#### Transcript Chapter 6 - Algebra I PAP

```Chapter 6
Probability
Section 6.1
Chance Experiments and Events
Chance Experiment
A chance experiment is any activity or
situation in which there is uncertainty
about which of two or more possible
outcomes will result.
Examples of Chance Experiments
Suppose two six-sided dice are rolled and
they both land on sixes.
Or a coin is flipped and it lands on heads.
Or record the color of the next 20 cars to pass
an intersection.
Sample Space
The collection of all possible outcomes
of a chance experiment is the sample
space for the experiment.
Example of Sample Space
An experiment is to be performed to study
student preferences in the food line in the
cafeteria. Specifically, the staff wants to
analyze the effect of the student’s gender
on the preferred food line (burger, salad or
main entrée).
Sample Space Example - continued
The sample space consists of the following six
possible outcomes.
1. A male choosing the burger line.
2. A female choosing the burger line.
3. A male choosing the salad line.
4. A female choosing the salad line.
5. A male choosing the main entrée line.
6. A female choosing the main entrée line.
Example - continued
The sample space could be represented by using set
notation and ordered pairs.
sample space = {(male, burger), (female, burger),
(female, main entree)}
If we use M to stand for male, F for female, B for
burger, S for salad and E for main entrée the notation
could be simplified to
sample space = {MB, FB, MS, FS, ME, FE}
Example - continued
Yet another way of illustrating the sample space
would be using a picture called a “tree” diagram.
Burger
Main Entree
Male
Outcome (Female, Burger)
Burger
Female
Main Entree
This “tree” has two sets of “branches” corresponding to the two bits of
information gathered. To identify any particular outcome of the sample
space, you traverse the tree by first selecting a branch corresponding to
gender and then a branch corresponding to the choice of food line.
Suppose a six-sided die is rolled. The
possible outcomes are that the die could land
with 1 dot up or 2, 3, 4, 5, or 6 dots up.
S = {1, 2, 3, 4, 5, 6}
The sum of the
probabilities
thethe
“S” stands for sample space. We use set
notation tooflist
This would be
an example
a outcomes
sample
space.
the sample
outcomes
of the of
sample
space. in
space equals ONE.
Suppose two coins are flipped. The sample
space would be:
S = {HH, HT, TH, TT}
Where H = heads and T = tails
We can also use a tree diagram to represent a sample
space.
H
H
T
H
T
T
HT
branches out to
show an outcome.
Events
An event is any collection of outcomes from the
sample space of a chance experiment.
A simple event is an event consisting of exactly one
outcome.
If we look at the lunch line example and use the following sample
space description {MB, FB, MS, FS, ME, FE}
The event that the student selected is male is given by
male = {MB, MS, ME}
The event that the preferred food line is the burger line is given by
burger = {MB, FB}
The event that the person selected is a female that prefers the
salad line is {FS}. This is an example of a simple event.
Suppose a six-sided die is rolled. The
outcome that the die would land on an even
number would be
E = {2, 4, 6}
This would be an example of an event.
We typically use capital letters to denote an event.
Venn Diagrams
A Venn Diagram is an informal picture that is
used to identify relationships.
The collection of all possible outcomes of a
chance experiment are represented as the
interior of a rectangle.
The rectangle represents the
sample space and the
event A.
Forming New Events
Let A and B denote two events.
The event not A consists of all experimental outcomes
that are not in event A. Not A is sometimes called the
complement of A and is usually denoted by AC or A’.
represents the event
not A.
Complementary Event Example
Suppose a six-sided die is rolled. The event
that the die would land on an even number
would be
E = {2, 4, 6}
The superscript “C”
What
would the event
stands for complement
be of the
landing on an even number?
EC = {1, 3, 5}
The sum of the
probabilities of
complementary
die
NOT
events equals
ONE.
These complementary events can be shown
on a Venn Diagram.
C = {1, 3, 5}
E = {2, 4, 6}Let
and
E
the circle represent event E.
Let the rectangle represent the
sample space.
event not E.
The union of A or B - consists of all outcomes that
are in at least one of the two events, that is, in A or
in B or in both.
A or B = A ∪ B
Consider a
The
bride
takes
all the union
marriage
or
union
This
is similar
to
her
stuff
of
two
people
– B.
of&Athe
and
groom
allof B are put
when
people
All of two
Atakes
and
all
hismarry,
stuff &
theydo
put
what
together!
together!
theyit do
with their
possessions ?
This symbol means
And live happily
“union”
ever after!
Suppose a six-sided die is rolled. The event
that the die would land on an even number
would be E = {2, 4, 6}
The event that the die would land on a prime
number would be P = {2, 3, 5}
What would be the event E or P happening?
E or P = {2, 3, 4, 5, 6}
This is an example of the union of two events.
A or B = A ∪ B
Let A and B denote two events.
The shaded area represents the event A U B.
Let’s revisit rolling a die and getting an even or
a prime number . . .
E or P = {2, 3, 4, 5, 6}
E or P would be any number
Another way
to represent
in either
circle. this is with a Venn Diagram.
Even number
Why is the number 1
outside the circles?
Prime number
3
4
2
6
1
5
A and B = A ∩ B
Let A and B denote two events.
This symbol means
“intersection”
The intersection of A and B - consists of all outcomes
that are in both of the events
Suppose a six-sided die is rolled. The event that the
die would land on an even number would be
E = {2, 4, 6}
The event that the die would land on a prime number
would be
P = {2, 3, 5}
What would be the event E and P happening?
E and P = {2}
This is an example of the intersection of two
events.
Let’s revisit rolling a die and getting an even or
a prime number . . .
E and P would
be ONLY
the
E and
P = {2}
middle part that the circles
To represent
with a Venn Diagram:
havethis
in common
3
4
2
6
1
5
Mutually Exclusive/Disjoint
Two events that have no common
outcomes are said to be disjoint or
mutually exclusive.
