Transcript Chapter 3 - Houston County Schools

Chapter
3
Probability
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All rights reserved.
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Group Project
• Task: Sample a minimum of 30 male and 30 female students
for their shoe size
1. Find the mean of each sample
2. Find the Sample Standard Deviation
3. Create a dot-plot with your results. Then trace the top of the
dot-plot to create a bell chart
4. Assuming your data is normally distributed, assume you
want to know where 68% of the sample lies. What shoe size
would be the smallest, and what shoe size would be the
largest for 68% of the sample?
5. You have 15 minutes to gather your sample. Then return to
class and complete your work
Larson/Farber 5th edition
2
Chapter Outline
• 3.1 Basic Concepts of Probability
• 3.2 Conditional Probability and the Multiplication
Rule
• 3.3 The Addition Rule
• 3.4 Additional Topics in Probability and Counting
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Section 3.1
Basic Concepts of Probability
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Section 3.1 Objectives
•
•
•
•
Identify the sample space of a probability experiment
Identify simple events
Use the Fundamental Counting Principle
Distinguish among classical probability, empirical
probability, and subjective probability
• Determine the probability of the complement of an
event
• Use a tree diagram and the Fundamental Counting
Principle to find probabilities
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Probability Experiments
Probability experiment
• An action, or trial, through which specific results (counts,
measurements, or responses) are obtained.
Outcome
• The result of a single trial in a probability experiment.
Sample Space
• The set of all possible outcomes of a probability
experiment.
Event
• Consists of one or more outcomes and is a subset of the
sample space.
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Probability Experiments
• Probability experiment: Roll a die
• Outcome: {3}
• Sample space: {1, 2, 3, 4, 5, 6}
• Event: {Die is even}={2, 4, 6}
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Example: Identifying the Sample Space
A probability experiment consists of tossing a coin and
then rolling a six-sided die. Describe the sample space.
Solution:
There are two possible outcomes when tossing a coin:
a head (H) or a tail (T). For each of these, there are six
possible outcomes when rolling a die: 1, 2, 3, 4, 5, or
6. One way to list outcomes for actions occurring in a
sequence is to use a tree diagram.
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Solution: Identifying the Sample Space
Tree diagram:
H1 H2 H3 H4 H5 H6
T1 T2 T3 T4 T5 T6
The sample space has 12 outcomes:
{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
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Simple Events
Simple event
• An event that consists of a single outcome.
 e.g. “Tossing heads and rolling a 3” {H3}
• An event that consists of more than one outcome is
not a simple event.
 e.g. “Tossing heads and rolling an even number”
{H2, H4, H6}
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Example: Identifying Simple Events
Determine whether the event is simple or not.
• You roll a six-sided die. Event B is rolling at least a 4.
Solution:
Not simple (event B has three outcomes: rolling a 4, a 5,
or a 6)
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Fundamental Counting Principle
Fundamental Counting Principle
• If one event can occur in m ways and a second event
can occur in n ways, the number of ways the two
events can occur in sequence is m•n.
• Can be extended for any number of events occurring
in sequence.
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Example: Fundamental Counting
Principle
You are purchasing a new car. The possible
manufacturers, car sizes, and colors are listed.
Manufacturer: Ford, GM, Honda
Car size: compact, midsize
Color: white (W), red (R), black (B), green (G)
How many different ways can you select one
manufacturer, one car size, and one color? Use a tree
diagram to check your result.
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Solution: Fundamental Counting
Principle
There are three choices of manufacturers, two car sizes,
and four colors.
Using the Fundamental Counting Principle:
3 ∙ 2 ∙ 4 = 24 ways
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Example
•
•
•
•
•
•
At a local restaurant, for breakfast you can get:
3 kinds of meat (ham/bacon/sausage)
Eggs 3 different ways (over easy/fried/scrambled)
3 different drink choices (coffee/juice/milk)
3 kinds of bread (pancakes/waffles/crepes)
How many different ways can you eat breakfast
there?
Types of Probability
Classical (theoretical) Probability
• Each outcome in a sample space is equally likely.
Number of outcomes in event E
P( E ) 
Number of outcomes in sample space
• Often referred to as “the desired outcome divided by
the total outcome”
• This is theoretical because it doesn’t account for skill,
weather, rounded dice etc. Just a mathematical
probability
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Example: Finding Classical Probabilities
You roll a six-sided die. Find the probability of each
event.
1. Event A: rolling a 3
2. Event B: rolling a 7
3. Event C: rolling a number less than 5
Solution:
Sample space: {1, 2, 3, 4, 5, 6}
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Solution: Finding Classical Probabilities
1. Event A: rolling a 3
Event A = {3}
1
P (rolling a 3)   0.167
6
2. Event B: rolling a 7
0
P(rolling a 7)   0
6
Event B= { } (7 is not in
the sample space)
3. Event C: rolling a number less than 5
Event C = {1, 2, 3, 4}
4
P(rolling a number less than 5)   0.667
6
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Example
• What is the probability of selecting the following
from a deck of cards:
• Seven of diamonds
• A diamond
• A seven
• A diamond, heart, or spade
Types of Probability
Empirical (statistical) Probability
• Based on observations obtained from probability
experiments.
• Relative frequency of an event.
Frequency of event E f
P( E ) 

