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Chapter 6
Probability
The Study of Randomness
Lesson 6-1
The Idea of Probability
Randomness

Behavior is random if, while individual
outcomes are uncertain, for a large number
of repetitions, outcomes are regularly
distributed.

Example: If I roll a die once, I can’t predict with
certainty what number it will land on, but if I roll
sixty times, I can expect it to land on 1 ten
times, 2 ten times, 3 ten times, etc.
Probability Models

The probability of an outcome is the
proportion of times the outcome would
occur for a large number of repetitions.

Example: The probability of a die landing on 4 is
the proportion of times a die lands on 4 for a
large number of repetitions.
Chance Behavior is unpredictable in the
short run but has a regular and
predictable pattern in the long run.
Tossing a Coin
Example – Page 335, #6.10
A recent opinion poll showed that about 73% of married
women agree that their husbands do at least their fair share
of household chores. Suppose that this is exactly true.
Choosing a married woman at random that has probability
0.73 of getting one who agrees that her husband does his
share.
A). Simulate drawing 20 women, then 80 women, then
320. What proportion agree in each case? We expect
(but because of chance variation we can’t be sure) that
proportion will be closer to 0.73 in longer runs of trials.
Example – Page 335, #6.10
A). Use Randint and let 0 – Agree and 1 – disagree.
12 women disagreed and
8 women agreed
8
 0.40
20
Example – Page 335, #6.10
A). Use Randint and let 0 – Agree and 1 – disagree.
Sample 80
36 women disagreed and
44 women agreed
44
 0.55
80
Sample 320
167 women disagreed and
153 women agreed
153
 0.49
320
Example – Page 335, #6.10
B). Simulate drawing 20 women 10 times and record the
percents in each trial who agreed. Then simulate
drawing 320 women 10 times and again record the
10 percents. Which set of 10 results is less variable?
We expect the results of 320 trials to be more
predictable (less variable) than the results of 20 trials.
That is “long-run regularity” showing itself.
Example – Page 335, #6.10
20 Women
320 Women
Trail 1
0.50
.51
Trial 2
.70
.54
Trial 3
.54
.55
Trial 4
.55
.53
Trial 5
.45
.54
Trial 6
.55
.52
Trial 7
.45
.48
Trial 8
.50
.50
Trail 9
.35
.46
Trial 10
.65
.45
Example – Page 334, #6.4
Probability is a measure of how likely an event is to occur.
Match one of the probabilities that follow with each
statement about an event. (The probability is usually a
much more exact measure of likelihood than is the verbal
statement.)
Example – Page 334, #6.4
0,
0.01, 0.03,
0.6,
0.99,
1
0
A). This event is impossible. It can never occur.
1
B). This event is certain. It will occur on every trial of
the random phenomenon.
0.01
C). This event is very unlikely, but it will occur once in
a while in a long sequence of trials.
0.6
D). This event will occur more often than not.
Lesson 6-2, Part 1
Probability Models
Addition/Complements
Sample Space (S)

The set of all possible outcomes of an
event is the sample space S of the event.

Example: For the event “roll a die and observe
what number it lands on” the sample space
contains all possible numbers the die could
land on.
S  1, 2,3, 4,5,6
Event

An event is an outcome (or a set of outcomes)
from a sample space



Example 1: When flipping three coins, an event may be
getting the result HTH. In this case, the event is one
outcome from the sample space.
Example 2: When flipping three coins, an event may be
getting two tails. In this case, the event is a set of
outcomes (HTT, TTH, THT) from the sample space
An event is usually denoted by a capital letter.


Example: Call getting two tails event A.
The probability of event A is denoted P(A).
Identifying Outcomes


Tree Diagrams
Multiplication Rule

One task a number of ways and a second
task b number of ways, then both tasks
can be done a x b number of ways.
Tree Diagram
Toss a coin and rolling a dice
Possible outcomes
2  6  12
With or Without Replacement
Assume you are drawing index cards with the digits 1 – 10.
You are picking 3 cards.
With Replacement
10  10  10  1000
Without Replacement
10  9  8  720
different ways to arrange
a 3-digit number
Example – Page 341, #6.14
For each of the following, use a tree diagram or the
multiplication principal to determine the number of
outcomes in the sample space. Then write the sample
space using set notation.
A). Toss 2 coins.
2 2  4
S  HH, HT ,TT ,TH 
Example – Page 341, #6.14
B). Toss 3 coins.
2 2 2  8
S  HHH, HHT , HTT ,TTH, HTH,TTT ,THT ,THH 
Example – Page 341, #6.14
C). Toss 4 coins.
2  2  2  2  16
HHHH, HHHT , HHTH, HTHH,THHH, HHTT ,