A and B are mutually
exclusive events
Suppose a six-sided die is rolled. Consider
the following 2 events:
A = {2}
B = {6}
On a single die roll, is it possible for A and
B to happen at the same time?
These events are mutually exclusive.
Mutually exclusive (or disjoint) events -two
events have no outcomes in common; two events
that NEVER happen simultaneously
A Venn Diagram for the roll of a six-sided die
and the following two events:
A = {2} B = {6}
A and B are mutuallyThe
exclusive
intersection of A and B is
(disjoint) since they have no empty!
outcomes in common
3
4
6
2
1
5
Mutually Exclusive for more than 2 events
Let A1, A2, …, Ak denote k events
1. The event A1 or A2 or … or Ak consists of all
outcomes in at least one of the individual events
A1, A2, …, Ak. ( i. e. A1 U A2 U … U Ak)
2. The event A1 and A2 and … and Ak consists of all
outcomes that are simultaneously in every one of
the individual events A1, A2, …, Ak.
( i. e. A1 ∩ A2 ∩ … ∩ Ak)
These k events are mutually exclusive if no two
of them have any common outcomes.
Practice with Venn Diagrams
On the following four slides you will find Venn
Diagrams representing the students at your
school. Some students are enrolled in Statistics,
some in Calculus, and some in Computer Science.
For the next four slides, indicate what relationships
the shaded regions represent. (Use complement,
intersection, and union)
Statistics
Calculus
Computer Science
Calculus or Computer Science
Statistics
Calculus
Computer Science
(Statistics or Computer Science) and not Calculus
Statistics
Calculus
Com Sci
Computer Science
Statistics and Computer Science and not Calculus
Statistics
Calculus
Computer Science
Statistics and not (Computer Science or Calculus)
Section 6.2
Definition of Probability
What is Probability?
Three different approaches to probability
Probability – Classical Approach
If a chance experiment has k outcomes, all
equally likely, then each individual outcome
has the probability 1/k and the probability of
an event E is
number of outcomes favorable to E
P(E) 
number of outcomes in the sample space
On some football teams, the honor of calling the
toss at the beginning of the football game is
determined by random selection.
Suppose this week a member of the 11-player
offensive team will be selected to call the toss.
There are five interior linemen on the offensive
team.
If event L is defined as the event that an interior
linemen is selected to call the toss, what is the
probability of L?
P(L) =
5
11
Probability - Example
Consider the experiment consisting of rolling two fair
dice and observing the sum of the up faces. A sample
space description is given by
{(1, 1), (1, 2), … , (6, 6)}
where the pair (1, 2) means 1 is the up face of the 1st
die and 2 is the up face of the 2nd die. This sample
space consists of 36 equally likely outcomes.
Let E stand for the event that the sum is 6.
Event E is given by E={(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}.
The event consists of 5 outcomes, so
5
P(E) 
 0.1389
36
Consider an archer shooting arrows at a
target.
The probability of getting a bulls’ eye should
be the ratio of the area of the inner circle to
the area of the entire target.
What if a very experienced
archer were shooting the
arrows? Would the
probability of a bull’s eye still
be the same?
The classical approach doesn’t work for
every situation.
2) Subjective Approach
The problem with a subjective approach is that different people
canprobabilities
be interpreted
a personal
could Probability
assign different
to the as
same
outcome based
measure on
of their
the strength
belief that a particular
subjectiveor
viewpoints.
outcome will occur.
Example: An airline passenger may report that
her probability of being placed on standby
(denied a seat) due to overbooking is 0.1. She
arrived at this through personal experience and
observation of events.
3) Relative Frequency Approach
The probability of event E, denoted by P(E), is defined
to be the value approached by the relative frequency
of occurrence of E in a very long series of trials of a
chance experiment. Thus, if the number of trials is
quite large,
number of times E occurs
P(E) 
number of trials
Consider flipping a coin and recording the relative
When the number of
coin flips is small, there
is a lot of variability in
the relative frequency
in this graph).
What do you notice in
the graph at the right?
Consider flipping a coin and recording
The graph at the right
shows the relative
frequency when the
coin is flipped a large
number of times.
What do you notice in
this graph at the right?
Relative Frequency Approach
In many “real-life” processes and chance
the probability
a certain
Orexperiments,
in other words,
after a of
large
number
outcome or event is unknown, but never the
trials,
the relative
frequency
less this
probability
can be estimated
approaches
the
true probability.
reasonably
well from
observation.
The
justification is the Law of Large Numbers.
of
Law of Large Numbers: As the number of
repetitions of a chance experiment increases, the
chance that the relative frequency of occurrence
for an event will differ from the true probability of
the event by more than any very small number
approaches zero.
Relative Frequency Approach
Consider the chance experiment of rolling a “fair” die. We
would like to investigate the probability of getting a “1” for the
up face of the die. The die was rolled and after each roll the up
face was recorded and then the proportion of times that a 1
turned up was calculated and plotted.
Repeated Rolls of a Fair Die
Proportion of 1's
0.200
0.180
1/6
Relative Frequency
0.160
0.140
0.120
0.100
0.080
0.060
0.040
0.020
0.000
0
200
400
600
800
1000
# of Rolls
1200
1400
1600
1800
2000
Classical Approach
The process was simulated again and this time the
results were similar. Notice that the proportion of 1’s
seems to stabilize and in the long run gets closer to
the “theoretical” value of 1/6.
Repeated Rolls of a Fair Die
Proportion of 1's
0.350
Relative Frequency
0.300
0.250
0.200
1/6
0.150
0.100
0.050
0.000
0
200
400
600
800
1000
# of Rolls
1200
1400
1600
1800
2000
Methods for Determining Probability
1. The classical approach: Appropriate for experiments that can be
described with equally likely outcomes.
2. The subjective approach: Probabilities represent an individual’s
judgment based on facts combined with personal evaluation of other
information.