Total frequency
n
• This is still just another way of saying “What I
wanted to observe divided by the total observations”
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Example: Finding Empirical Probabilities
1. A company is conducting a telephone survey of
randomly selected individuals to get their overall
impressions of the past decade (2000s). So far, 1504
people have been surveyed. What is the probability
that the next person surveyed has a positive overall
impression of the 2000s? (Source: Princeton Survey
Research Associates International)
Response
Number of times, f
Positive
406
Negative
752
Neither
316
Don’t know
30
Σf = 1504
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Solution: Finding Empirical Probabilities
Response
event
Number of times, f
Positive
406
Negative
752
Neither
316
Don’t know
30
frequency
Σf = 320
f
406
P( positive)  
 0.270
n 1504
The probability of someone having a positive response
to the last decade is .27, or 27%
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Example
• You go fishing, with an equally likely chance to catch
one fish instead of another
Fish Type Number Caught -f
Bluegill
13
• This is how you did:
Redgill
17
10
• Remember, these are observed Crappy
∑f = 40
frequencies
• If you catch another fish, what is the probability that
it will be a Bluegill?
• P(bluegill) = 13/40 = .325
Law of Large Numbers
Law of Large Numbers
• As an experiment is repeated over and over, the
empirical probability of an event approaches the
theoretical (actual) probability of the event.
•Using the
probability simulator,
toss a coin 20 times
•We would expect to
see 50% heads and
50% tales. –Do you?
•Now toss it 300
times –is it closer to
what we expect?
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Types of Probability
Subjective Probability
• Intuition, educated guesses, and estimates.
• e.g. A doctor may feel a patient has a 90% chance of a
full recovery.
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Example: Classifying Types of Probability
Classify the statement as an example of classical,
empirical, or subjective probability.
1. The probability that you will get the flu this
year is 0.1.
Solution:
Subjective probability (most likely an educated guess)
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Example: Classifying Types of Probability
Classify the statement as an example of classical,
empirical, or subjective probability.
2. The probability that a voter chosen at random will be
younger than 35 years old is 0.3.
Solution:
Empirical probability (most likely based on a survey)
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Example: Classifying Types of Probability
Classify the statement as an example of classical,
empirical, or subjective probability.
3. The probability of winning a 1000-ticket raffle with
1
one ticket is 1000 .
Solution:
Classical probability (equally likely outcomes)
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Range of Probabilities Rule
Range of probabilities rule
• The probability of an event E is between 0 and 1,
inclusive.
• 0 ≤ P(E) ≤ 1
Impossible
Unlikely
Even
chance
[
0
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Likely
Certain
]
0.5
1
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Complementary Events
Complement of event E
• The set of all outcomes in a sample space that are not
included in event E.
• Denoted E ′ (E prime)
• P(E) + P(E ′) = 1
• P(E) = 1 – P(E ′)
E′
• P(E ′) = 1 – P(E)
E
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Example: Probability of the Complement
of an Event
You survey a sample of 1000 employees at a company
and record the age of each. Find the probability of
randomly choosing an employee who is not between 25
and 34 years old.
Employee ages Frequency, f
15 to 24
54
25 to 34
366
35 to 44
233
45 to 54
180
55 to 64
125
65 and over
42
Σf = 1000
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Solution: Probability of the Complement
of an Event
• Use empirical probability to
find P(age 25 to 34)
f
366
P(age 25 to 34)  
 0.366
n 1000
• Use the complement rule
366
P(age is not 25 to 34)  1 
1000
634