S  HTHT , HTTH,THTH,TTHH,THHT , HTTT , 
THTT ,TTHT ,TTTH,TTTT



Example – Page 342, #6.18
Suppose you select a card from a standard deck of 52
playing cards. In how many ways can the selected card be:
A). A red card?
26
C). A queen and a heart?
1
D). A queen or a heart?
B). A heart?
13
16
E). A queen that is not a heart?
3
Probability Rules


The probability of any event is between 0 and 1.
 A probability of 0 indicates the event will never
occur.
 A probability of 1 indicates the event will always
occur
 0 ≤ P(A) ≤ 1
If S is the sample space, then P(S) = 1.
 The sum of the probabilities of all possible
outcomes must be 1
Example – Page 348, #6.22
The New York Times (August 21, 1989) reported a poll that interviewed a
random sample of 1025 women. The married women in the sample were
asked whether their husbands did their fair share of household chores.
Here are the results:
Outcome
Probability
Does more than his fair share
0.12
Does his fair share
0.61
Does less than his fair share
?
These proportions are probabilities for the random phenomenon of
choosing a married women at random and asking her opinion.
Example – Page 348, #6.22
Outcome
Probability
Does more than his fair share
0.12
Does his fair share
0.61
Does less than his fair share
?
A). What must be the probability that women chosen says
that her husband does less than his fair share? Why?
All three probabilities must add up to 1
1  (0.12  0.61)  0.27
Example – Page 348, #6.22
Outcome
Probability
Does more than his fair share
0.12
Does his fair share
0.61
Does less than his fair share
?
B). The event “I think my husband does at least his fair
share” contains the first two outcomes. What is the
probability.
(0.12  0.61)  0.73
Complement

The complement of an event A, denoted
by Ac , is the set of outcomes that are not
in A.


P(Ac) = 1 – P(A)
Example: When flipping two coins, the
probability of getting two heads is 0.25.

The probability of not getting two heads is
1 – 0.25 = 0.75
Complement
S
AC
A
The sample space S
P (S )  1
The set A and its complement
C
P (A)  1  P (A )
Addition Rule



If events A and B are disjoint if they have no
outcomes in common.
Events like this, that can’t occur together, are
called disjoint or mutually exclusive.
For two disjoint events A and B, the probability
that one or the other occurs is the sum of the
probabilities of the two events.

P(A or B) = P ( A  B )  P ( A)  P (B )
Addition Rule

Example: Let event A be rolling a die and
landing on an even number, and event B be
rolling a die and landing on an odd number.


The outcomes for A are {2 , 4, 6} and the
outcomes for B are {1, 3 , 5}. These events are
disjoint because they have no outcomes in
common.
The probability of A or B (landing on either even
or an odd number)

P(A or B) = P(A) + P(B)
Disjoint
Two disjoint sets, A and B
S
B
A
We write P(A or B) as P(A  B).
The symbol  means “union,”
representing the outcomes in
event A or event B (or both).
Lesson 6-2, Part 2
Probability Models
Independence
Independent

Events A and B are independent
because the probability A does not
change the probability of B.

Example: Roll a yellow die and a red die.
Event A is the yellow die landing on an
even number, and event B is the red die
landing on an odd number.
Multiplication Rule

For two independent events A and B, the
probability that both A and B occur is the
product of the probabilities of the two
events.