3. The relative frequency approach: An estimate is based on an
accumulation of experimental results. This estimate, usually derived
empirically, presumes a replicable chance experiment.
Section 6-3
Basic Properties of Probability
Fundamental Properties of Probability
Property 1. Legitimate Values
For any event E, 0 < P(E) < 1
Property 2. Sample space
If S is the sample space, P(S) = 1
Fundamental Properties of Probability
If two events E and F are disjoint, P(E or F) = P(E) + P(F)
Property 4. Complement
For any event E, P(E) + P(not E) = 1
Equally Likely Outcomes
Consider an experiment that can result in any
one of N possible outcomes. Denote the
corresponding simple events by O1, O2,… On.
If these simple events are equally likely to
occur, then
1
1
1
1. P(O1 )  ,P(O2 )  , ,P(ON ) 
N
N
N
2. For any event E,
number of outcomes in E
P(E) 
N
Die Example
Suppose you roll a six-sided die once. Let E
be the event that you roll an even number.
P(E) = P(2 or 4 or 6) = 3/6
Number of outcomes in E
Over N
Card Example
Consider the experiment consisting of randomly picking
a card from an ordinary deck of playing cards (52 card
deck).
Let A stand for the event that the card chosen is a King.
The sample space is given by
The sample space is given by S =
S = {A, K,,2, A , K  , , 2, A,, 2, A,…, 2}
and consists of 52 equally likely outcomes.
and consists of 52 equally likely outcomes.
The
eventby
is given by
The event
is given
A={K, K, K, K}
and consists of 4 outcomes, so
and consists of 4 outcomes, so
4
1
P(A) 
  0.0769
52 13
Dice Example
Consider the experiment consisting of rolling two
fair dice and observing the sum of the up faces.
Let E stand for the event that the sum is 7.
The sample space is given by
S={(1 ,1), (1, 2), … , (6, 6)}
and consists of 36 equally likely outcomes.
The event E is given by
E={(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}
and consists of 6 outcomes, so
6
P(E) 
36
Another Dice Example
Consider the experiment consisting of rolling two
fair dice and observing the sum of the up faces.
Let F stand for the event that the sum is 11.
The sample space is given by
S={(1 ,1), (1, 2), … , (6, 6)}
and consists of 36 equally likely outcomes.
The event F is given by
F={(5, 6), (6, 5)}
2
1
and consists of 2 outcomes, so P(F) = 36 = 18 ≈ 0.0556
Events
If events E1, E2, . . ., Ek are disjoint (mutually exclusive)
events, then
P(E1 or E2 or . . . or Ek) = P(E1) + P(E2) + . . . + P(Ek)
In words, the probability that any of these k disjoint events
occurs is the sum of the probabilities of the individual
events.
A large auto center sells cars made by many
different manufacturers. Three of these are
Honda, Nissan, and Toyota. Consider a chance
experiment that consist of observing the make
of the next car sold. Suppose that P(H) = 0.25,
P(N) = 0.18, P(T) = 0.14.
Are these disjoint events?
P(H or N or T) =
yes
.25 + .18+ .14 = .57
P(not (H or N or T)) =
1 - .57 = .43
Dice Example
Consider the experiment consisting of rolling
two fair dice and observing the sum of the up
faces.
Let E stand for the event that the sum is 7 and
F stand for the event that the sum is 11.
6
2
P(E) 
& P(F) 
36
36
Since E and F are disjoint events
6
2
8
2
P(E F)  P(E)  P(F) 


= ≈ 0.2222
9
36 36 36
Section 6-4
Conditional Probability
Motion Sickness Example
A study was performed to look at the relationship
between motion sickness and seat position in a bus.
The following table summarizes the data.
Seat Position in Bus
Front Middle Back
Nausea
58
166
193
No Nausea 870
1163
806
Motion Sickness Example
Let’s use the symbols N, NC, F, M, B to stand for
the events Nausea, No Nausea, Front, Middle
and Back respectively.
Seat Position in Bus
Front Middle Back
Nausea
58
166
193
No Nausea 870
1163
806
Total
928
1329
999
Total
417
2839
3256
Motion Sickness Example
Computing the probability that an individual
in the study gets nausea we have
P(N) =
417
3256
= 0.1281
Seat Position in Bus
Front Middle Back
Nausea
58
166
193
No Nausea 870
1163
806
Total
928
1329
999
Total
417
2839
3256
Motion Sickness Example
Other probabilities are easily calculated by
dividing the numbers in the cells by 3256 to
get
P(N)
P(N and F)
Front
0.0178
0.2672
0.2850
Nausea
No Nausea
Total
P(F)
Seat Position in Bus
Middle
0.0510
0.3572
0.4082
Back
0.0593
0.2475
0.3068
P(M and NC)
Total
0.1281
0.8719
1.0000
Conditional Probability
Let E and F be two events with P(F) > 0.
The conditional probability of the event
E given that the event F has occurred,
denoted by P(E|F), is
P(E F)
P(E | F) 
P(F)
Conditional Probability
If we want to see if nausea is related to seat
position we might want to calculate the
probability that “a person got nausea given
he/she sat in the front seat.”
We usually indicate such a conditional probability
with the notation P(N | F).
P(N | F) stands for the “probability of N given F.
Motion Sickness Example
The event “a person got nausea given
he/she sat in the front seat” is an example of
what is called a conditional probability.
Of the 928 people who sat in the front, 58 got
nausea so the probability that “a person got nausea
given he/she sat in the front seat is
58
 0.0625
928
Job Satisfaction Example
A survey of job satisfaction of teachers was
taken, giving the following results
L
E
V
E
L
Job Satisfaction
Satisfied Unsatisfied Total
College
74
43
117
High School
224
171
395
Elementary
126
140
266
Total
424
354
778
Job Satisfaction Example
If all the cells are divided by the total number
surveyed, 778, the resulting table is a table of
empirically derived probabilities.