 0.634
1000
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Employee ages
Frequency, f
15 to 24
54
25 to 34
366
35 to 44
233
45 to 54
180
55 to 64
125
65 and over
42
Σf = 1000
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Example
• The probability of rolling a 2 or a 5 on a 6 sided die is
P(2 or 5) = 2/6 or 1/3
• What is the complement of this?
• P(not 2 or 5) = 4/6 or 2/3
Example: Probability Using a Tree
Diagram
A probability experiment consists of tossing a coin and
spinning the spinner shown. The spinner is equally
likely to land on each number. Use a tree diagram to
find the probability of tossing a tail and spinning an odd
number.
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Solution: Probability Using a Tree
Diagram
Tree Diagram:
H
T
1 2 3 4 5 6 7 8
1 2 3 4 5 6 7 8
H1 H2 H3 H4 H5 H6 H7 H8
T1 T2 T3 T4 T5 T6 T7 T8
4 1
  0.25
P(tossing a tail and spinning an odd number) =
16 4
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Example: Probability Using the
Fundamental Counting Principle
Your college identification number consists of 8 digits.
Each digit can be 0 through 9 and each digit can be
repeated. What is the probability of getting your college
identification number when randomly generating eight
digits?
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Solution: Probability Using the
Fundamental Counting Principle
• Each digit can be repeated
• There are 10 choices for each of the 8 digits
• Using the Fundamental Counting Principle, there are
10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10
= 108 = 100,000,000 possible identification numbers
• Only one of those numbers corresponds to your ID
number
1
P(your ID number) =
100, 000, 000
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Example
Use the Fundamental
Counting Principle
• What are the total possible combinations to create
Phone Numbers?
• An area code cannot begin with a 1 or a 0
• A local number cannot begin with a 1 or a 0
• Here are the combinations:
X
X
X
X
X
X
Choices 2-9
X
X
X
X
•It is expected that by the
year 2020, some areas
may run out of numbers
•What is the solution?
Choices 0-9
We have 8 x 10 x 10 x 8 x 10 x 10
possible phone numbers
x 10 x 10 x 10 x 10 = 6,400,000,000
Section 3.1 Summary
• Identified the sample space of a probability
experiment
• Identified simple events
• Used the Fundamental Counting Principle
• Distinguished among classical probability, empirical
probability, and subjective probability
• Determined the probability of the complement of an
event
• Used a tree diagram and the Fundamental Counting
Principle to find probabilities
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Assignment
•
•
•
•
•
•
Page 138 1-6 Oral
Page 138 11-14
Page 139 21-28
Page 140 37-42
Page 141 55-62
Read about “Odds” on page 143 and answer
questions 76 and 77.
Larson/Farber 5th edition
40
Chapter 3 Quiz 1
1.
2.
3.
4.
Name a type of probability
Name a 2nd type of probability
Name the 3rd type of probability
Using the Fundamental Counting Rule, how many different ways can you dress if
you have 3 pair of pants, 5 types of shirts, and a choice between tennis shoes and
boots
5. You purchase a new car. You have 3 manufacturers –Ford, Chevy, or Dodge to
choose from. You have 4 colors –red, blue, black, and white. You can also choose
between a 2 door and a 4 door. How many different ways can you select one
manufacturer, one car size, and one color?
6. Using the information from question 5 create a tree diagram to calculate the
probability of purchasing a white 4 door Ford Taurus.
7. True/False: The probability of any event can be demonstrated as: -1 ≤ P(E) ≤ 1
8. What type of probability is this: “The probability that a voter chosen at random
will be younger than 35 years old is 0.3.”
9. True/False: Subjective Probability is often based on a feeling.
10. True/False: This is a simple event: “You roll a six-sided die. Event B is rolling at
41
least a 4”
Section 3.2
Conditional Probability and the
Multiplication Rule
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Section 3.2 Objectives
• Determine conditional probabilities
• Distinguish between independent and dependent
events
• Use the Multiplication Rule to find the probability of
two events occurring in sequence
• Use the Multiplication Rule to find conditional
probabilities
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Conditional Probability
Conditional Probability
• The probability of an event occurring, given that
another event has already occurred
• Denoted P(B | A) (read “probability of B, given A”)
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Example: Finding Conditional
Probabilities
Two cards are selected in sequence from a standard
deck. Find the probability that the second card is a
queen, given that the first card is a king. (Assume that
the king is not replaced.)
Solution:
Because the first card is a king and is not replaced, the
remaining deck has 51 cards, 4 of which are queens.
4
P( B | A)  P(2 card is a Queen |1 card is a King ) 
 0.078
51
nd
st
NOTE: We are only calculating the QUEEN. Later we
will calculate the probability of drawing BOTH
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Example: Finding Conditional
Probabilities
The table shows the results of a study in which
researchers examined a child’s IQ and the presence of a
specific gene in the child. Find the probability that a
child has a high IQ, given that the child has the gene.
Gene
Present
Gene not
present
Total
High IQ
33
19
52
Normal IQ
39
11
50
Total
72
30
102
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Solution: Finding Conditional
Probabilities
There are 72 children who have the gene. So, the
sample space consists of these 72 children.
Gene
Present
Gene not
present
Total
High IQ
33
19
52
Normal IQ
39
11
50
Total
72
30
102
Of these, 33 have a high IQ.
33
P(B | A)  P(high IQ | gene present ) 
 0.458
72
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Independent and Dependent Events
Independent events
• The occurrence of one of the events does not affect
the probability of the occurrence of the other event
• P(B | A) = P(B) or P(A | B) = P(A)
• Events that are not independent are dependent
•In words: If the probability of B given A is still calculated as the
probability of B, then A has not affected B. Therefore B is
independent of A.
•For example, the probability of rolling a 5 after flipping a heads.
Rolling a 5 is not affected by whether the coin is heads or tails.
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Example: Independent and Dependent
Events
Decide whether the events are independent or dependent.
1. Selecting a king from a standard deck (A), not
replacing it, and then selecting a queen from the deck
(B).
Solution:
P( B | A)  P(2nd card is a Queen |1st card is a King ) 
P ( B )  P (Queen) 
4
52
4
51
Dependent (the occurrence of A changes the probability
of the occurrence of B)
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Example: Independent and Dependent
Events
Decide whether the events are independent or dependent.
2. Tossing a coin and getting a head (A), and then
rolling a six-sided die and obtaining a 6 (B).
Solution:
P( B | A)  P(rolling a 6 | head on coin) 
P( B)  P(rolling a 6) 
1
6
1
6
Independent (the occurrence of A does not change the
probability of the occurrence of B)
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The Multiplication Rule
Multiplication rule for the probability of A and B
• The probability that two events A and B will occur in
sequence is
 P(A and B) = P(A) ∙ P(B | A)
• For independent events the rule can be simplified to
 P(A and B) = P(A) ∙ P(B)
 Can be extended for any number of independent
events
 Remember this: If you see “AND” then you use
the Multiplication Rule
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Example: Using the Multiplication Rule
Two cards are selected, without replacing the first card,
from a standard deck. Find the probability of selecting a
king and then selecting a queen.
Solution:
Because the first card is not replaced, the events are
dependent.
P( K and Q)  P( K )  P(Q | K )
4 4
 
52 51
16

 0.006
2652
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Example: Using the Multiplication Rule
A coin is tossed and a die is rolled. Find the probability
of getting a head and then rolling a 6.
Solution:
The outcome of the coin does not affect the probability
of rolling a 6 on the die. These two events are
NOTE:
independent.
P( H and 6)  P( H )  P(6)
1 1
 