P(A and B) = P ( A  B )  P ( A )  P (B )
Multiplication Rule

Example: The probability that a yellow die
lands on an even number and the red die
lands on an odd number is:


P(A and B) = P(A) · P(B) = ½ x ½ = ¼
If events A and B are independent,


then their complements, Ac and Bc are also
independent, and
Ac independent of B
Intersection
Two sets A and B are not
disjoint.
B
A and B
A
We write P(A and B) as
P(A  B). The symbol  means
“intersection,” representing
the outcomes that are in both
event A and event B.
Example – Page 349, #6.26
Example 6.10 (page 345) states that the first digits of numbers in
legitimate records often follow a distribution know as Benford’s Law.
Here is the distribution:
1st Digit
1
2
3
Probability
0.301
0.176 0.125
4
5
6
7
8
0.097
0.079
0.067
0.058
0.051 0.046
P(A) = P(first digit is 1) = 0.301
P(B) = P(first digit is 6 or greater) = 0.222
P(C) = P(first digit is odd) = 0.609
P(D) = P(first digit is less than 4) = 0.602
9
Example – Page 349, #6.26
1st Digit
1
2
3
Probability
0.301
0.176 0.125
4
5
6
7
8
9
0.097
0.079
0.067
0.058
0.051 0.046
P(A) = P(first digit is 1) = 0.301
P(B) = P(first digit is 6 or greater) = 0.222
P(C) = P(first digit is odd) = 0.609
P(D) = P(first digit is less than 4) = 0.602
A). P(D) = P(1, 2, or 3) = 0.301 + 0.176 + 0.125 = 0.602
B). P (B  D )  P (B )  P (D )  0.222  0.602  0.824
C). P (D C )  1  P (D )  1  0.602  0.398
Example – Page 349, #6.26
1st Digit
1
2
3
Probability
0.301
0.176 0.125
4
5
6
7
8
9
0.097
0.079
0.067
0.058
0.051 0.046
P(A) = P(first digit is 1) = 0.301
P(B) = P(first digit is 6 or greater) = 0.222
P(C) = P(first digit is odd) = 0.609
P(D) = P(first digit is less than 4) = 0.602
D). P (C  D )  P(1 and 3)  P (1)  P (3)  0.301  0.125  0.426
E). P (B  C )  P(7 or 9)  P (7)  P (9)  0.058  0.046  0.104
Example – Page 354, #28
An automobile manufacturer buys computer chips from
a supplier. The supplier sends a shipment containing
5% defective chips. Each chip chosen from this shipment
has probability 0.05 of being defective, and each
automobile uses 12 chips selected independently. What
is the probability that all 12 chips in a car will work properly?
P(all chips work properly) = 1 – P(defective chips)
 1  .05    0.95   0.5404
12
12
Example – Page 354, #6.30
Choose at random a U.S. resident at least 25 years of age. We
are interested in the events
A = {The person chosen completed 4 years of college}
B = {The person chosen is 55 years old or older}
Example – Page 354, #6.30
Government data recorded in table 4.6 on page 241 allows
us to assign probabilities to these events.
Example – Page 354, #6.30
A). Explain why P(A) = 0.256
P (A) 
44845
 0.256
175230
Example – Page 354, #6.30
B). Find P(B).
56008
P (B ) 
 0.320
175230
Example – Page 354, #6.30
C). Find the probability that the person is at least 55 years old
and had 4 years of college education, P(A and B). Are
events A and B independent?
Example – Page 354, #6.30
A and B are not independent
10596
 0.0604 because P(A and B)  P (A)  P (B )
P(A and B) 
175230
0.0604  0.256(0.320)
0.0604  0.08192
Example – Page 355, #6.32
An athlete suspected of having used steroids is given two tests that
operate independently of each other. Test A has a probability 0.9 of
being positive if steroids have been used. Test B has probability 0.8
of being positive if steroids have been used. What is the probability
that neither test is positive if steroids have used?
C
C
P(neither test is positive)  P  A   P B

 (1  0.9)(1  0.8)
 0.02
Lesson 6-3, Part 1
General Probability Rules
Basic Properties of Probability



0 ≤ P(A) ≤ 1 for any event A
P(S) = 1
Complement Rule


For any event A, P(Ac) = 1 – P(A)
Addition Rule

Events A and B are disjoint


P(A or B) = P(A  B) = P(A) + P(B)
Multiplication Rule

Events A and B are independent

P(A and B) = P(A  B) = P(A) · P(B)
Addition Rule – Not Disjoint


The union of two or more events that at
least one of those occurs.
The addition Rule for the Union to Two Events:



P(A or B) = P(A) + P(B) – P(A and B)
P(A  B) = P(A) +P(B) – P(A  B)
Mutually Exclusive

Two events are mutually exclusive:
 If the P(A and B) =0
Addition Rule – Not Disjoint
P(B)