College
High School
Elementary
Total
Job Satisfaction
Satisfied
Unsatisfied
0.0951
0.0553
0.2879
0.2198
0.1620
0.1799
0.5450
0.4550
Total
0.1504
0.5077
0.3419
1.0000
College
High School
Elementary
Total
Job Satisfaction
Satisfied
Unsatisfied
0.0951
0.0553
0.2879
0.2198
0.1620
0.1799
0.5450
0.4550
Total
0.1504
0.5077
0.3419
1.0000
For convenience, let C stand for the event that the teacher
teaches college, S stand for the teacher being satisfied and so
on. Let’s look at some probabilities and what they mean.
P(C) = 0.1504 is the proportion of teachers who are college teachers
P(C ∩ 𝑆) = 0.0951 is the proportion of teachers who are college teachers
and who are satisfied with their job
P(S) = 0.5450 is the proportion of teachers who are satisfied with
their job
College
High School
Elementary
Total
Job Satisfaction
Satisfied
Unsatisfied
0.0951
0.0553
0.2879
0.2198
0.1620
0.1799
0.5450
0.4550
Total
0.1504
0.5077
0.3419
1.0000
The proportion of teachers who are
college teachers given they are satisfied is
P (C ∩ S)
P(C l S) =
P(𝑆)
0.0951
=
0.5450
= 0.1745
Restated: This is the proportion of satisfied that are
college teachers.
College
High School
Elementary
Total
Job Satisfaction
Satisfied
Unsatisfied
0.0951
0.0553
0.2879
0.2198
0.1620
0.1799
0.5450
0.4550
Total
0.1504
0.5077
0.3419
1.0000
The proportion of teachers who are
satisfied given they are college teachers is
P(S l C) = P(S ∩ C) = P(C ∩ S)
P(C)
P(C)
=
0.0951
0.1504
= 0.6323
Restated: This is the proportion of college teachers
that are satisfied.
Section 6-5
Independence
Independence
•Two events E and F are said to be independent
if P(E|F) = P(E).
•If E and F are not independent, they are said to
be dependent events.
•If P(E|F) = P(E), it is also true that P(F|E) = P(F).
Sometimes the knowledge that one event has
occurred does NOT change our assessment of
the likelihood that another event occurs.
Consider the genetic trait, hitch
hiker’s thumb, which is the ability to
bend the last joint of the thumb
back at an angle of 60° or more.
Whether or not an offspring has hitch hiker’s thumb is
determined by two random events: which gene is
contributed by the father and which gene is contributed
by the mother. Which gene is contributed by the father
does NOT affect which gene is contributed by the mother
These are independent events.
Let’s consider a bank that offers
different types of loans:
The bank offers both adjustable-rate and fixed-rate loans on
single-family dwellings, condominiums and multifamily
dwellings. The following table, called a joint-probability
table, displays probabilities based upon the bank’s long-run
loaning practices.
Single Family
Condo
Multifamily
Total
.40
.21
.09
.70
Fixed
Total
.10
.50
.09
.30
.11
.20
.30
Bank Loans Continued
Single Family
Condo
Multifamily
Total
.40
.21
.09
.70
Fixed
Total
.10
.50
.09
.30
.11
.20
.30
P(Adjustable loan | Condo) = 0.21/0.30 = 0.70
Knowing that the loan is for a condominium
does not change the probability that it is an
adjustable-rate loan. Therefore, the event that
a randomly selected loan is adjustable and the
event that a randomly selected loan is for a
condo are independent.
Job Satisfaction Revisited
College
High School
Elementary
Total
Job Satisfaction
Satisfied
Unsatisfied
0.0951
0.0553
0.2879
0.2198
0.1620
0.1799
0.5450
0.4550
Total
0.1504
0.5077
0.3419
1.0000
Are C and S independent or dependent events?
P(C S) 0.095
P(C)  0.150 and P(C | S) 

 0.175
P(S)
0.545
P(C|S)  P(C) so C and S are dependent events.
Multiplication Rule for Two Independent
Events
Two events E and F are independent, if and only if,
P(E  F)  P(E)  P(F)
Hitch Hiker’s Thumb
Revisited
Suppose that there is a 0.10 probability that a
parent will pass along the hitch hiker’s thumb gene
to their offspring.
are independent
events,
WhatSince
is thethese
probability
that a child
willwe
have a hitch
multiply the probabilities together.
hiker’sjust
thumb?
Thismom
wouldAND
happen
the mother
P(H+ from
H+iffrom
hitch hiker’s gene (H+) AND if the father
0.1contributes
× 0.1 = 0.01
a hitch hiker’s gene (H+).
TV/VCR Example
Consider the person who purchases from two different
manufacturers a TV and a VCR. Suppose we define
the events A and B by
A = event the TV doesn’t work properly
B = event the VCR doesn’t work properly
Suppose P(A) = 0.01 and P(B) = 0.02.
What is the probability that the TV and VCR
do not work properly?
If we assume that the events A and B are independent, then
P (A and B) = P(A ∩ B) = P(A) ● P(B) = (0.01)(0.02) = 0.0002
Job Satisfaction Revisited
Consider the teacher satisfaction survey
College
High School
Elementary
Total
Job Satisfaction
Satisfied
Unsatisfied
0.0951
0.0553
0.2879
0.2198
0.1620
0.1799
0.5450
0.4550
Total
0.1504
0.5077
0.3419
1.0000
P(C) = 0.1504, P(S) = 0.5450 and P(C ∩ S) = 0.0951
Since P(C)P(S) = (0.1504)(0.5450) = 0.081968
P(C ∩ S) = 0.0951 which shows P(C ∩ S) ≠ P(C)P(S)
This shows that C and S are dependent events.
Multiplication Rule for k Independent Events
Events E1, E2, . . ., Ek are independent if knowledge
that some number of the events have occurred does
not change the probabilities that any particular one
or more of the other events occurred.