2 6
1
  0.083
12
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•Previously, we used a tree
diagram
•Now we see that this is similar
to the Fundamental Counting
Principle, except that we are
calculating probabilities, not
total possible outcomes
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Example: Using the Multiplication Rule
The probability that a particular knee surgery is
successful is 0.85. Find the probability that three knee
surgeries are successful.
Solution:
The probability that each knee surgery is successful is
0.85. The chance for success for one surgery is
independent of the chances for the other surgeries.
P(3 surgeries are successful) = (0.85)(0.85)(0.85)
≈ 0.614
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Example: Using the Multiplication Rule
Find the probability that none of the three knee
surgeries is successful.
Solution:
Because the probability of success for one surgery is
0.85. The probability of failure for one surgery is
1 – 0.85 = 0.15
P(none of the 3 surgeries is successful) = (0.15)(0.15)(0.15)
≈ 0.003
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Example: Using the Multiplication Rule
Find the probability that at least one of the three knee
surgeries is successful.
Solution:
“At least one” means one or more. The complement to
the event “at least one is successful” is the event “none
are successful.” Using the complement rule
P(at least 1 is successful) = 1 – P(none are successful)
≈ 1 – 0.003
= 0.997
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Chapter 3 Quiz 2
T/F: “P(B | A)” reads as “the probability of B, not given A”
What is the probability of getting a queen if you have already pulled 2 queens
previously from a standard deck (without replacement)?
3. Using the previous question, what is the probability of getting a queen if you DO
replace the 2 queens?
4. What is the probability of rolling a 6 on a six sided die, if you have already rolled a
6 three times in a row?
5. The probability of winning a game of cards is 1/1000. The probability of winning a
game of basketball is 1/1000. What is the probability of winning both games?
6. T/F: Pulling a queen from a deck of standard cards is independent from flipping a
coin and getting heads.
7. State the multiplication rule for independent events.
8. The probability of success is .90. What is the probability of success 4 times in a
row?
9. The probability of failure is .20. What is the probability of failure 3 times in a row?
10. The probability of success is .90. What is the probability of failure?
1.
2.
Larson/Farber 5th edition
57
Roulette Wheel
Larson/Farber 5th edition
58
Example
• A U.S. roulette wheel has 38 numbered slots (1 through 36, 0,
and 00)
• Of these, 18 are black, 18 are red, and 2 are green
• A dealer spins the ball one direction, and the wheel the other.
The ball can land with equal (and independent) probability on
any one of the 38 numbered spots
• What is the possibility of the ball landing on red twice in a
row?
• There are 38 possible choices, with 18 red, therefore the
probability of landing red is 18/38, or 9/19
• Doing it twice in a row would be 9/19 x 9/19 = 81/361, or .224
• What is the probability of landing green twice in a row?
What is the probability of having 9 girls in a row?
½ x ½ x ½ x ½ x ½ x ½ x ½ x ½ x ½ = (1/2)9
Example
• According to the National Hurricane Center, the
probability of S. Florida getting hit by a hurricane in a
single year is 5/19
1. What is the probability of getting hit by a hurricane 3
years in a row?
2. What is the probability of not being hit by a hurricane
in the next ten years
3. What is the probability of getting hit at least once in the
next 10 years?
Example
1. P(hurricane and hurricane and hurricane) = P(hurricane x hurricane
x hurricane) = 5/19 x 5/19 x 5/19 = 125/6859 = .018 or 1.8%
2. P(no hurricane 10 years) = 1-P(hurricane) or 1-5/19 = 14/19 or .737
-this is one year, so we multiply it times 10, or 14/19 x 14/19…. Or
(14/19)10 = (.737)10 = .047, or 4.7%
3. P(at least one hurricane in 10 years) = 1-P(no hurricanes in 10
years) = 1- .047 = .953
• There is a .953, or 95.3% probability of S. Florida getting hit by a
hurricane in the next 10 years, but only a 1.8% probability that it
will happen 3 years in a row
• The P(event happening at least once) = 1-P(event does not
happen)
Chocolate Covered Cherries 
• You love chocolate covered cherries
• There are 5 in a box of 20 the rest are chocolates that
look just the same
• What is the probability that you will get 2 in a row?
• Well, for the first cherry, the probability is a simple
5/20
• What about the second cherry? Looking only at that
cherry, what is the probability of getting a chocolate
covered cherry the 2nd time?
Cherries
• Since we ate a cherry, there are only 4 left
• Since we ate a chocolate, there are only 19 left
• Therefore, the probability of getting a cherry the 2nd
time is 4/19
• So, what is the probability of getting a cherry two
times in a row is ¼ x 4/19 = 1/19
Example: Using the Multiplication Rule to
Find Probabilities
More than 15,000 U.S. medical school seniors applied to
residency programs in 2009. Of those, 93% were matched
with residency positions. Eighty-two percent of the seniors
matched with residency positions were matched with one of
their top three choices. Medical students electronically rank
the residency programs in their order of preference, and
program directors across the United States do the same. The
term “match” refers to the process where a student’s
preference list and a program director’s preference list
overlap, resulting in the placement of the student for a
residency position. (Source: National Resident Matching
Program)
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(continued)
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Example: Using the Multiplication Rule to
Find Probabilities
1. Find the probability that a randomly selected senior was
matched with a residency position and it was one of the
senior’s top three choices.
Solution:
A = {matched with residency position}
B = {matched with one of top three choices}
P(A) = 0.93 and P(B | A) = 0.82
P(A and B) = P(A)∙P(B | A) = (0.93)(0.82) ≈ 0.763
dependent events
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Example: Using the Multiplication Rule to
Find Probabilities
2. Find the probability that a randomly selected senior who
was matched with a residency position did not get
matched with one of the senior’s top three choices.
Solution:
Use the complement:
P(B′ | A) = 1 – P(B | A)
= 1 – 0.82 = 0.18
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Section 3.2 Summary
• Determined conditional probabilities
• Distinguished between independent and dependent
events
• Used the Multiplication Rule to find the probability
of two events occurring in sequence
• Used the Multiplication Rule to find conditional
probabilities
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Chapter 3 Bell Ringer 1 (on your own)
• You win a trip to Madrid Spain, and have 2 extra
tickets
• 10 “cousins” miraculously show up to accompany
you
• You write their names on 10 cards, and place the
cards in a hat, and select one name
• Then you select a second name without replacement
• If 3 of the cousins speak Spanish, find the probability
of selecting 2 Spanish-speaking cousins
• 3/10 x 2/9 = 6/90 = 1/15 or .067
Assignment
• Page 150 7-12 (no explanation)
• Page 151-154 17-31 odd (no explanation)
Larson/Farber 5th edition
69
Chapter 3 Quiz 3 (20 points)
• Solve Problem 20 on Page 152 (on your own)
• Solve all parts
• 30 minutes
Larson/Farber 5th edition
70
Section 3.3
Addition Rule
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Section 3.3 Objectives
• Determine if two events are mutually exclusive
• Use the Addition Rule to find the probability of two
events
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Mutually Exclusive Events
You need to know this because if they are not
Mutually Exclusive, then they are an “AND”
problem, and we would use the multiplication rule
Mutually exclusive
• Two events A and B cannot occur at the same time
A
B
A and B are mutually
exclusive
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A
B
A and B are not mutually
exclusive
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Example: Mutually Exclusive Events
Decide if the events are mutually exclusive.
Event A: Roll a 3 on a die.
Event B: Roll a 4 on a die.
Solution:
Mutually exclusive (The first event has one outcome, a
3. The second event also has one outcome, a 4. These
outcomes cannot occur at the same time.)
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Example: Mutually Exclusive Events
Decide if the events are mutually exclusive.
Event A: Randomly select a male student.
Event B: Randomly select a nursing major.
Solution:
Not mutually exclusive (The student can be a male
nursing major.)
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The Addition Rule
Addition rule for the probability of A or B
• The probability that events A or B will occur is
 P(A or B) = P(A) + P(B) – P(A and B)
• For mutually exclusive events A and B, the rule can
This is where they overlap, so
be simplified to
you subtract this value so that
you are not counting it twice
 P(A or B) = P(A) + P(B)
 Can be extended to any number of mutually
exclusive events
 NOTE: When you see “Or” then you use the
Addition Rule
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Example: Using the Addition Rule
You select a card from a standard deck. Find the
probability that the card is a 4 or an ace.
Solution:
The events are mutually exclusive (if the card is a 4, it
cannot be an ace)
Deck of 52 Cards
P(4 or ace)  P(4)  P(ace)
4
4
 