P(A)
P(A or B) = P(A) +P(B) – P(A and B)
P P(A  B) = P(A) +P(B) – P(A  B)
Two events are mutually exclusive if the
P(A and B) = 0
Example 1
Addition Rule – Not Disjoint
The accompanying table represents the blood groups and
RH types of 100 people.
RH Types of
100 People
RH
A
Positive 35
Negative 5
Total
40
B
8
2
10
Group
AB
4
1
5
O
39
6
45
Total
86
14
100
Based on the table what is the probability a person will
have group B blood or be RH Positive?
Example 1
Addition Rule – Not Disjoint
RH Types of
100 People
RH
A
B
Group
AB
O
Total
Positive 35
Negative 5
8
2
4
1
39
6
86
14
Total
10
5
45
100
40
P(B Blood or RH +) = P(B Blood) + P(RH +) – P(B Blood and RH +)

10
86
8


 0.88
100 100 100
Example 2
Addition Rule – Mutually Exclusive
A study of credit card fraud was conducted by Master Card
International, and the table below is based on the results”
Method of Fraud
Number
Stolen Card
243
Counterfeit Card
85
Mail/Phone Order
52
Other
46
Total
426
Based on the table, what is the
probability that fraud resulted
from a stolen card or a mail/phone
order ?
Example 2
Addition Rule – Mutually Exclusive
Method of Fraud
Number
Stolen Card
243
Counterfeit Card
85
Mail/Phone Order
52
Other
46
Total
426
P(stolen or Mail/Phone)= P(stolen card) + P(Mail/Phone) – P(Stolen and Mail/Phone)
243 52


 0  0.69
426 426
Example – Page 365, #6.46
Call a household prosperous if its income exceeds $100,000. Call the
household educated if the household completed college. Select an
American household at random, and let A be the event that the selected
household is prosperous and B the event that it is educated. According
the Census Bureau, P(A) = 0.134, P(B) = 0.254, and the joint probability
that a household is both prosperous and educated is P(A and B) = 0.080.
What is the probability P(A or B) that the household selected is either
prosperous or educated?
P(A or B) = P(A) + P(B) – P(A and B)
= 0.134 + 0.254 – 0.080
= 0.308
Example – Page 364, #6.48
Consolidated builders has bid on two large construction projects. The
company president believes that the probability of winning the first
contract (event A) is 0.6, that the probability of winning the second (event
B) is 0.5, and that the joint probability of winning both jobs (event {A and
B}) is 0.3. What is the probability of event {A or B} that Consolidated will
win at least one of the jobs?
P(at least one contract)
= P(A or B) = P(A) + P(B) – P(A and B)
= 0.6 + 0.5 – 0.3
= 0.80
Example – Page 365, #6.52
Musical styles other than rock and pop are becoming more
popular. A survey of college students finds that 40% like
country music, 30% like gospel music, and 10% like both.
A) Make a Venn Diagram with these results.
Neither
0.4
Gospel Only
0.2
Both
0.1
Country Only
0.30
Example – Page 365, #6.52
Musical styles other than rock and pop are becoming more
popular. A survey of college students finds that 40% like
country music, 30% like gospel music, and 10% like both.
B) What percent of college students like country but not
gospel?
P(country but not gospel) = P(C) – P(C and G)
 0.40  .10
 0.30
Example – Page 365, #6.52
Musical styles other than rock and pop are becoming more
popular. A survey of college students finds that 40% like
country music, 30% like gospel music, and 10% like both.
C) What percent like neither country nor gospel.
P(neither country or gospel) = 1 – P(C or G)
= 1 – [P(C) + P(G) – P(C and G)]
 1  [.40  .30  .10]
 0.40
Basic Properties of Probability



0 ≤ P(A) ≤ 1 for any event A
P(S) = 1
Complement Rule


For any event A, P(Ac) = 1 – P(A)
Addition Rule

Events A and B are not disjoint


Events A and B are disjoint


P(A or B) = P(A  B) = P(A) + P(B) – P(A and B)
P(A or B) = P(A  B) = P(A) + P(B)
Multiplication Rule

Events A and B are independent

P(A and B) = P(A  B) = P(A) · P(B)
Conditional Probability
Notation


The probability that event B occurs if we know for
certain that event A will occur is called
conditional probability.
The conditional probability of B given A is:


P(B|A) which reads “the probability of event B given
event A has occurred.”
If events A and B are independent, then
knowing that event A will occur does not change
the probability of B.