P (E1  E2  ...  Ek )  P (E1 )  P (E2 )  ...  P (Ek )
This relationship remains valid if one or more of the
events are replaced by their complement (not E).
Suppose that a desktop computer system consist of a monitor, a
mouse, a keyboard, the computer processor itself, and storage
devices such as a disk drive. Most computer system problems
due to manufacturer defects occur soon in the system’s lifetime.
Purchasers of new computer systems are advised to turn their
computers on as soon as they are purchased and then to let the
computer run for a few hours to see if any problems occur.
Let
E1 = event that a newly purchased monitor is not defective
E2 = event that a newly purchased mouse is not defective
E3 = event that a newly purchased disk drive is not defective
E4 = event that a newly purchased processor is not defective
Suppose the four events are independent with
P(E1) = P(E2) = .98
P(E3) = .94
P(E4) = .99
Let
E1 = event that a newly purchased monitor is not defective
E2 = event that a newly purchased mouse is not defective
E3 = event that a newly purchased disk drive is not defective
E4 = event that a newly purchased processor is not defective
Suppose the four events are independent with
In the long run, 89% of such
P(E1) = P(E2) = .98 P(E3) = .94 P(E4) = .99
systems will run properly
when tested shortly after
purchase.
What is the probability that none of these
components
are defective?
P (E1  E2  E 3  E 4 ) 
(.98)(.98)(.94)(.99) = .8937
Let
E1 = event that a newly purchased monitor is not defective
E2 = event that a newly purchased mouse is not defective
E3 = event that a newly purchased disk drive is not defective
E4 = event that a newly purchased processor is not defective
Suppose the four events are independent with
P(E1) = P(E2) = .98 P(E3) = .94 P(E4) = .99
What is the probability that all these components will run
properly except the monitor?
C
P (E1
 E2  E3  E 4 ) 
(.02)(.98)(.94)(.99) = .018
Sampling Schemes
•Sampling is with replacement if, once selected, an individual or
object is put back into the population before the next selection.
•Sampling is without replacement if, once selected, an
individual or object is not returned to the population prior to
subsequent selections.
Card Example
Suppose we are going to select three cards from an ordinary
deck of cards. Consider the events:
E1 = event that the first card is a king
E2 = event that the second card is a king
E3 = event that the third card is a king.
Card Example – With Replacement
If we select the first card and then place it
back in the deck before we select the
second, and so on, the sampling will be
with replacement.
4
P(E1 )  P(E2 )  P(E3 ) 
52
P(E1
E2
E3 )  P(E1 )P(E2 )P(E3 )
4 4 4

 0.000455
52 52 52
Card Example – Without Replacement
If we select the cards in the usual manner
without replacing them in the deck, the
sampling will be without replacement.
4
3
2
P(E1 )  , P(E2 )  , P(E3 ) 
52
51
50
P(E1
E2
E3 )  P(E1 )P(E2 )P(E3 )
4 3 2

 0.000181
52 51 50
Suppose I will pick two cards from a standard deck.
This can be done two ways:
Sampling
without
replacement
– thethe
events
typically
1)Pick
a card
at random,
replace
card,arethen
pick a
dependent events.
second card
2) Pick a card at random, do NOT replace, then pick a
Sampling with replacement – the events are typically
second card.
independent events.
If I pick two cards from a standard deck without
replacement,
what isofthe
probability
select
Probability
given I drew
ontwo
the
first card.
Are the events E1 = first card is a spade and E2 = second card is
NO
P(E1 and E2) =
P(E1) × P(E2|E1) =
1 12 1


4 51 17
Jury Selection
Suppose a jury pool in a city contains 12000 potential
jurors and 3000 of them are African American women.
Consider the events
E1 = event that the first juror selected is an African American
woman
E2 = event that the second juror selected is an African
American woman
E3 = event that the third juror selected is an African American
woman
E4 = event that the fourth juror selected is an African
American woman
Jury Selection Continued
Clearly the sampling will be without
replacement so
3000
2999
P(E1 ) 
, P(E2 ) 
,
12000
11999
2998
2997
P(E3 ) 
,P(E 4 ) 
11998
11997
So P(E1
E2
E3
E4 )
3000 2999 2998 2997

 0.003900
12000 11999 11998 11997
Jury Selection Continued
If we “treat” the Events E1, E2, E3 and
E4 as being with replacement
(independent) we would get
3000
P(E1 )  P(E 2 )  P(E3 ) 
 0.25
12000
So P(E1
E2
E3
E4 )  (0.25)(0.25)(0.25)(0.25)
 0.003906
Jury Selection Summary
•Notice the result calculated by sampling without
replacement is 0.003900 and the result calculated by
sampling with replacement is 0.003906. These
results are substantially the same.
•Clearly when the number of items is large and the
number in the sample is small, the two methods give
essentially the same result.
Important Observation
• If a random sample of size n is taken from a
population of size N, the theoretical probabilities of
successive selections calculated on the basis of
sampling with replacement and on the basis of sample
without replacement differ by insignificant amounts
under suitable circumstances.
•Typically independence is assumed for the purposes
of calculating probabilities when the sample size n is
less than 5% of the population size N.
Suppose the manufacturer of a certain
brand of light bulbs made 10,000 of these
bulbs and 500 are defective. You randomly
pick a package of two such bulbs off the shelf of a
store. What is the probability that both bulbs are
defective?
Are the events E1 = the first bulb is defective and
E2 = the second bulb is defective independent?
thisprobability
question, let’s
explore thea
What would
be the
of selecting
probabilities of these two events.
defective light bulb?
500/10,000 = .05
Light Bulbs Continued
What would be the probability of selecting a
Ifdefective
a random light
sample
of size n is taken from a population of
bulb?
size N, then the outcomes of selecting successive items
= .05 can be treated as
from the population500/10,000
without replacement
independent when the sample size n is at most 5% of the
Having selected population
one defective
bulb, what is the
size N.
probability of selecting another without
replacement? 499/9999 = .0499
These values are so close to each other that when
rounded to three decimal places they are both .050. For
all practical purposes, we can treat them as being
independent.
