Transcript Chapter 1: Statistics

Chapter 4: Probability
Limiting Relative Frequency
0.6
Relative Frequency
0.5
0.4
0.3
0.2
0.1
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200
400
600
800 Trials 1000
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.
4.1: The Nature of Probability
Example: Consider an experiment in which we roll two sixsided fair dice and record the number of threes face up. The
only possible outcomes are 0 three’s, 1 three, or 2 three’s.
Here are the results after 100 and after 1000 rolls.
100 Rolls
Outcome Frequency
0
80
1
19
2
1
1000 Rolls
Outcome Frequency
0
690
1
282
2
28
We can express these results (from the 1000 rolls) in terms of
relative frequencies and display the results using a histogram.
0
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7
0
.
6
0
.
5
0
.
4
RelativFrquncy
0
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3
0
.
2
0
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1
0
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0
0
1
2
T
h
r
e
e
'
s
F
a
c
e
U
p
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.
1. Use random number tables.
2. Use a computer to randomly generate number values
representing the various experimental outcomes.
3. Key to either method is to maintain the probabilities.
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
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?
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).
Note:
(a) This does not mean exactly one head will occur in every
two tosses of the coin.
(b) In the long run, the proportion of times that a head will
occur is approximately 1/2.
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.
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
Relative Frequency:
0.7
0.6
0.5
0.4
0.3
Expected value = 1/2
0.2
0.1
0
0
5
10
15
Trial
20
Cumulative Relative Frequency:
0.6
Expected value = 1/2
0.55
0.5
0.45
0.4
0
5
10
15
20
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)
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.
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.
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.
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
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}
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.
Outcome
F
FFF
F
F
P
FFP
P
FP
P
P
S = { FFF, FFP, FP, P }
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
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
Note:
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.
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.
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.
3. In symbols:
n( A )
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.
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 H’s and T’s
A typical outcome: HHTTH
There are 32 possible outcomes, all equally likely.
A = {HTTTT, THTTT, TTHTT, TTTHT, TTTTH}
n( A ) 5
P(A ) 

n( S ) 32
B = {HHHHH}
P( B) 
n( B) 1

n( S ) 32
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.
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.
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
Note:
1. The probability is zero if the event cannot occur.
2. The probability is one if the event occurs every time (a
sure thing).
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
1
P( EC) 
5
1 4

P(TJ )  4  P( EC)  4  
 5 5
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.
2. The probability of event A is
a
P(A ) 
a b
3. The probability that event A will not occur is
b
P (A does not occur ) 
a b
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
3
3
P( fail) 

11  3 14
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”).
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.”
Note:
1. P(A )  P(A )  1
2. P ( A )  1  P ( A )
for any event A
3. Every event A has a complementary event A
4. Complementary probabilities are very useful when the
question asks for the probability of “at least one.”
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 = {at least one head in 5 tosses},
B = {at most 3 heads in 5 tosses}?
P(A )  1  P(A )
 1  P(0 heads in 5 tosses)
1 31
 1 
32 32
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
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.
Solution:
P(Automatic Transmission)
155  170
325 65



75  85  155  170 485 97
P(2 - door )