P(A|B) = P(A) or P(B|A) = P(B)
Conditional Probability
Independent
Example: When flipping a coin twice, what is the probability
of getting heads on the second flip if the first flip was head?
Event A: getting head on first flip
Event B: getting head on second flip
Events A and B are independent since the outcome of the
first flip does not change the probability of the second flip.
P (B | A )  P (B ) 
1
2
Multiplication Rule
Not Independent


The intersection of two or more events is that all
of those events occur.
The probability that two events, A and B both
occur is the probability that event A occurs
multiplied by the probability that event B also
occurs – that is by the probability that event B
occurs given that event A occurs.


P(A and B) = P(A) · P(B|A)
P(A  B) = P(A) · P(B|A)
Conditional Probability

To find the probability of event B given the
event A, we restrict our attention to the
outcomes in A. We then find in what
fraction of those outcomes B also occurred.

P(B|A) =

P(B|A) =
P(A and B)
P(A)
P(A  B)
P(A)
Example
Conditional Probability
RH Types of
100 People
RH
Group
A
B
AB
O
Total
Positive
35
8
4
39
86
Negative
5
2
1
6
14
Total
40
10
5
45
100
What is the probability that a person will be RH positive given
they have blood type O?
39
P(O and RH +)
39
 100 
 0.87
P(RH +|O Blood) =
45 45
P(O blood)
100
Example
Are the Events Disjoint? Independent?
Police report that 78% of drivers are given a breath test,
36% a blood test, and 22% both tests
1. Are giving a DWI suspect a blood test and a breath test
mutually exclusive?
2. Are giving the two tests independent?
State the events we’re interest in.
Let A = {suspect is given a breath test}.
Let B = {suspect is given a blood test}.
Example
Are the Events Disjoint? Independent?
Police report that 78% of drivers are given a breath test,
36% a blood test, and 22% both tests
State the events we’re interest in.
Let A = {suspect is given a breath test}.
Let B = {suspect is given a blood test}.
State the given probabilities.
P(A) = 0.78
P(B) = 0.36
P(A and B) = P(A  B) = 0.22
Example
Are the Events Disjoint? Independent?
Let A = {suspect is given a breath test}.
Let B = {suspect is given a blood test}.
P(A) = 0.78
P(B) = 0.36
P(A and B) = P(A  B) = 0.22
1. Are giving a DWI suspect a blood test and a breath test
mutually exclusive?
Disjoint events cannot both happen at the same time, so
check to see if P(A and B) = 0.
P(A and B) = 0.22, the events cannot be mutually
exclusive.
22% of all suspects get both tests, so a breath test
and blood test are not disjoint events.
Lesson 6-3, Part 2
General Probability Rules
Conditional Probability
Example
Are the Events Disjoint? Independent?
Let A = {suspect is given a breath test}.
Let B = {suspect is given a blood test}.
P(A) = 0.78
P(B) = 0.36
P(A and B) = P(A  B) = 0.22
2. Are the two tests independent?
Does getting a breath test change the probability of
getting a blood test? That is, does P(B|A) = P(B)?
P ( A  B ) 0.22
P (B | A ) 

 0.28
P (A)
0.78
P (B )  0.36
Because the two probabilities are not the same, the two
events are not independent.
Example
Are the Events Disjoint? Independent?
Overall, 36% of the drivers get blood tests, but only 22%
of those get a breath test too. Since suspects who get
a breath test are less likely to have a blood test, the two
events are not independent.
Example – Page 369, #6.54
Choose an adult American woman at random. Table 6.1 describes the
population from which we draw. Use the information in that table to
answer the following questions.
A). What is the probability that the women chosen is
65 years old or older?
B). What is the conditional probability that the women
chosen is married, given that she is 65 or over?
C). How many women are both married and in the over 65
age group? What is the probability that the women we
choose is a married women at least 65 years old?
Example – Page 369, #6.54
18669
 0.18
A). P( 65 years old or older) =
103870
Example – Page 369, #6.54
B). P(M|65) =
8270
P(65 and M)
8270
103870