Light Bulbs Continued
What is the probability that both bulbs are
defective?
Are the selections independent?
We can assume independence.
P(defectiv e  defective)  (0.05)(0.05) = .0025
Section 6-6
Some General Probability Rules
Since the intersection is added in twice, we subtract
out the intersection.
For any two events E and F,
P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
E
F
Musical styles other than rock and pop
are becoming more popular. A survey of
college students finds that the
probability they like country music is .40.
The probability that they liked jazz is .30
and that they liked both is .10. What is
the probability that they like country or
jazz?
P (Country ∪ Jazz) = P(C) + P(J) – P(C ∩ J)
=
.4 + .3
-
.1 = .6
Job Satisfaction Example
Consider the teacher satisfaction survey
College
High School
Elementary
Total
Job Satisfaction
Satisfied
Unsatisfied
0.0951
0.0553
0.2879
0.2198
0.1620
0.1799
0.5450
0.4550
Total
0.1504
0.5077
0.3419
1.0000
Find the probability of a teacher teaching at college level or being satisfied.
P(C ∪ S) = P(C) + P(S) – P(C ∩ S)
= 0.1504 + 0.5450 – 0.0951
= 0.6003
Here is a process to use when calculating the union
of two or more events.
P(E ∪ F)
Yes
mutually exclusive?”
In some problems, the intersection of the
two events is given (see
No previous
example).
P(E) + P(F)
In some problems, the P(E) + P(F) – P(E ∩ F)
intersection of the two
events is not given, but we
know that the events are
If independent
independent.
P(E) ● P(F)
Suppose two six-sided dice are rolled (one white
and one red). What is the probability that the
white die lands on 6 or the red die landsHow
on can
1?
Let
A = white die landing on 6
you find the
probability
of A and B?
B = red die landing on 1
Are A and B disjoint? NO, independent events
cannot be disjoint
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
1 1  1  1  11
     
6 6  6  6  36
General Multiplication Rule
For any two events E and F,
P(E F)  P(E | F)P(F)
From symmetry we also have
P(E F)  P(F | E)P(E)
Here is a process to use when calculating the
intersection of two or more events.
events independent?”
P(A ∩ B)
Yes
No
P(A) ● P(B)
P(B l A) ● P(A)
Secretary Example
18% of all employees in a large company are
secretaries and furthermore, 35% of the secretaries are
male . If an employee from this company is randomly
selected, what is the probability the employee will be a
secretary and also male.
Let E stand for the event the employee is male.
Let F stand for the event the employee is a secretary.
The question can be answered by noting that
P(F) = 0.18 and P(E|F) = 0.35 so
P(E F)  P(E | F)P(F)  (0.35)(0.18)  0.063
There are seven girls and eight boys in a math class.
The teacher selects two students at random to
answer questions on the board. What is the
probability that both students are girls?
Are these events independent? NO
7 6
P( G1 ∩ G2) =
  .2
15 14
Light Bulbs Continued
A certain brand of light bulbs are defective five
percent of the time. You randomly pick a package of
two such bulbs off the shelf of a store. What is the
probability that exactly one bulb is defective?
Let D1 = first light bulb is defective
D2 = second light bulb is defective
P(exactly one defective) = P((D1 ∩ D2C) ∪ (D1C ∩ D2))
= (.05)(.95) + (.95)(.05) = .0950
An electronics store sells DVD players made by one
of two brands. Customers can also purchase
extended warranties
for theinDVD
The
This can happen
one ofplayer.
two ways:
1) They
purchased the
extended
following
probabilities
are
given: warranty and Brand 1
DVD player
Let
B1 = event that brand 1 is purchased
OR
2)BThey
purchased
the extended
warranty and Brand 2
that brand
2 is purchased
2 = event
DVD player
E = event that extended warranty is purchased
P(B1) = .7
P(B2) = .3
P(E|B1) = .2
P(E|B2) = .4
If a DVD customer is selected at random, what is
the probability that they purchased the extended
warranty?
E = (E ∩ B1) ∪ (E ∩ B2)
Law of Total Probabilities
If B1 and B2 are disjoint events with probabilities
P(B1) + P(B2) = 1, for any event E
P(E) = P(E ∩ B1) + P(E ∩ B2)
= P(E l B1) ● P(B1) + P(E l B2) ●P(B2)
More generally B1, B2, …, Bk are disjoint events with probabilities
P(B1) + P(B2) + … + P(Bk) = 1, for any event E
P(E) = P(E ∩ B1) + P(E ∩ B2) + … + P(E ∩ Bk)
= P(E l B1) ● P(B1) + P(E l B2) ●P(B2) +…+ P(E l Bk) ●P(Bk)
DVD Player Example Continued
Let
B1 = event that brand 1 is purchased
B2 = event that brand 2 is purchased
E = event that extended warranty is purchased
P(B1) = .7
P(B2) = .3
P(E|B1) = .2
P(E|B2) = .4
If a DVD customer is selected at random, what is
These are disjoint events
the probability that they purchased the extended
warranty?
Use the General Multiplication Rule:
E = (E ∩ B1) ∪ (E ∩ B2)
P(E) = P(E ∩ B1) + P(E ∩ B2)
P(E) = P(E l B1) ● P(B1) + P(E l B2) ●P(B2)
DVD Player Example Continued
Let
B1 = event that brand 1 is purchased
B2 = event that brand 2 is purchased
E = event that extended warranty is purchased
P(B1) = .7
P(B2) = .3
P(E|B1) = .2
P(E|B2) = .4
If a DVD customer is selected at random, what is
the probability that they purchased the extended
warranty?
P(E) = P(E l B1) ● P(B1) + P(E l B2) ●P(B2)
P(E) = (.2)(.7) + (.4)(.3) = .26
This is an example of the Law of Total Probabilities.
