates in the State of Altered require 2 letters
followed by 4 digits. No two letters can be the same, but it
is ok to repeat digits. How many different license plates
can me made?
26 ∙ 25 ∙ 10 ∙ 10 ∙ 10 ∙ 10
= 6,500,000
Probability example #2 : Roll the Dice
+
 Simple
You are rolling two fair, six-sided dice – one that’s red and one that’s green.
Sample
Space
36
Outcomes
Since the dice are fair, each
outcome is equally likely.
Each outcome has
probability 1/36.
Probability Rules
How many possible outcomes are there?
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 the “sum is not 5”. Find P(B).
P(B) = 1 – 4/36
= 32/36
+
Suppose event A is defined as the “sum of 5”. Find P(A).
Probability Example #3: Distance Learning
+
 Simple
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:
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
Simple Probability Example #4
Canada has 2 official languages, English and French. Choosing a Canadian
at random in a recent survey gave the following distribution of responses:
English
French
Asian/Pacific
Other
0.63
0.22
0.06
?
What is the probability of ‘Other’? Why?
0.09, because the entire P(sample space) = 1
What is the probability that a Canadian’s “mother tongue” is not English?
1 – P (English) = 1 - 0.63 = .37
What is the probability that a Canadian’s mother tongue is either
French or Asian Pacific?
P(Fr) + P(AP) = 0.22+0.06 = .28
#5: Two-Way Tables
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.
+
 Example
Tree Diagrams - When there are many options, you may
want to use a TREE DIAGRAM to help
you find all the different combinations of
outcomes.
Simple Probability Example #6 – Four different people are
each flipping a coin
(4 distinct coin flips). What is the probability of getting...
a) At least 2 heads?
b) Either 3 heads or 3 tails?
Example #6 – Four different people are each flipping a coin
(4 distinct coin flips). What is the probability of getting...
H
H
T
H
T
H
T
H
H
T
T
H
T
T
H
T
H
T
H
T
H
T
HHHH
HHHT
HHTH
HHTT
HTHH
HTHT
HTTH
HTTT
H
T
H
T
H
T
H
T
THHH
THHT
THTH
THTT
TTHH
TTHT
TTTH
TTTT
a) At least 2 heads?
b) Either exactly 3 heads or 3 tails?
Simple Probability Example #7 – In a family of 3 children…
a) What is the probability exactly two are boys?
b) What is the probability at least one is a girl?
G
G
B
B
G
B
G
GGG
B
G
B
G
B
G
GGB
B
GBG
GBB
BGG
BGB
BBG
BBB
Venn Diagrams
• NOT Disjoint: P(A ∩ B)
• Disjoint: P(A ∪ B)

There are 174 total people.

You have 90 total males. 19 of them have their ears pierced.

You have 103 total people with their ears pierced.

Define event A as males

Define event B as pierced ears.
+
Venn Diagrams
Venn Diagrams Example #1
In an apartment complex, 40% of residents read the USA
Today, while 25% of residents read the New York Times. Five
percent of residents read both. Suppose we select an
apartment resident at random and record which of the two
papers the person reads. Find the probability the person reads
at least 1 of the 2 papers. Find the probability the person
doesn’t read either paper.
Read USA Today
35%
Read NY Times
5%
20%
Topic D: Venn Diagrams Example #1
Find the probability the person reads at least 1 of the 2 papers.
P(A or B) = P(A) + P(B) - P(A ∩ B)
= .40 + .25 – .05
= .60
P(only A or only B or both)
= .35 + .05 + .20
= .60
Find the probability the person doesn’t read either paper.
1 – P( at least one) = P (none)
1 - .60 = .40
.40
Read USA Today
35%
Read NY Times
5%
20%
Venn Diagrams Example #2
After observing several pizza orders, I discover the following
pizza topping requests: 67 people wanted sausage on their
pizza, 55 wanted pepperoni, 50 wanted mushrooms, 22 wanted
sausage and mushrooms, 18 wanted pepperoni and
mushrooms, 15 wanted sausage and pepperoni, and 10 wanted
all three. What is the probability that a randomly selected
person wants
a)
b)
c)
d)
only sausage?
pepperoni and mushrooms, but no sausage?
pepperoni or sausage?
Neither pepperoni nor sausage nor mushrooms?
Pepperoni
Sausage
40
32
5
pepperoni and
mushrooms, but no
sausage?
10
12
only sausage?
8
20
23
pepperoni or sausage?
Mushrooms
d) Neither pepperoni nor
sausage nor mushrooms?
+ 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)
+ Homework:
# 39, 40, 45, 46, 49, 50, 51–54, 56
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
notes on Venn Diagrams (Useful
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.
The Venn diagram below illustrates why.
Probability Rules
Resource)
+
 Additional
Notes on Venn Diagrams (useful

Because Venn diagrams have uses in other branches of
mathematics, some standard vocabulary and notation have
been developed.
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
resource)
+
 Additional
notes on Venn Diagrams (useful
The intersection of events A and B (A ∩ B) is the set of all outcomes
in both events A and B.
Probability Rules
resource)
+
 Additional
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.
Notes on Venn Diagrams (useful
resource)
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.
+
 Additional