Transcript Chapter 4 ~ Probability Limiting Relative Frequency 0.6 0.5 0.4 Relative Frequency 0.3 0.2 0.10 Trials Chapter Goals • Learn the basic concepts of probability • Learn the rules that apply to the.

Chapter 4 ~ Probability
Limiting Relative Frequency
0.6
0.5
0.4
Relative
Frequency
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0.2
0.1
0
0
200
400
600
800
1000
1200
Trials
1
Chapter Goals
• Learn the basic concepts of probability
• Learn the rules that apply to the probability of both
simple and compound events
• In order to make inferences, we need to study
sample results in situations in which the population
is known
2
4.1 ~ The Nature of Probability
Example: Consider an experiment in which we roll two sixsided fair dice and record the number of 3s face
up. The only possible outcomes are zero 3s, one
3, or two 3s. Here are the results after 100 rolls,
and after 1000 rolls:
100 Rolls
Outcome Frequency
0
80
1
19
2
1
1000 Rolls
Outcome Frequency
0
690
1
282
2
28
3
Using a Histogram
• We can express these results (from the 1000 rolls) in terms of
relative frequencies and display the results using a histogram:
0.7
0.6
0.5
Relative
Frequency
0.4
0.3
0.2
0.1
0.0
0
1
2
Three’s Face Up
4
Continuing the Experiment
• If we continue this experiment for several thousand
more rolls:
1. The frequencies will have approximately a 25:10:1
ratio in totals
2. The relative frequencies will settle down
Note: We can simulate many probability experiments:

Use random number tables

Use a computer to randomly generate number values
representing the various experimental outcomes

Key to either method is to maintain the probabilities
5
4.2 ~ Probability of Events
Probability that an Event Will Occur: The relative
frequency with which that event can be expected to occur
• The probability of an event may be obtained in three
different ways:
– Empirically
– Theoretically
– Subjectively
6
Experimental or Empirical Probability
Experimental or Empirical Probability:
1. The observed relative frequency with which an
event occurs
2. Prime notation is used to denote empirical
probabilities:
n( A )
P (A ) 
n
3. n(A): number of times the event A has occurred
4. n: number of times the experiment is attempted