Soccer Example
A soccer team wins 60% of its games when it scores the first goal,
and 10% of its games when the opposing team scores first. If the
team scores the first goal about 30% of the time, what fraction of
the games does it win? Let W be the event that the team wins, and
SF be the event that it scores first.
By the law of total probability:
P(W) = P(W l SF) ● P(SF) + P(W l SFc) ● P(SFc)
= (0.6)(0.3) + (0.1)(0.7)
= 0.25
Chess Example
You enter a chess tournament where your probability of winning a game
is 0.3 against beginners, 0.4 against intermediate-level players, and 0.5
against the advanced-level players. Half of the players are beginners,
one-fourth are intermediate-level, and one-fourth are advanced-level.
You play a game against a randomly chosen opponent. What is the
probability of wining?
Let B = beginners
I = intermediate-level
W = event of winning
Chess Example Continued
Let B = beginners
I = intermediate-level
W = event of winning
P(B) = 0.5
P(W l B) = 0.3
P(I) = 0.25
P(A) = 0.25
P(W l I ) = 0.4
P(W l A) = 0.5
By the law of total probability, the probability of winning is:
P(W) = P(W l B)P(B) + P(W l I) P(I) + P(W l A) P(A)
P(W) = (0.3)(0.5) + (0.4)(0.25) + (0.5)(0.25)
P(W) = 0.375
Four-sided Die Example
We roll a fair four-sided die. If the result is 1 or 2, we roll once more but
otherwise, we stop. What is the probability that the sum total of our rolls is
at least 4?
Let Ai be the event that the result of the first roll is I, and note that P(Ai) =
¼ for each i. Let B be the event that the sum total is at least 4. Given the
event A1, the sum total will be at least 4 if the second roll results in 3 or 4,
which happens with probability ½. Similarly, given the event A2, the sum
total will be at least 4 if the second roll results in 2, 3, or 4, which happens
with probability ¾. Also, given the event A3, we stop and the sum total
remains below 4. Therefore,
Four-sided Die Example Continued
Therefore,
P(B l A1) = ½
(1, 3) (1, 4)
P(B l A2) = ¾
P(B l A3) = 0
(2, 2) (2, 3) (2, 4)
P(B l A4) = 1
(4)
By the law of total probability,
P(B) = P(B l A1)P(A1) + P(B l A2)P(A2) + P(B l A3)P(A3) + P(B l A4)P(A4)
= (0.5)(0.25) + (0.75)(0.25) + (0) (0.25) + (1) (0.25)
= 0.5625
Bayes’ Rule (Theorem)
A formula discovered by the Reverend Thomas Bayes,
an English Presbyterian minister, to solve what he
called “converse” problems.
Let’s examine the following problem before looking at
the formula . . .
Lyme’s disease is the leading tick-borne disease in the United
States and England. Diagnosis of the disease is difficult and is aided
by a test that detects particular antibodies in the blood. The
article, “Laboratory Consideration in the Diagnosis and
Management of Lyme Borreliosis”, American Journal of Clinical
Pathology, 1993, used the following notations:
+ represents a positive result on a blood test
- represents a negative result on a blood test
L represents the patient actually has Lymes
LC represents the patient doesn’t have Lymes
The article gave the following probabilities:
P(L) = .00207
P(LC) = .99723
P(+|L) = .937
P(-|L) = .063
P(+|LC) = .03
P(-|LC) = .97
Lyme’s Disease Example Continued
The article gave the following probabilities:
P(L) = .00207
P(LC) = .99723
P(+|L) = .937
P(-|L) = .063
P(+|LC) = .03
P(-|LC) = .97
Bayes’ converse problem poses this question:
“Given that a patient tests positive, what is the
probability that he or she really has the disease?”
written: P(L|+)
This question is of primary concern in medical
diagnosis problems!
Lyme’s Disease Continued
TheUsing
article gave
the following
the Law
of Totalprobabilities:
Probabilities, the denominator
P(L) = .00207
P(LC) = .99723
becomes
C)P(LC).
P(+|L) = .937
= .063
P(+|L)P(L) P(-|L)
+ P(+|L
P(+|LC) = .03
P(-|LC) = .97
Bayes reasoned as follows:
P (L   )
P (L |  ) 
P ( )
Substitute values:
PSince
(  | L )  P (L )

C
C
P (  |PL(L) P()L) P (P(
 |LL
) )  P (L )
we can use P(+|L)
P(L) for
.937(.×00207
) the numerator.

.937(.00207)  .03(.99793)
 .0596
Bayes’ Rule
If B1 and B2 are disjoint events with
P(B1) + P(B2) = 1, then for any even E
P(E l B1)●P(B1)
P(B1 l E) =
P(E l B1)●P(B1)+P(E l B2)●P(B2)
More generally, if B1, B2, …, Bk are disjoint events with
P(B1) + P(B2) + …+ P(Bk) = 1, then for any even E
P(Bk l E) =
P(E l B1)●P(B1)
P(E l B1)●P(B1)+P(E l B2)●P(B2)+ …+P(E l Bk)●P(Bk)
A company that makes radios, uses three different
subcontractors (A, B and C) to supply on switches used in
assembling a radio. 50% of the switches come from
company A, 35% of the switches come from company B
and 15% of the switches come from company C.
Furthermore, it is known that 1% of the switches that
company A supplies are defective, 2% of the switches that
company B supplies are defective and 5% of the switches
that company C supplies are defective.
If a radio from this company was inspected and failed the
inspection because of a defective on switch, what are the
probabilities that that switch came from each of the
suppliers?