52 52
8

 0.154
52
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4♣
4♥
4♠
4♦
A♣
A♠ A♥
A♦
44 other cards
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Example: Using the Addition Rule
You roll a die. Find the probability of rolling a number
less than 3 or rolling an odd number.
Solution:
The events are not mutually exclusive (1 is an
outcome of both events)
Roll a Die
4
Odd
3
5
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6
Less than
1 three
2
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Solution: Using the Addition Rule
Roll a Die
4
Odd
3
5
6
Less than
1 three
2
P(less than 3 or odd )
 P(less than 3)  P(odd )  P(less than 3 and odd )
2 3 1 4
     0.667
6 6 6 6
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OR probabilities with events that are NOT
Mutually Exclusive
• What is the probability of selecting a diamond or a
picture card (Jack, Queen, King)?
• Normally, we would begin by adding probabilities
P(diamond) + P(Face Card) =
• 13/52 + 12/52 = 25/52 –about 48%
• This, however, calculates the probability of selecting
one or the other, but in reality you could have a
diamond face card –these are not mutually
exclusive
Not Mutually Exclusive
• There are, in fact 3 cards which are in both categories
–Diamonds, and Face Cards
• Therefore, we are actually counting these cards twice
• Therefore, we will subtract the probability of getting
a diamond AND a face card –which is 3/52
Diamond
Face Card
Diamond/Face Card
• We now have 13/52 + 12/52 – 3/52 = (13+12-3)/52 =
22/52 = 11/26 –about 42%
Example: Using the Addition Rule
The frequency distribution shows
the volume of sales (in dollars)
and the number of months in
which a sales representative
reached each sales level during
the past three years. If this sales
pattern continues, what is the
probability that the sales
representative will sell between
$75,000 and $124,999 next
month?
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Sales volume ($)
Months
0–24,999
3
25,000–49,999
5
50,000–74,999
6
75,000–99,999
7
100,000–124,999
9
125,000–149,999
2
150,000–174,999
3
175,000–199,999
1
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Solution: Using the Addition Rule
• A = monthly sales between
$75,000 and $99,999
• B = monthly sales between
$100,000 and $124,999
• A and B are mutually exclusive
P( A or B)  P( A)  P( B)
7 9
 
36 36
16

 0.444
36
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Sales volume ($)
Months
0–24,999
3
25,000–49,999
5
50,000–74,999
6
75,000–99,999
7
100,000–124,999
9
125,000–149,999
2
150,000–174,999
3
175,000–199,999
1
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Example: Using the Addition Rule
A blood bank catalogs the types of blood, including
positive or negative Rh-factor, given by donors during
the last five days. A donor is selected at random. Find
the probability that the donor has type O or type A
blood.
Type O
Type A
Type B
Type AB
Total
Rh-Positive
156
139
37
12
344
Rh-Negative
28
25
8
4
65
184
164
45
16
409
Total
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Solution: Using the Addition Rule
The events are mutually exclusive (a donor cannot have
type O blood and type A blood)
Type O
Type A
Type B
Type AB
Total
Rh-Positive
156
139
37
12
344
Rh-Negative
28
25
8
4
65
184
164
45
16
409
Total
P(type O or type A)  P(type O )  P(type A)
184 164


409 409
348

 0.851
409
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Example: Using the Addition Rule
Find the probability that the donor has type B blood or
is Rh-negative.
Type O
Type A
Type B
Type AB
Total
Rh-Positive
156
139
37
12
344
Rh-Negative
28
25
8
4
65
184
164
45
16
409
Total
Solution:
The events are not mutually exclusive (a donor can have
type B blood and be Rh-negative)
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Solution: Using the Addition Rule
Type O
Type A
Type B
Type AB
Total
Rh-Positive
156
139
37
12
344
Rh-Negative
28
25
8
4
65
184
164
45
16
409
Total
P(type B or Rh  neg)
 P(type B)  P(Rh  neg)  P(type B and Rh  neg)
45
65
8
102




