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Chapter 7
Probability
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 1
7.1
The Nature of Probability
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 2
Definitions



An experiment is a controlled operation that
yields a set of results.
The possible results of an experiment are called
its outcomes.
An event is a subcollection of the outcomes of
an experiment.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 3
Definitions continued


Empirical probability is the relative frequency
of occurrence of an event and is determined by
actual observations of an experiment.
Theoretical probability is determined through
a study of the possible outcomes that can occur
for the given experiment.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 4
Empirical Probability
number of times
event E has occurred
P(E) 
total number of times the
experiment has been performed


Example: In 100 tosses of a fair die, 19
landed showing a 3. Find the empirical
probability of the die landing showing a 3.
Let E be the event of the die landing
showing a 3.
19
P(E) 
 0.19
100
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 5
The Law of Large Numbers

The law of large numbers states that
probability statements apply in practice to a
large number of trials, not to a single trial. It is
the relative frequency over the long run that is
accurately predictable, not individual events or
precise totals.
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Slide 12 - 6
7.2
Theoretical Probability
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Slide 12 - 7
Equally likely outcomes


If each outcome of an experiment has the same
chance of occurring as any other outcome, they
are said to be equally likely outcomes.
For equally likely outcomes, the probability of
Event E may be calculated with the following
formula.
number of outcomes favorable to E
P(E) 
total number of possible outcomes
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 8
Example






A die is rolled. Find the probability of rolling
a) a 2.
b) an odd number.
c) a number less than 4.
d) an 8.
e) a number less than 9.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 9
Solutions: There are six equally likely
outcomes: 1, 2, 3, 4, 5, and 6.

a)
number of outcomes that will result in a 2 1
P(2) 

total number of possible outcomes
6


b) There are three ways an odd number can
occur 1, 3 or 5.
3 1
P(odd)  
6 2
c) Three numbers are less than 4.
3 1
P(number less than 4)  
6 2
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 10
Solutions: There are six equally likely
outcomes: 1, 2, 3, 4, 5, and 6. continued

d) There are no outcomes that will result in
an 8.
0
P(number greater than 8)   0
6

e) All outcomes are less than 9. The event
must occur and the probability is 1.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 11
Important Facts




The probability of an event that cannot occur is 0.
The probability of an event that must occur is 1.
Every probability is a number between 0 and 1
inclusive; that is, 0  P(E) 1.
The sum of the probabilities of all possible
outcomes of an experiment is 1.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 12
Example

A standard deck of cards is well shuffled. Find
the probability that the card is selected.
a) a 10.
b) not a 10.
c) a heart.
d) an ace, one or 2.
e) diamond and spade.
f) a card greater than 4 and less than 7.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 13
Example continued
a) a 10
There are four 10’s in
a deck of 52 cards.
4
1
P(10) 

52 13
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b) not a 10
P(not a 10)  1 P(10)
1
 1
13
12

13
Slide 12 - 14
Example continued
c) a heart
There are 13 hearts
in the deck.
13 1
P(heart) 

52 4
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d) an ace, 2 or 3
There are 4 aces, 4
twos and 4 threes, or
a total of 12 cards.
12 3
P(A, 1 or 2) 

52 13
Slide 12 - 15
Example continued
d) diamond and spade
The word and means
both events must
occur. This is not
possible.
P(diamond and spade)
0

0
52
Copyright © 2009 Pearson Education, Inc.
e) a card greater than 4
and less than 7
The cards greater
than 4 and less than
7 are 5’s, and 6’s.

P 5 or 6 

P > 4 and < 7 
8
2


52 13
Slide 12 - 16
7.3
Odds
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 17
Odds Against

Odds against event 

P event fails to occur

P event occurs



P success 


P failure
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 18
Example: Odds Against


Example: Find the odds against rolling a 5 on
one roll of a die.

1
P 5 
6


5
P fails to roll a 5 
6
5
5 6 5
odds against
6
   
rolling a 5
1 6 1 1
6
The odds against rolling a 5 are 5:1.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 19
Odds in Favor

Odds in favor of event 

P event occurs

P success 

P failure 
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
P event fails to occur

Slide 12 - 20
Example

Find the odds in favor of landing on blue in one
spin of the spinner.