Example - continued
Define the events
S1 = event that the on switch came from subcontractor A
S2 = event that the on switch came from subcontractor B
S3 = event that the on switch came from subcontractor C
D = event that the on switch was defective
From the problem statement we have
P(S1) = 0.5, P(S2) = 0.35 , P(S3) = 0.15
P(D|S1) =0.01, P(D|S2) =0.02, P(D|S3) =0.05
P(Switch
came
from
supplier
givenit itwas
wasdefective) =
P(Switch
came
from
supplier
AA
given
defective) =
P(D l S1)●P(S1)
P(D | S )
P(SP(S
=
1 l D)
| D)  P(D l S1)●P(S1)+P(D l S2)●P(S2)+ …+P(D lS3)●P(S3)
P(D | S )P(S )  P(D | S )P(S )  P(D | S )P(S )
1
1
1
1
2
2
3
3
(.5)(.01)
(.5)(.01)  (.35)(.02)  (.15)(.05)
.005
.005


 .256
.005  .007  .0075 .0195

Similarly,
P(S3 | D) 
(.35)(.02)
(.15)(.05)
 .359 and P(S2 | D) 
 .385
.0195
.0195
These calculations show that 25.6% of the
on defective switches are supplied by
subcontractor A, 35.9% of the defective on
switches are supplied by subcontractor B
and 38.5% of the defective on switches are
supplied by subcontractor C.
Even though subcontractor C supplies only a small
proportion (15%) of the switches, it supplies a
reasonably large proportion of the defective
switches (38.9%).
AIDS Example
Just for the heck of it Bob decides to take a test for AIDS and it comes
back positive.
• The test is 99% effective (1% FP and FN).
• Suppose 0.3% of the population in Bob’s “bracket” has AIDS.
What is the probability that he has AIDS?
• A1 = people in Bob’s bracket with AIDS
• A2 = people in Bob’s bracket without AIDS
• B = people in Bob’s bracket who would test positive
AIDS Example Continued
• A1 = people in Bob’s bracket with AIDS
• A2 = people in Bob’s bracket without AIDS
• B = people in Bob’s bracket who would test positive
P(A1) = 0.003
P(BlA1) = 0.99
P(A2) = 0.997
P(BlA2) = 0.01
AIDS Example Continued
By Bayes’ rule
P(A1lB) =
______(0.99)(0.003)______
(0.99)(0.003) + (0.01)(0.997)
=
_____0.00297_____
0.00297 + 0.00997
=
0.2295
Voter Example
In a certain county 60% of registered voters are Republicans, 30%
are Democrats, and 10% are Independents. When those voters
opposed it, 65% of the Democrats opposed it, and 55% of the
Independents opposed it. What is the probability that a randomly
selected voter in this county opposes increased military spending?
Let R = registered republicans
D = registered democrats
I = registered independents
B = registered voters opposing increased military spending
Voter Example Continued
A registered voter from our county writes a letter to the local
paper, arguing against increased military spending. What is the
probability that this voter is a Democrat?
Let R = registered republicans
D = registered democrats
I = registered independents
B = registered voters opposing increased military spending
P(R) = 0.60
P(BlR) = 0.4
P(D) = 0.30
P(BlD) = 0.65
P(I) = .10
P(BlI) = 0.55
Voter Example Continued
P(R) = 0.60
P(D) = 0.30
P(I) = .10
P(B l R) = 0.4
P(B l D) = 0.65
P(B l I) = 0.55
By the total probability theorem:
P(B) = P(B l R) ● P(R) + P(B l D) ● P(D) + P(B l I) ● P(I)
= (0.4)(0.6) + (0.65)(0.3) + (0.55)(0.1)
= 0.49
Voter Example Continued
A registered voter from our county writes a letter to the local paper,
arguing against increased military spending. What is the probability that
this voter is a Democrat?
• Presumably that is P(D l B), so by Bayes’ theorem:
P(D l B) =
__________(0.65)(0.3)__________
(0.4)(0.6) + (0.65)(0.3) + (0.55)(0.1)
=
0.195
0.49
=
0.3980
Section 6-7
Estimating Probabilities Empirically using
Simulation
Simulation
1. Design a method that uses a random mechanism
(such
as a random
generator
table,
Simulation
provides anumber
means of
estimatingor
probabilities
tossing
a coin
or die,
to represent
an
when we
are unable
to etc.)
determine
them analytically
or it
observation.
Be sure
that thethem
important
is impractical
to estimate
empirically by
characteristics of theobservation.
actual process are preserved.
2. Generate an observation using the method in step
1 and determine if the outcome of interest has
occurred.
3. Repeat step 2 a large number of times
4. Calculate the estimated probability by dividing the
number of observations of the outcome of interest
by the total number of observations generated.
Suppose that couples who wanted children were to
continue having children until a boy was born. Would
this change the proportion of boys in the population?
We will use simulation to estimate the proportion of
boys in the population if couples were to continue
having children until a boy was born.
1)We will use a single-random digit to represent a
child, where odd digits represent a male birth and
even digits represent a female birth.
2) Select random digits from a random digit table until
a male is selected and record the number of boys
and girls.
3) Repeat step 2 a large number of times.
Boy Simulation Continued
Below
are four
from
thenumber
random
at the
Continue
this rows
process
a large
of digit
times table
(at least
back of our textbook. 100 trials).
RowCalculate the proportion of boys out of the number of
children born.
6 0 9 3 8 7 6 7 9 9 5 6 2 5 6 5 8 4 2 6 4
7 4 1 0 1 0 2 2 0 4 7 5 1 1 9 4 7 9 7 5 1
8 6 4 7 3 6 3 4 5 1 2 3 1 1 8 0 0 4 8 2 0
Notice that with only 10 trials, the proportion of
9 8 0 2 8 boys
7 9 is310/22,
8 4 which
0 4 is
2 close
0 8to90.5!
1 2 3 3 2
Trial 1: girl, boy
Trial 5: boy
Trial 9: girl, boy
Trial 2: boy
Trial 6: boy
Trial 3: girl, boy
Trial 7: boy
Trial 10: girl, girl, girl,
girl, girl, girl, boy
Trial 4: girl, boy
Trial 8: girl, girl, boy
```