 0.249
409 409 409 409
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Summary
Type of Probability and
Probability Rules
Formula
Classical Probability
P(E) = number of outcomes E divided by
Number of outcomes in sample space
Empirical Probability
P(E) = Frequency of Event E divided by
Total Frequency
Complementary Events P(E) + P(E') = 1, P(E) = 1 - P(E'), P(E') = 1-P(E)
Multiplication Rule
P(A and B) = P(A) *P(B|A)
P(A and B) = P(A) * P(B) --> Independent events
Addition Rule
P(A or B) = P(A) + P(B) - P(A and B)
P(A or B) = P(A) + P(B) Mutually Exclusive
Events
f
n
Assignment
• Page 161 9-12, 17-26
Larson/Farber 5th edition
89
Assignment
• Page 152-154 22-32 Even (no explanation)
Larson/Farber 5th edition
90
Section 3.3 Summary
• Determined if two events are mutually exclusive
• Used the Addition Rule to find the probability of two
events
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Chapter 3 Quiz 4 25 points
• On your own:
• On page 167, do problem 3 all 5 parts
Larson/Farber 5th edition
92
Section 3.4
Additional Topics in Probability and
Counting
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Section 3.4 Objectives
• Determine the number of ways a group of objects can
be arranged in order
• Determine the number of ways to choose several
objects from a group without regard to order
• Use the counting principles to find probabilities
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Permutations
Permutation
• An ordered arrangement of objects
• The number of different permutations of n distinct
objects is n! (n factorial)
 n! = n∙(n – 1)∙(n – 2)∙(n – 3)∙ ∙ ∙3∙2 ∙1
 0! = 1 this keeps us from getting zero for an
answer every time we calculate any factorial 
 Examples:
• 6! = 6∙5∙4∙3∙2∙1 = 720
• 4! = 4∙3∙2∙1 = 24
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Factorial Notation
•
•
•
•
•
•
•
•
1! = 1
NOTE: By
2! = 1 x 2 = 2
definition, 0! = 1
3! = 1 x 2 x 3 = 6
4! = 1 x 2 x 3 x 4 = 24
5! = 1 x 2 x 3 x 4 x 5 = 120
What can we conclude about the next factorial?
6! will be 5! x 6
Therefore, we can break factorials down into part, in a
similar manner to exponents
• For example, 8! = 6! x 7 x 8
Reducing Factorials
• Solve 8!/5!
• This can be rewritten as (8 x 7 x 6 x 5!)/5!
• Therefore, we can cancel out the 5!, and we are left
with 8 x 7 x 6 = 336
• Solve 26!/21!
• This is 21! x 22 x 23 x 24 x 25 x 26/21!
 This is reduced to 22 x 23 x 24 x 25 x 26 =
7893600
• 500!/499! = ?
Permutations
• Suppose you had to organize a Wayne’s World III
concert, and you had 4 bands
• How many different possible combinations (or
permutations) are there to decide what order they play
in?
• This is different than asking how many ways you can
wear t-shirts and the pants, because those are two
separate items, and can be worn in every combination
• Here, after using one band, the combinations drop by
one choice.
• So how do we do it?
Fundamental Counting Principle
• You still use the Fundamental Counting Principle, but
you have to pay attention 
• Originally, you have 4 choices of bands to start with
• After the first band plays, you have 3 choices
• After that, two choices
• And finally, you have just the one choice of which
band to play
• Therefore, we calculate 4 x 3 x 2 x 1 = 24  or 4!
• There are 24 different ways to arrange your concert
Example: Permutation of n Objects
The objective of a 9 x 9 Sudoku number
puzzle is to fill the grid so that each
row, each column, and each 3 x 3 grid
contain the digits 1 to 9. How many
different ways can the first row of a
blank 9 x 9 Sudoku grid be filled?
Solution:
The number of permutations is
9!= 9∙8∙7∙6∙5∙4∙3∙2∙1 = 362,880 ways
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Permutations
Permutation of n objects taken r at a time
• The number of different permutations of n distinct
objects taken r at a time
■
n!
n Pr 
(n  r )!
where r ≤ n
• In English: You have a certain number of choices (n
objects) and you don’t choose them all. You choose
several of them (r at a time)
• Remember: A permutation is an ordered
arrangement of objects
• I personally call this “N Pick R”
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Permutations
• We call each ordered arrangement a “Permutation”
– Permutations state that no item is used more than once –
each arrangement is unique
– The order of arrangement makes a difference
• Again, suppose you have a 4 band concert:
• Suppose you decide that the Rolling Stones will perform last,
no matter what
• Then your choice of first group is 3
• Your choice of second group is 2
• Your choice of third group is 1
• Your choice of fourth group is 1
• Using the F.C.P., we calculate 3 x 2 x 1 x 1 = 6 permutations
Example
• Suppose you have U2, the Rolling Stones, Aerosmith, and
Metallica performing
• You decide that U2 will open, and the Rolling Stones will
close
• How many permutations do you have?
• 1 for the 1st
• 2 for the 2nd
• 1 for the 3rd
• 1 for the 4th
• Therefore 1 x 2 x 1 x 1 = just two permutations
A Look at the Formula for Permutations
• Suppose you have 13 players on your little league team, but
only 9 can bat at a time
• How many permutations are there for you to pick the batting
order?
• 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 (note, we stop here,
because we have 9 batters)
• = 259,459,200
• We can rewrite this as:
• 13 x 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
•
4x3x2x1
• Or, 13!/4!
n!
• Or 13! (13-9)!
n Pr 
(n  r )!
• This is the equation:
Example: Finding nPr
Find the number of ways of forming four-digit codes in
which no digit is repeated.
Solution:
• You need to select 4 digits from a group of 10
• n = 10, r = 4
Say “10 pick 4”
10!
10!
10
P4 
(10  4)!