Question: What happens to the observed relative
frequency as n increases?
7
Example
 Example: Consider tossing a fair coin. Define the event H
as the occurrence of a head. What is the probability
of the event H, P(H)?
1. In a single toss of the coin, there are two possible outcomes
2. Since the coin is fair, each outcome (side) should have an equally
likely chance of occurring
3. Intuitively, P(H) = 1/2 (the expected relative frequency)
Notes:
 This does not mean exactly one head will occur in every two
tosses of the coin
 In the long run, the proportion of times that a head will occur is
approximately 1/2
8
Long-Run Behavior
To illustrate the long-run behavior:
1. Consider an experiment in which we toss the coin several
times and record the number of heads
2. A trial is a set of 10 tosses
3. Graph the relative frequency and cumulative relative
frequency of occurrence of a head
4. A cumulative graph demonstrates the idea of long-run
behavior
5. This cumulative graph suggests a stabilizing, or settling
down, effect on the observed cumulative probability
6. This stabilizing effect, or long-term average value, is
often referred to as the law of large numbers
9
Experiment
• Experimental results of tossing a coin 10 times each trial:
Trial
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Number of
Relative
Cumulative
Heads Observed Frequency Relative Frequency
5
5/10
5/10 = 0.5000
4
4/10
9/20 = 0.4500
4
4/10
13/30 = 0.4333
5
5/10
18/40 = 0.4500
6
6/10
24/50 = 0.4800
7
7/10
28/60 = 0.4667
6
6/10
34/70 = 0.4857
4
4/10
38/80 = 0.4750
7
7/10
45/90 = 0.5000
3
3/10
48/100 = 0.4800
4
4/10
52/110 = 0.4727
6
6/10
58/120 = 0.4838
7
7/10
65/130 = 0.5000
4
4/10
69/140 = 0.4929
3
3/10
72/150 = 0.4800
7
7/10
79/160 = 0.4938
6
6/10
85/170 = 0.5000
3
3/10
88/180 = 0.4889
6
6/10
94/190 = 0.4947
4
4/10
98/200 = 0.4900
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Relative Frequency
0.8
0.7
0.6
0.5
0.4
0.3
0.2
Expected value = 1/2
0.1
0
0
5
10
15
20
25
Trial
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Cumulative Relative Frequency
0.6
Expected value = 1/2
0.55
0.5
0.45
0.4
0
5
10
15
20
25
Trial
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Law of Large Numbers
Law of Large Numbers: If the number of times an experiment
is repeated is increased, the ratio of the number of successful
occurrences to the number of trials will tend to approach the
theoretical probability of the outcome for an individual trial
– Interpretation: The law of large numbers says: the larger
the number of experimental trials n, the closer the empirical
probability P(A) is expected to be to the true probability
P(A)
– In symbols: As n  ,
P' (A)  P(A)
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4.3 ~ Simple Sample Spaces
• We need to talk about data collection and
experimentation more precisely
• With many activities, like tossing a coin, rolling a die,
selecting a card, there is uncertainty as to what will
happen
• We will study and characterize this uncertainty
14
Experiment & Outcome
Experiment: Any process that yields a result or an
observation
Outcome: A particular result of an experiment
 Example: Suppose we select two students at random and ask
each if they have a car on campus:
1. A list of possible outcomes: (Y, Y), (Y, N), (N, Y), (N, N)
2. This is called ordered pair notation
3. The outcomes may be displayed using a tree diagram
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Tree Diagram
Student 1
Student 2
Outcomes
Y
Y, Y
N
Y, N
Y
N, Y
N
N, N
Y
N
1. This diagram consists of four branches: 2 first generation
branches and 4 second generation branches
2. Each branch shows a possible outcome
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Sample Space & Event
Sample Space: The set of all possible outcomes of an experiment.
The sample space is typically called S and may take any number of
forms: a list, a tree diagram, a lattice grid system, etc. The
individual outcomes in a sample space are called sample points.
n(S) is the number of sample points in the sample space.
Event: any subset of the sample space. If A is an event, then n(A) is
the number of sample points that belong to A
 Example: For the student car example above:
S = { (Y, Y), (Y, N), (N, Y), (N, N) }
n(S) = 4
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Examples
 Example: An experiment consists of two trials. The first is
tossing a penny and observing a head or a tail;
the second is rolling a die and observing a 1, 2,
3, 4, 5, or 6. Construct the sample space:
S = { H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6 }
 Example: Three voters are randomly selected and asked if
they favor an increase in property taxes for road
construction in the county. Construct the
sample space:
S = { NNN, NNY, NYN, NYY, YNN, YNY, YYN, YYY}
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Example
 Example: An experiment consists of selecting electronic parts from
an assembly line and testing each to see if it passes
inspection (P) or fails (F). The experiment terminates as
soon as one acceptable part is found or after three parts are
tested. Construct the sample space:
F
Outcome
FFF
F
F
P
P
P
FFP
FP
P
S = { FFF, FFP, FP, P }
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Example
 Example: The 1200 students at a local university have
been cross tabulated according to resident
and college status:
Arts and Sciences
Resident
600
Nonresident
175
Business
280
145
– The experiment consists of selecting one student at
random from the entire student body
– n(S) = 1200
20
Example
 Example: On the way to work, some employees at a certain
company stop for a bagel and/or a cup of coffee.
The accompanying Venn diagram summarizes the
behavior of the employees for a randomly selected
work day:
Coffee
Bagel
32
18
16
11
– The experiment consists of selecting one employee at random
– n(S) = 77
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Notes
1. The outcomes in a sample space can never overlap
2. All possible outcomes must be represented
3. These two characteristics are called
mutually exclusive and all inclusive
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4.4 ~ Rules of Probability
• Consider the concept of probability and relate it to
the sample space
Recall: the probability of an event is the relative
frequency with which the event could be expected to
occur, the long-term average
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Equally Likely Events
1. In a sample space, suppose all sample points are equally likely to
occur
2. The probability of an event A is the ratio of the number of
sample points in A to the number of sample points in S
n( A )
3. In symbols: P(A ) 
n( S )
4. This formula gives a theoretical probability value of event
A’s occurrence
5. The use of this formula requires the existence of a sample
space in which each outcome is equally likely
24
Example
 Example: A fair coin is tossed 5 times, and a head (H) or a tail
(T) is recorded each time. What is the probability of:
A = {exactly one head in 5 tosses}, and
B = {exactly 5 heads}?
•
•
•
•
The outcomes consist of a sequence of 5 Hs and Ts
A typical outcome includes a mixture of Hs and Ts, like: HHTTH
There are 32 possible outcomes, all equally likely
A = {HTTTT, THTTT, TTHTT, TTTHT, TTTTH}
n( A ) 5
P(A ) 