 0.443
18669
18669
P(65)
103870
Example – Page 369, #6.54
8270
 0.08
C). P(M and 65) =
103870
Example – Page 370, #6.56
Choose an employed person at random. Let A be the event that the
person chosen is a women and B the event that the person holds a
managerial or professional job. Government data tell us that P(A) = 0.46
and the probability of managerial and professional jobs among women
is P(B|A) = 0.32. Find the probability that a randomly chosen employed
person is a women holding a managerial or professional position.
P(Women and Managerial/Professional Position)
P(A and B) = P(A) · P(B|A) = 0.46  0.32  0.1472
Example – Page 370, #6.58
A poker player holds a flush when all 5 cards in the hand belong
to the same suit. We will find the probability of a flush when 5
cards are dealt. Remember that a deck contains 52 cards, 13 of
each suit, and that when the deck is well shuffled, each card dealt
is equally likely to be any of those that remain in the deck.
Example – Page 370, #6.58
A). We will concentrate on spades. What is the probability that the first
card dealt is a spade? What is the conditional probability that the
second card is a spade, given that the first is a spade?
P(1st
P(2nd
13
 0.25
card Spade) =
52
card
spade|1st
12
 0.2353
card spade) =
51
Example – Page 370, #6.58
B). Continue to count the remaining cards to find the conditional
probability of a spade on the third, fourth, and fifth card, given
that all previous cards are spades.
11
 0.22
card
card spade) =
50
10
th
rd
 0.2041
P(4 card spade|3 card spade) =
49
9
th
th
 0.1875
P(5 card spade|4 card spade) =
48
P(3rd
spade|2nd
Example – Page 370, #6.58
C). The probability of being dealt 5 spades is the product of the five
probabilities you have found? Why? What is this probability?
P(5 spades) = 0.25  0.2353  0.22  0.2041  0.1875  0.0004952
The product of these conditional probabilities gives
the probability of a flush in spades by the extended
multiplication rule: We must draw a spade then another,
and then a third, fourth, and fifth.
Example – Page 370, #6.58
D). The probability of being dealt 5 hearts, or 5 diamonds or 5 clubs is
the same as the probability of being dealt 5 spades. What is
the probability of being dealt a flush?
P(flush) = (0.0004952)4  0.001981
Example – Page 378, #6.64
Enzyme immunoassay (EIA) tests are used to screen blood specimens for
the presence of antibodies to HIV, the virus that causes AIDS. Antibodies
indicate the presence of the virus. The test is quite accurate but is not
always correct. Here are approximate probabilities of positive and
negative EIA outcomes when the blood tested does and does not
actually contain antibodies to HIV.
Test result
Positive
Negative
Antibodies Present
0.9985
0.0015
Antibodies Absent
0.006
0.994
Suppose that 1% of a large population carries antibodies to HIV in
their blood.
Example – Page 378, #6.64
Test result
Positive
Negative
Antibodies Present
0.9985
0.0015
Antibodies Absent
0.006
0.994
A). Draw a tree diagram for selecting a person from this
population (outcomes: antibodies present or absent)
and for testing his or her blood (outcomes: EIA positive
or negative).
Example – Page 378, #6.64
Test result
Positive
Negative
Antibodies Present
0.9985
0.0015
Antibodies Absent
0.006
0.994
0.01
Subject
0.99
Antibody
Present
Antibody
Absent
0.9985
0.0015
0.006
0.994
EIA +
EIA –
EIA +
EIA –
Example – Page 378, #6.64
B). What is the probability that the EIA is positive for an randomly
chosen person from this population?
P(Test +) = P(antibody and Test +) + P(no antibody and Test +) =
(0.01)(0.9985)  (0.99)(0.006)  0.016
0.01
Subject
0.99
Antibody
Present
Antibody
Absent
0.9985
0.0015
0.006
0.994
EIA +
EIA –
EIA +
EIA –
Example – Page 378, #6.64
C). What is the probability that a person has the antibody given that the
EIA test is positive?
P(Antibody|Test +) =
0.01
Subject
0.99
P(Test + and Antibody)  (0.9985)(0.01)  0.624
0.016
P(Test +)
Antibody
Present
Antibody
Absent
0.9985
0.0015
0.006
0.994
EIA +
EIA –
EIA +
EIA –