Download 5.2 Notes Part 2

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Chapter 5: Probability: What are the Chances?
Section 5.2
Probability Rules
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) Each student is equally likely to be chosen. 103 students have
pierced ears. So, P(pierced ears) = P(B) = 103/178.
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
(b) We want to find P(male and pierced ears), that is, P(A and B).
Look at the intersection of the “Male” row and “Yes” column. There
are 19 males with pierced ears. So, P(A and B) = 19/178.
(c) We want to find P(male or pierced ears), that is, P(A or B). There
are 90 males in the class and 103 individuals with pierced ears.
However, 19 males have pierced ears – don’t count them twice!
P(A or B) = (19 + 71 + 84)/178. So, P(A or B) = 174/178
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
Probability Rules
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)
or
P(A U B) = P(A) + P(B) – P(A ∩ B)
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General Addition Rule for Two Events
+ Checkpoint

A standard deck of playing cards (with joker removed) consists
of 52 cards in four suits – clubs, diamonds, hearts and spades.
Each suit has 13 cards, with denominations ace, 2, 3, 4, 5, 6, 7,
8, 9, 10, jack, queen, and king. The jack, queen, and king are
referred to as “face cards.” Imagine that we shuffle the deck
thoroughly and deal one card. Let’s define events A: getting a
face card and B: getting a heart.
1) Make a two-way table that displays the sample
space.
Heart
Non-heart
Face Card
3
9
Non-face card
10
30
+ Checkpoint
2) Find P(A and B)
3
P(face card and heart) 
52
3) Explain why P(A or B) ≠ P(A) + P(B). Then use the
general addition rule to find P(A or B).
Since A and B are not disjoint, this is not equal to the sum
of the probability of A and the probability of B.
P(A or B) = P(A) + P(B) – P(A and B) = 0.423
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
+ Example – Venn Diagrams, TwoWay Tables and Probability

According to the National Center for Health Statistics, in
December 2008, 78% of U.S. 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.
1) Make a two-way table that displays the sample space of
this chance process.
Cell Phone No Cell Phone
Total
Landline
0.60
0.18
0.78
No Landline
0.20
0.02
0.22
Total
0.80
0.20
1.00
+ 2) Construct a Venn Diagram to represent the outcomes of
this chance process.
+ 3) Find the probability that the household has at least one of
the two types of phones.
To find the probability that the household has at least one of
the two types of phones, we need to find the probability that
the household has a landline, a cell phone, or both.
P(A U B) = P(A) + P(B) – P(A ∩ B)
= 0.78 + 0.80 – 0.60
= 0.98
There is a 98% chance that the household has at least one of
the two types of phones.
4) Find the probability that the household has a cell phone
only.
P(cell phone only )  P( A  B)  0.20
c
+ Calculators & Probability
Many
probability problems involve
simple computations that you can do
on your calculator. It may be tempting
to just write down your final answer
without show the supporting work.
Don’t do it!!! A “naked answer,” even
if it’s correct, will usually earn you no
credit on a free-response question.
+ 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
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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 A
total number of outcomes in sample space
+ 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.
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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)