6!
10  9  8  7  6  5  4  3  2  1

6  5  4  3  2 1
 5040 ways
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Example: Finding nPr
Forty-three race cars started the 2007 Daytona 500.
How many ways can the cars finish first, second, and
third?
Solution:
• You need to select 3 cars from a group of 43
• n = 43, r = 3
43!
43!

43 P3 
(43  3)! 40!
 43  42  41
 74, 046 ways
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Distinguishable Permutations
Distinguishable Permutations
• The number of distinguishable permutations of n
objects where n1 are of one type, n2 are of another
type, and so on
n!
■
n1 ! n2 ! n3 !   nk !
where n1 + n2 + n3 +∙∙∙+ nk = n
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Example: Distinguishable Permutations
A building contractor is planning to develop a
subdivision that consists of 6 one-story houses, 4 twostory houses, and 2 split-level houses. In how many
distinguishable ways can the houses be arranged?
Solution:
• There are 12 houses in the subdivision
• (6+4+2)
• n = 12, n1 = 6, n2 = 4, n3 = 2
12!
6! 4! 2!
 13,860 distinguishable ways
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Example
• How many distinct ways can the letters in the name
Mississippi be arranged?
• n=11
• There are 4 “I”s so n1 = 4
• There are 4 “S”s so n2 = 4
• There are 2 “P”s so n3= 2
• n!/n1!*n2!*n3!...
• 11!/4!∙4!∙2! = 11x10x9x8x7x6x5x4!/4!x4x3x2x1x2x1
(cancel out the 4!)
• 11x10x9x8x7x6x5/4x3x2x1x2x1
• = 34,650 possible combinations
On the Calculator
• Find the number of ways of forming 3 digit codes in
which no digit is repeated
• We use: nPr = n!/(n-r)!
• What is n? What is r?
• How do we do this on the calculator?
• Type in 10, then go to  math  PRB  nPr and
then hit enter…
Bell Ringer
• You and a friend decide to list the top 10 bands of all
time
• You decide to order them as well, from 1 to 10
• To make it a little easier, you decide that you will
only choose from 25 bands which you agree on
1. How many possible top ten lists can you make?
2. What if you both agree that Nickelback is the best
band of all time  how many top ten band lists are
there now? 25P10 = 1.186 x 1013 25P9 = 7.414 x 1011
11
P
=
4.744
x
10
24 9
Combinations
• Suppose you named 5 influential people from the past
30 years, and asked your friend to identify which 3
were the most influential
• One friend names three, another names the same
three, but in a different order
• To us, it does not matter what order they name them,
because we are interested in the names, not the order
they are named
• This is a Combination
Combinations
• A Combination of items occurs when:
 The items are selected from the same group
 No item is used more than once
 The order of items does not matter
• Permutations problems have situations in which the
ORDER of ITEMS matters
• Combinations problems have situations in which the
ORDER of ITEMS makes no difference
Combinations
Combination of n objects taken r at a time
• A selection of r objects from a group of n objects
without regard to order
n!
■ C 
n
r
( n  r )! r !
Remember, this was the equation for Permutations:
■
n!
n Pr 
(n  r )!
© 2012 Pearson Education, Inc. All rights reserved.
where r ≤ n
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Example: Combinations
A state’s department of transportation plans to develop a
new section of interstate highway and receives 16 bids
for the project. The state plans to hire four of the
bidding companies. How many different combinations
of four companies can be selected from the 16 bidding
companies?
Solution:
• You need to select 4 companies from a group of 16
• n = 16, r = 4
• Order is not important
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Solution: Combinations
I call this “N Choose R”
16!
16 C4 
(16  4)!4!
16!

12!4!
16 15 14 13 12!

12! 4  3  2 1
 1820 different combinations
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Determine the Difference
• Six students are running for student president, VP, and treasurer. The
greatest number of votes will be president, least number is treasurer. How
many possible outcomes are there for these positions?
• Six people are on the board of supervisors for the park committee. A 3 man
committee is needed to study the possibility of expanding the park. How
many different committees could be formed from the six?
• Baskin Robbins offers 31 different flavors of ice cream. One of their items
is a bowl of 3 scoops of ice cream, each a different flavor. How many such
bowls are possible?
• 1) Order matters, this is a permutation problem
• 2) Order does not matter, this is a combination problem
• 3) Order does not matter, this is a combination problem
A Closer Look of Combinations
For Combinations: nCr = n!/(n-r)!r!
•Using
Consider
letters A,B,C,D
this the
equation,
we get:
•nCr
What
is the number
of permutations,
= n!/(n-r)!r!
= 4C3
= 4!/(4-3)!3!taken
= 4 3 at a
time?
combinations
• nPr = n!/(n-r)! = 4P3 = 24
• Here they are:
-Note that while there are 24
Permutations, there are only 4 different
Combinations of letters
-Therefore, each column of 6
Permutations represents one
Combination
ABC
ACB
BAC
BCA
CAB
CBA
ABD
ADB
BAD
BDA
DAB
DBA
ACD
ADC
CAD
CDA
DAC
DCA
BCD
BDC
CBD
CDB
DBC
CDB
ABC ABD ACD BCD
Poker
•
•
•
•
There are 52 cards in a deck
A standard poker hand is 5 cards
The order in which you are dealt does not matter
How many different 5 card poker hands are possible?
• nCr = 52C5 = 2,598,960
Example
• Find the probability of being dealt 5 diamonds from a
standard deck
• It does not matter what order the cards are dealt
• The top of our ratio would be 13
C5
C5
• The bottom would be 52
• Therefore, we have 1287/2,598,960 = .0005, or .05
percent  this is half of one tenth of one percent
Combining the Fundamental Counting
Principle with Combinations
• Remember, the FCP basically says multiply the
number of options from the first choice times the
number of options from the second choice and so on
• What if you have different combinations of options?
• Calculate these combinations, and multiply them in
the same manner
Example
• Suppose we have 54 Republicans and 46 Democrats
in the US Senate (total 100)
• Furthermore, suppose we want to form a committee
of 3 Republicans and 2 Democrats
• How many possible committee combinations are
there?
• 54C3 x 46C2 or 24,804 x 1035 = 25,672,140
• There are over 25 million combinations
Example: Finding Probabilities
A student advisory board consists of 17 members. Three
members serve as the board’s chair, secretary, and
webmaster. Each member is equally likely to serve any
of the positions. What is the probability of selecting at
random the three members that hold each position?
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Solution: Finding Probabilities
• There is only one favorable outcome
• There are
17!
17 P3 
(17  3)!
17!