n( S ) 32
n( B) 1
P( B) 

• B = {HHHHH}
n( S ) 32
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Subjective Probability
1. Suppose the sample space elements are not equally
likely, and empirical probabilities cannot be used
2. Only method available for assigning probabilities may
be personal judgment
3. These probability assignments are called subjective
probabilities
4. Personal judgment of the probability is expressed by
comparing the likelihood among the various outcomes
26
Basic Probability Ideas
1. Probability represents a relative frequency
2. P(A) is the ratio of the number of times an event can be
expected to occur divided by the number of trials
3. The numerator of the probability ratio must be a
positive number or zero
4. The denominator of the probability ratio must be a
positive number (greater than zero)
5. The number of times an event can be expected to occur
in n trials is always less than or equal to the total
number of trials, n
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Properties
1. The probability of any event A is between 0 and 1:
0  P(A)  1
2. The sum of the probabilities of all outcomes in the
sample space is 1:
 P(A)  1
all outcomes
Notes:
 The probability is zero if the event cannot occur
 The probability is one if the event occurs every time
(a sure thing)
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Example
 Example: On the way to work Bob’s personal judgment is that he
is four times more likely to get caught in a traffic jam
(TJ) than have an easy commute (EC). What values
should be assigned to P(TJ) and P(EC)?
P(TJ )  4  P( EC)
P(TJ )  P( EC)  1
4  P( EC)  P( EC)  1
5  P( EC)  1
P( EC) 
1
5
1 4

P(TJ )  4  P( EC)  4  
 5 5
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Odds
Odds: another way of expressing probabilities
• If the odds in favor of an event A are a to b, then:
1. The odds against A are b to a
a
2. The probability of event A is: P(A ) 
a b
3. The probability that event A will not occur is
b
P (A does not occur ) 
a b
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Example
 Example: The odds in favor of you passing an introductory
statistics class are 11 to 3. Find the probability you
will pass and the probability you will fail.
• Using the preceding notation: a = 11 and b = 3:
P( pass) 
11
11

11  3 14
P( fail) 
3
3

11  3 14
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Complement of An Event
Complement of an Event: The set of all sample points in the
sample space that do not belong to event A. The complement of
event A is denoted by A (read “A complement”).
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Example
 Example:
1. The complement of the event “success” is “failure”
2. The complement of the event “rain” is “no rain”
3. The complement of the event “at least 3 patients recover” out
of 5 patients is “2 or fewer recover”
Notes:
 P ( A )  P ( A )  1 for any event A
 P(A )  1  P(A )
 Every event A has a complementary event A
 Complementary probabilities are very useful when the question
asks for the probability of “at least one.”
33
Example
 Example: A fair coin is tossed 5 times, and a head(H) or a tail
(T) is recorded each time. What is the probability of
1) A = {at least one head in 5 tosses}
2) B = {at most 3 heads in 5 tosses}
Solutions:
1) P(A )  1  P(A )
 1  P(0 heads in 5 tosses)
1
31
 1