3
P blue 
8


5
P not blue 
8
3
3 8 3
8
odds in favor    
5 8 5 5
8
The odds in favor of spinning blue are 3:5.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 21
Probability from Odds



Example: The odds against spinning a blue on
a certain spinner are 4:3. Find the probability
that
a) a blue is spun.
b) a blue is not spun.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 22
Solution

Since the odds are 4:3 the denominators must
be 4 + 3 = 7.

The probabilities ratios are:


4
P blue 
7


3
P not blue 
7
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Slide 12 - 23
7.4
Expected Value (Expectation)
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Slide 12 - 24
Expected Value


E  P1  A1  P2  A2  P3  A3  ...  Pn  An
The symbol P1 represents the probability that
the first event will occur, and A1 represents the
net amount won or lost if the first event occurs.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 25
Example

Teresa is taking a multiple-choice test in which
there are four possible answers for each
question. The instructor indicated that she will
be awarded 3 points for each correct answer
and she will lose 1 point for each incorrect
answer and no points will be awarded or
subtracted for answers left blank.
 If Teresa does not know the correct answer to
a question, is it to her advantage or
disadvantage to guess?
 If she can eliminate one of the possible
choices, is it to her advantage or
disadvantage to guess at the answer?
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 26
Solution

Expected value if Teresa guesses.


1
P guesses correctly 
4


3
P guesses incorrectly 
4
  
1
3
E  3  1
4
4
3 3
  0
4 4
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 27
Solution continued—eliminate a choice



1
P guesses correctly 
3


2
P guesses incorrectly 
3
  
1
2
E  3  1
3
3
2 1
 1 
3 3
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Slide 12 - 28
Example: Winning a Prize


When Calvin Winters attends a tree farm event,
he is given a free ticket for the $75 door prize. A
total of 150 tickets will be given out. Determine
his expectation of winning the door prize.
 

1
149
E
75 
0
150
150
1

2
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 29
Example

When Calvin Winters attends a tree farm event,
he is given the opportunity to purchase a ticket
for the $75 door prize. The cost of the ticket is
$3, and 150 tickets will be sold. Determine
Calvin’s expectation if he purchases one ticket.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 30
Solution
 
 
1
149
E
75 
3
150
150
75 447


150 150
372

150
 2.48

Calvin’s expectation is $2.48 when he
purchases one ticket.
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Slide 12 - 31
Fair Price
Fair price = expected value + cost to play
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Slide 12 - 32
Example



Suppose you are playing a game in which you
spin the pointer shown in the figure, and you are
awarded the amount shown under the pointer. If
is costs $10 to play the game, determine
a) the expectation of the
$10
$2
person who plays the
$15
$20
game.
$10
$2
b) the fair price to play the
game.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 33
Solution

Amt. Shown
on Wheel
$2
$10
$15
$20
Probability
3/8
3/8
1/8
1/8
Amount
Won/Lost
$8
$0
$5
$10
       
3
3
1
1
E  $8  $0  $5  $10
8
8
8
8
24
5 10

0 
8
8 8
9

 1.125  $1.13
8
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 34
Solution

Fair price = expectation + cost to play
= $1.13 + $10
= $8.87
Thus, the fair price is about $8.87.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 35
7.5
Tree Diagrams
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Slide 12 - 36
Counting Principle

If a first experiment can be performed in M
distinct ways and a second experiment can be
performed in N distinct ways, then the two
experiments in that specific order can be
performed in M • N distinct ways.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 37
Definitions

Sample space: A list of all possible outcomes
of an experiment.

Sample point: Each individual outcome in the
sample space.

Tree diagrams are helpful in determining
sample spaces.
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Slide 12 - 38
Example

a)
b)
c)
d)
Two balls are to be selected without
replacement from a bag that contains one
purple, one blue, and one green ball.
Use the counting principle to determine the
number of points in the sample space.
Construct a tree diagram and list the sample
space.
Find the probability that one blue ball is
selected.
Find the probability that a purple ball followed
by a green ball is selected.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 39
Solutions
a) 3 • 2 = 6 ways
b)
B PB
P
G PG
BP
B
P
G
G
P
BG
GP
B
GB
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

c)
4 2
P blue  
6 3
d)
P Purple,Green

1
P P,G
6

Slide 12 - 40
P(event happening at least once)
 event happening
 event does 
P
 1 P 

 at least once

 not happen
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Slide 12 - 41
7.6
Or and And Problems
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Slide 12 - 42
Or Problems