 17 16 15  4080
14!
ways the three positions can be filled
1
P( selecting the 3 members) 
 0.0002
4080
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Example: Finding Probabilities
You have 11 letters consisting of one M, four I’s, four
S’s, and two P’s. If the letters are randomly arranged in
order, what is the probability that the arrangement spells
the word Mississippi?
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Solution: Finding Probabilities
• There is only one favorable outcome
• There are
11!
 34, 650
1! 4! 4! 2!
11 letters with 1,4,4, and 2
like letters
distinguishable permutations of the given letters
1
P( Mississippi)
 0.00003
34650
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Example: Finding Probabilities
A food manufacturer is analyzing a sample of 400 corn
kernels for the presence of a toxin. In this sample, three
kernels have dangerously high levels of the toxin. If
four kernels are randomly selected from the sample,
what is the probability that exactly one kernel contains a
dangerously high level of the toxin?
Hint: this is 2 separate probabilities. One is the
probability of choosing exactly one kernel that is
toxic, and the 2nd probability is choosing exactly
3 kernels that are not toxic.  We want both,
not one or the other, so it’s an “And” problem
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Solution: Finding Probabilities
• The possible number of ways of choosing one toxic
kernel out of three toxic kernels is
3C1 = 3
• The possible number of ways of choosing three
nontoxic kernels from 397 nontoxic kernels is
397C3 = 10,349,790
• Using the Multiplication Rule, the number of ways of
choosing one toxic kernel and three nontoxic kernels
is
3C1 ∙ 397C3 = 3 ∙ 10,349,790 3 = 31,049,370
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Solution: Finding Probabilities
• The number of possible ways of choosing 4 kernels
from 400 kernels is
400C4 = 1,050,739,900
• The probability of selecting exactly 1 toxic kernel is
C  C
3 1 397 3
P(1 toxic kernel) 
C
400 4
31,049,370

 0.030
1,050,739,900
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Section 3.4 Summary
• Determined the number of ways a group of objects
can be arranged in order
• Determined the number of ways to choose several
objects from a group without regard to order
• Used the counting principles to find probabilities
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Assignment
• Page 174-176 7-47 Odd
Larson/Farber 5th edition
131
Chapter 3 BellRinger 2
• The access code for a car’s security system consists
of 4 digits
• Each digit can be 0 through 9
• How many access codes are possible if:
 Each digit can only be used once and not repeated
 Each digit can be repeated
• 10 x 9 x 8 x 7 = 5040
• 104 = 10,000
Chapter 3 Quiz 5
• You are dealt 5 cards from the top of the deck.
1. What is the probability of being dealt 4 of a kind?
2. What is the probability of being dealt a full house (3
of a kind and 2 of a kind)?
3. What is the probability of being dealt 3 of a kind
(the other two cards do not matter)?
4. What is the probability of being dealt a flush (all the
same suit)?
5. What is the probability of being dealt 2 Kings, a
Jack, a Ten, and an Ace?
Larson/Farber 5th edition
133
Chapter 3 Final Exam Extra Credit
• Suppose you are playing 5 card stud. In this game, you get dealt 5 cards
and that is it –you bet on what you are dealt
• The highest hand in Poker is a Royal Flush. In a Royal Flush, you have a
10, Jack, Queen, King and Ace all of the same suit –so it’s also a straight
1. What is the probability that you are dealt a Royal Flush in Hearts?
5 points, show your equation, show your answer
• Hint: You are the only player
• Hint: These cards are dealt without replacement
• Suppose you are missing one of those cards (For example, you get a 6
instead of a jack, and all you need is a jack to get a Royal Flush).
2. If you were given an additional card, what is the probability of drawing
the card that you are missing? (just that card)
5 Points, show your equation, show your answer
• Hint: You are still the only player
• Hint: These cards are still dealt without replacement
Larson/Farber 5th edition
134
Chapter 3 Final Exam EC (2 pts each)
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•
1.
•
2.
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You have 30 marbles in a bag -10 red, 10 blue, 7 green 2 yellow, 1 black
You are trying to avoid pulling a black marble
What is the probability of pulling a black marble on your first pull?
Starting over, you pull 3 red and 2 blue (without replacement)
Now what is the probability of pulling the black marble?
You get lucky, and you don’t pull a black marble. Instead you pull a yellow
one.
3. What is the probability of pulling a yellow marble, given that you’ve
already pulled 3 red and 2 blue without replacement?
4. Starting over, without replacement, what is the probability of pulling 2
red, then 2 blue in a row? (Red,Red,Blue,Blue)
5. Starting over, with replacement, what is the probability of pulling 3 green
in a row? (Green,Green,Green)
Larson/Farber 5th edition
135