32 32
2) P( B)  1  P( B)
 1  P (4 or 5 heads)
 1  ( P (4 heads)  P (5 heads))
5
1
6
26 13
 1  
   1 


 32 32 
32 32 16
34
Example
 Example: A local automobile dealer classifies purchases
by number of doors and transmission type. The
table below gives the number of each
classification.
Manual
Automatic
Transmission Transmission
2-door
75
155
4-door
85
170
If one customer is selected at random, find the probability that:
1) The selected individual purchased a car with automatic
transmission
2) The selected individual purchased a 2-door car
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Solutions
1) P(Automatic Transmission)
155  170
325 65



75  85  155  170 485 97
2) P(2 - door )
75  155
230 46



75  85  155  170 485 97
36
4.5 ~ Mutually Exclusive Events & the Addition Rule
Compound Events: formed by combining several simple
events:
• The probability that either event A or event B will
occur: P(A or B)
• The probability that both events A and B will
occur: P(A and B)
• The probability that event A will occur given that event
B has occurred: P(A | B)
37
Mutually Exclusive Events
Mutually Exclusive Events: Events defined in such a way
that the occurrence of one event precludes the occurrence
of any of the other events. (In short, if one of them
happens, the others cannot happen.)
Notes:

Complementary events are also mutually exclusive

Mutually exclusive events are not necessarily
complementary
38
Example
 Example: The following table summarizes visitors to a
local amusement park:
Male
Female
Total
All-Day
Pass
1200
900
2100
Half-Day
Pass
800
700
1500
Total
2000
1600
3600
One visitor from this group is selected at random:
1) Define the event A as “the visitor purchased an all-day pass”
2) Define the event B as “the visitor selected purchased a half-day
pass”
3) Define the event C as “the visitor selected is female”
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Solutions
1) The events A and B are mutually exclusive
2) The events A and C are not mutually exclusive. The intersection
of A and C can be seen in the table above or in the Venn diagram
below:
3)
A
C
1200
900
700
800
40
General Addition Rule
General Addition Rule: Let A and B be two events defined in a
sample space S:
P(A or B)  P(A)  P( B)  P(A and B)
• Illustration:
A
B
Note: If two events A and B are mutually exclusive: P(A and B)  0
41
Special Addition Rule
Special Addition Rule: Let A and B be two events defined in a
sample space. If A and B are mutually exclusive events, then:
P(A or B)  P(A)  P( B)
• This can be expanded to consider more than two mutually
exclusive events:
P(A or B or C…) = P(A) + P(B) + … + P(E)
B
A
D
C
42
Example
 Example: All employees at a certain company are
classified as only one of the following:
manager (A), service (B), sales (C), or staff (D).
It is known that P(A) = 0.15, P(B) = 0.40,
P(C) = 0.25, and P(D) = 0.20
P ( A )  1  P ( A )  1  0.15  0.85
P( A and B )  0 (A and B are mutually exclusive )
P ( B or C )  P ( B)  P ( C )  0.40  0.25  0.65
P ( A or B or C )  P ( A )  P ( B)  P ( C )  0.15 + 0.40 + 0.25 = 0.80
43
Example
 Example: A consumer is selected at random. The
probability the consumer has tried a snack food
(F) is 0.5, tried a new soft drink (D) is 0.6, and
tried both the snack food and the soft drink is 0.2
P(Tried the snack food or the soft drink )
 P( F or D)  P( F)  P( D)  P( F and D)  0.5 + 0.6  0.2 = 0.9
P ( Not tried the snack food )  P ( F)  1  P ( F)  1  0.5 = 0.5
P(Tried neither the snack food nor the soft drink )
 P[( F or D)]  1  P( F or D)  1  0.9 = 0.1
P ( Tried only the soft drink )  P ( D )  P ( F and D )  0.6  0.2 = 0.4
44
4.6 ~ Independence, the Multiplication
Rule, & Conditional Probability
Independent Events: Two events A and B are
independent events if the occurrence (or nonoccurrence) of
one does not affect the probability assigned to the
occurrence of the other.
Note: If two events are not independent, they are dependent
45
Conditional Probability
1. Sometimes two events are related in such a way that the probability of
one depends upon whether the second event has occurred
2. Partial information may be relevant to the probability assignment
Conditional Probability: The symbol P(A | B) represents the probability
that A will occur given B has occurred. This is called conditional
probability.
P(A and B)
1. P(A| B) 
P( B)
2. Given B has occurred, the relevant sample space is no
longer S, but B (reduced sample space)
3. A has occurred if and only if the event A and B has occurred
46
Independent Events
Independent Events: Two events A and B are independent events if:
P(A | B) = P(A) or
P(B | A) = P(B)
Notes:

If A and B are independent, the occurrence of B does not affect
the occurrence of A

If A and B are independent, then so are:
A and B
A and B
A and B
47
Example
 Example: Consider the experiment in which a single fair die is
rolled: S = {1, 2, 3, 4, 5, 6 }. Define the following
events:
A = “a 1 occurs,” B = “an odd number occurs,”
and C = “an even number occurs”
P(A| B) 
P(A and B) 1 / 6 1


P( B)
3/ 6 3
P(A|C) 
P(A and C)
0

0
P(C)
3/ 6
P( B and A) 1 / 6
P( B|A) 

1
P(A )
1/ 6
48
Example
 Example: In a sample of 1200 residents, each person was asked if he
or she favored building a new town playground. The
responses are summarized in the table below:
Age
Less than 30 (Y)
30 to 50 (M)
More than 50 (O)
Total
Favor (F)
250
600
100
950
Oppose
50
75
125
250
Total
300
675
225
1200
If one resident is selected at random, what is the probability the resident will:
1) Favor the new playground?
2) Favor the playground if the person selected is less than 30?
3) Favor the playground if the person selected is more than 50?
4) Are the events F and M independent?
49
Solutions
950 19

1) P( F) 
1200 24
P( F and Y) 250 / 1200 5
2) P( F|Y) 


P(Y)
300 / 1200 6
3) P( F|O)  P( F and O)  100 / 1200  4
P(O)
225 / 1200
9
P( F and M) 600 / 1200 8
4) P( F| M) 

  P( F)
P( M)
675 / 1200 9
F and M are dependent
50
General Multiplication Rule
General Multiplication Rule: Let A and B be two events defined in
sample space S. Then:
P(A and B)  P(A)  P( B|A)
or
P(A and B)  P( B)  P(A| B)
Notes: How to recognize situations that result in the compound
event “and”:
 A followed by B
 A and B occurred simultaneously
 The intersection of A and B
 Both A and B
 A but not B (equivalent to A and not B)
51
Special Multiplication Rule
Special Multiplication Rule: Let A and B be two events defined in
sample space S. If A and B are independent events, then:
P(A and B)  P(A)  P( B)
This formula can be expanded. If A, B, C, …, G are independent events,
then P(A and B and C and ... and G)  P(A)  P( B)  P(C) P(G)
 Example: Suppose the event A is “Allen gets a cold this winter,” B is
“Bob gets a cold this winter,” and C is “Chris gets a cold this
winter.” P(A) = 0.15, P(B) = 0.25, P(C) = 0.3, and all three
events are independent. Find the probability that:
1. All three get colds this winter
2. Allen and Bob get a cold but Chris does not
3. None of the three gets a cold this winter
52
Solutions
1) P(All three get colds this winter )
 P(A and B and C)  P(A )  P( B)  P(C)
= (0.15)(0.25)(0.30) = 0.0113
2) P ( Allen and Bob get a cold, but Chris does not)
 P ( A and B and C )  P ( A )  P ( B)  P ( C )
= (0.15)(0.25)(0.70) = 0.0263
3) P ( None of the three gets a cold this winter)
 P ( A and B and C )  P ( A )  P ( B)  P ( C)
= (0.85)(0.75)(0.70) = 0.4463
53
Notes
1. Independence and mutually exclusive are two very different concepts
a. Mutually exclusive says the two events cannot occur
together, that is, they have no intersection
b. Independence says each event does not affect the other
event’s probability
2. P(A and B) = P(A) P(B) when A and B are independent
a. Since P(A) and P(B) are not zero, P(A and B) is nonzero
b. Thus, independent events have an intersection
3. Events cannot be both mutually exclusive and independent
a. If two events are independent, then they are not mutually exclusive
b. If two events are mutually exclusive, then they are not independent
54
4.7 ~ Combining the Rules of Probability
• Many probability problems can be represented by tree
diagrams
• Using the tree diagram, the addition and multiplication
rules are easy to apply
55
Example
 Example: A certain company uses three overnight delivery
services: A, B, and C. The probability of selecting
service A is 1/2, of selecting B is 3/10, and of selecting
C is 1/5. Suppose the event T is “on time delivery.”
P(T|A) = 9/10, P(T|B) = 7/10, and P(T|C) = 4/5. A
service is randomly selected to deliver a package
overnight. Construct a tree diagram representing this
experiment.
Notes:
 A set of branches that initiate from a single point has a total
probability of 1