P(A or B) = P(A) + P(B)  P(A and B)
Example: Each of the numbers 1, 2, 3, 4, 5, 6,
7, 8, 9, and 10 is written on a separate piece of
paper. The 10 pieces of paper are then placed
in a bowl and one is randomly selected. Find
the probability that the piece of paper selected
contains an even number or a number greater
than 5.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 43
Solution

P(A or B) = P(A) + P(B)  P(A and B)
 even or

P


 greater than 5
 even and

P even  P greater than 5  P 
 greater than 5

 

5
5
3
7




10 10 10 10

Thus, the probability of selecting an even
number or a number greater than 5 is 7/10.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 44
Example

Each of the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and
10 is written on a separate piece of paper. The
10 pieces of paper are then placed in a bowl
and one is randomly selected. Find the
probability that the piece of paper selected
contains a number less than 3 or a number
greater than 7.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 45
Solution


2
P less than 3 
10


3
P greater than 7 
10
There are no numbers that are both less than 3
and greater than 7. Therefore,
 less than 3 or  2
3
5 1
P


0


10 2
 greater than 7 10 10
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Slide 12 - 46
Mutually Exclusive

Two events A and B are mutually exclusive if
it is impossible for both events to occur
simultaneously.
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Slide 12 - 47
Example

One card is selected from a standard deck of
playing cards. Determine the probability of the
following events.
a) selecting a 3 or a jack
b) selecting a jack or a heart
c) selecting a picture card or a red card
d) selecting a red card or a black card
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 48
Solutions
a) 3 or a jack
 

4
4
P 3  P jack 

52 52
8
2


52 13
b) jack or a heart
 jack and
4 13 1
P jack  P heart  P 




 heart
 52 52 52

 

16 4


52 13
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 49
Solutions continued
c) picture card or red card
 picture &  12 26 6
P picture  P red  P 




 red card  52 52 52

  
d) red card or black card
26 26
P red  P black 

52 52
52

1
52
  
Copyright © 2009 Pearson Education, Inc.
32 8


52 13

Slide 12 - 50
And Problems


P(A and B) = P(A) • P(B)
Example: Two cards are to be selected with
replacement from a deck of cards. Find the
probability that two red cards will be selected.
     
P A  P B  P red  P red
26 26


52 52
1 1 1
  
2 2 4
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Slide 12 - 51
Example

Two cards are to be selected without
replacement from a deck of cards. Find the
probability that two red cards will be selected.
P  A   P  B   P  red  P  red
26 25


52 51
1 25 25
 

2 51 102
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Slide 12 - 52
Independent Events

Event A and Event B are independent events
if the occurrence of either event in no way
affects the probability of the occurrence of the
other event.

Experiments done with replacement will result in
independent events, and those done without
replacement will result in dependent events.
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Slide 12 - 53
Example

A package of 30 tulip bulbs contains 14 bulbs
for red flowers, 10 for yellow flowers, and 6 for
pink flowers. Three bulbs are randomly selected
and planted. Find the probability of each of the
following.
a.All three bulbs will produce pink flowers.
b.The first bulb selected will produce a red
flower, the second will produce a yellow
flower and the third will produce a red flower.
c. None of the bulbs will produce a yellow
flower.
d.At least one will produce yellow flowers.
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 54
Solution

30 tulip bulbs, 14 bulbs for red flowers,
10 for yellow flowers, and 6 for pink flowers.
a. All three bulbs will produce pink flowers.

 
 
 
P 3 pink  P pink 1  P pink 2  P pink 3

6 5 4



30 29 28
1

203
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 55
Solution
30 tulip bulbs, 14 bulbs for red flowers,
0010 for yellow flowers, and 6 for pink flowers.
b. The first bulb selected will produce a red flower,
the second will produce a yellow flower and the
third will produce a red flower.


   
  
P red, yellow, red  P red  P yellow  P red
14 10 13



30 29 28
13

174
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 56
Solution
30 tulip bulbs, 14 bulbs for red flowers,
0010 for yellow flowers, and 6 for pink flowers.

c. None of the bulbs will produce a yellow flower.
 none 
 first not   second not   third not 
P
 P
P 
P 



 yellow
 yellow   yellow
  yellow 
20 19 18



30 29 28
57

203
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 57
Solution

30 tulip bulbs, 14 bulbs for red flowers,
10 for yellow flowers, and 6 for pink flowers.
d. At least one will produce yellow flowers.
P(at least one yellow) = 1  P(no yellow)
57
 1
203
146

203
Copyright © 2009 Pearson Education, Inc.
Slide 12 - 58