75  155
230 46


75  85  155  170 485 97
4.5: Mutually Exclusive Events
and 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)
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.)
Note:
1. Complementary events are also mutually exclusive.
2. Mutually exclusive events are not necessarily
complementary.
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.”
Solution:
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.
A
C
1200
900
700
800
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
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 or ... or E)  P(A)  P( B)  P( E)
B
A
D
C
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) = .15, P(B) = .40,
P(C) = .25, and P(D) = .20
P ( A )  1  P ( A )  1.15 .85
P(A and B)  0 (A and B are mutually exclusive.)
P( B or C)  P( B)  P(C) .40.25 .65
P(A or B or C)  P(A)  P( B)  P(C) .15.40.25 .80
Example: A consumer is selected at random. The probability
the consumer has tried a snack food (F) is .5, tried a new soft
drink (D) is .6, and tried both the snack food and the soft
drink is .2.
P(Tried the snack food or the soft drink )
 P( F or D)  P( F)  P( D)  P( F and D) .5.6.2 .9
P ( Not tried the snack food )  P ( F)  1  P ( F)  1.5 .5
P(Tried neither the snack food nor the soft drink )
 P[( F or D)']  1  P( F or D)  1.9 .1
P(Tried only the soft drink)  P( D)  P( F and D) .6.2 .4
4.6: Independence, the
Multiplication Rule, and
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.
Background:
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.
Independent Events: Two events A and B are independent
events if
P(A | B) = P(A) or P(B | A) = P(B)
Note:
1. If A and B are independent, the occurrence of B does not
affect the occurrence of A.
2. If A and B are independent, then so are:
A and B
A and B
A and B
3. Independence cannot be shown on a Venn diagram.
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
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?
Solution:
950 19
P( F) 

1200 24
P( F and Y) 250 / 1200 5
P( F|Y) 


P(Y)
300 / 1200 6
P( F and O) 100 / 1200 4
P( F|O) 


P(O)
225 / 1200 9
P( F and M) 600 / 1200 8

  P( F)
P( M)
675 / 1200 9
F and M are dependent
P( F| M) 
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)
Note: How to recognize situations that result in the compound
event “and.”
1. A followed by B.
2. A and B occurred simultaneously.
3. The intersection of A and B.
4. Both A and B.
5. A but not B (equivalent to A and not B).
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) = .15, P(B) = .25, P(C) = .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.
Solution:
P (All three get colds this winter )
 P (A and B and C)  P (A )  P ( B)  P (C)
 (.15)(.25)(.30) .0113
P(Allen and Bob get a cold, but Chris does not )
 P (A and B and C)  P(A )  P( B)  P(C)
 (.15)(.25)(.70) .0263
P( None of the three gets a cold this winter )
 P(A and B and C)  P(A )  P( B)  P(C)
 (.85)(.75)(.70) .4463
Note:
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.
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.
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.
Note:
1. A set of branches that initiate from a single point has a
total probability 1.
2. 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.
The resulting tree diagram
Service
A
Delivery
9 / 10
T
1 / 10
1/ 2
T
7 / 10
3 / 10
B
T
3 / 10
T
4/5
1/ 5
C
T
1/ 5
T
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
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.
Solution:
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)
n(G or 2) 175

3. P(G or 2) 
n( S )
215
(total good or produced on Line 2 divided by total parts)
150 105 80
 P(G )  P(2)  P(G and 2) 


215 215 215
Example: This problem involves testing individuals for the
presence of a disease. Suppose the probability of having the
disease (D) is .001. If a person has the disease, the
probability of a positive test result (Pos) is .90. If a person
does not have the disease, the probability of a negative test
result (Neg) is .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.
Resulting tree diagram:
Test
Result
Disease
.001
.999
.90
Pos
.10
Neg
.05
Pos
.95
Neg
D
D
Solution:
1. P( Neg|D)  1  P( Pos|D)  1.90 .10
2. P( D and Pos)  P( D)  P( Pos|D)  (.001)(.90) .0009
3. P( Pos)  P( D and Pos)  P( D and Pos)
 P( D)  P( Pos|D)  P( D)  P( Pos| D)
 (.001)(.90)  (.999)(.05) .05085
Bayes’s Rule:
P(A k )  P(B | A k )
P(A k | B) 
 P ( A i )  P ( B | A i ) 
where A1, A2, …, An is an all-inclusive set of possible
outcomes given B.
Note: This rule allows us to revise (or adjust) the probabilities
assigned to events in accordance with new information.
Previous Example Continued: If a person has a positive test
result, find the probability they actually have the disease.
P(D and Pos)
P(D | Pos) 
P (Pos)
P (D)  P (Pos | D)

P (D)  P(Pos | D)  P(D)  P (Pos | D)
(.001)(.90)
.0009


 .0177
(.001)(.90)  (.999)(.05) .05085
Note: In this example, most positive test results are “false
positives.”