Each outcome for the experiment is represented by a branch that
begins at the common starting point and ends at the terminal
points at the right
56
Solution
• The resulting tree diagram:
9 / 10 T
A
1/ 2
3 / 10
1 / 10 T
7 / 10 T
B
3 / 10 T
1/ 5
4/5 T
C
1/ 5
T
57
Using the Tree Diagram
1. The probability of selecting service A and having the package
delivered on time:
1 9
9
P(A and T)  P(A )  P(T| A )   
2 10 20
2. The probability of having the package delivered on time:
P (T)  P (A and T)  P ( B and T)  P (C and T)
 P (A )  P (T| A )  P ( B)  P (T| B)  P (C)  P (T| C)
1 9
3 7
1 4
              
 2   10  10  10  5  5
9 21 4


20 100 25
41

50

58
Example
 Example: A manufacturer is testing the production of a new
product on two assembly lines. A random sample of
parts is selected and each part is inspected for defects.
The results are summarized in the table below:
Line 1 (1)
Line 2 (2)
Total
Good (G)
70
80
150
Defective (D)
40
25
65
Total
110
105
215
Suppose a part is selected at random:
1) Find the probability the part is defective
2) Find the probability the part is produced on Line 1
3) Find the probability the part is good or produced on Line 2
59
Solutions
65
1) P( D) 
215
(total defective divided by total number of parts)
110
2) P(1) 
215
(total produced by Line 1 divided by total number of parts)
3) P(G or 2)  n(G or 2)  175
n( S )
215
(total good or produced on Line 2 divided by total parts)
175
 P(G )  P(2)  P (G and 2)  150  105  80 
215 215 215 215
60
Example
 Example: This problem involves testing individuals for the
presence of a disease. Suppose the probability of
having the disease (D) is 0.001. If a person has the
disease, the probability of a positive test result (Pos) is
0.90. If a person does not have the disease, the
probability of a negative test result (Neg) is 0.95. For a
person selected at random:
1) Find the probability of a negative test result given the person has
the disease
2) Find the probability of having the disease and a positive test result
3) Find the probability of a positive test result
61
Resulting Tree Diagram
Test
Result
Disease
0.001
0.999
0.90
Pos
0.10
Neg
0.05
Pos
0.95
Neg
D
D
62
Solutions
1) P ( Neg| D )  1  P ( Pos| D )  1 0.90 = 0.10
2) P ( D and Pos )  P ( D )  P ( Pos| D )  (0.001)(0.90) = 0.0009
3) P( Pos)  P ( D and Pos)  P( D and Pos)
 P ( D)  P( Pos|D)  P( D)  P ( Pos| D)
 (0.001)(0.90) +(0.0009)(0.05) = 0.5085
63