AlfordChapter4.ppt

4.1 - 1
Copyright © 2010, 2007, 2004 Pearson Education, Inc. All Rights Reserved.
Chapter 4
Probability
4-1 Review and Preview
4-2 Basic Concepts of Probability
4-3 Addition Rule
4-4 Multiplication Rule: Basics
4-5 Multiplication Rule: Complements and
Conditional Probability
4-6 Counting

4.1 - 2
Section 4-1
Review and Preview

4.1 - 3
Review
Necessity of sound sampling methods.
Common measures of characteristics of
data
Mean
Standard deviation

4.1 - 4
Preview
Rare Event Rule for Inferential Statistics:
If, under a given assumption, the
probability of a particular observed
event is extremely small, we conclude
that the assumption is probably not
correct.
Statisticians use the rare event rule for
inferential statistics.

4.1 - 5
Section 4-2
Basic Concepts of
Probability

4.1 - 6
Part 1
Basics of Probability

4.1 - 7
Events and Sample Space
 Event
any collection of results or outcomes of a
procedure
 Simple Event
an outcome or an event that cannot be further
broken down into simpler components
 Sample Space
for a procedure consists of all possible simple
events; that is, the sample space consists of all
outcomes that cannot be broken down any
further

4.1 - 8
Example
 A pair of dice are rolled. The sample space has
36 simple events:
1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
where the pairs represent the numbers rolled on
each dice.
Which elements of the sample space correspond
to the event that the sum of each dice is 4?

4.1 - 9
Example
 Which elements of the sample space correspond
to the event that the sum of each dice is 4?
ANSWER:
3,1 2,2 1,3

4.1 - 10
Notation for
Probabilities
P - denotes a probability.
A, B, and C - denote specific events.
P(A) - denotes the probability of
event A occurring.

4.1 - 11
Basic Rules for
Computing Probability
Rule 1: Relative Frequency Approximation
of Probability
Conduct (or observe) a procedure, and count
the number of times event A actually occurs.
Based on these actual results, P(A) is
approximated as follows:
P(A) = # of times A occurred
# of times procedure was repeated

4.1 - 12
Example
Problem 20 on page 149
F = event of a false negative on polygraph
test
Thus this is not considered unusual since
it is more than 0.001 (see page 146). The
test is not highly accurate.
0918
.
0
98
9
)
( 

F
P

4.1 - 13
Rounding Off
Probabilities
When expressing the value of a probability,
either give the exact fraction or decimal or
round off final decimal results to three
significant digits. All digits are significant
except for the zeros that are included for proper
placement of the decimal point.
Example:
0.1254 has four significant digits
0.0013 has two significant digits

4.1 - 14
Example
F = event of a selecting a female senator
NOTE: total number of senators=100
Thus this does not agree with the claim
that men and women have an equal (50%)
chance of being selected as a senator.
160
.
0
100
16
16
84
16
)
( 



F
P

4.1 - 15
Example
A = event that Delta airlines passenger is
involuntarily bumped from a flight
Thus this is considered unusual since it is
less than 0.05 (see directions on page
149). Since probability is very low, getting
bumped from a flight on Delta is not a
serious problem.
000195
.
0
15378
3
)
( 

A
P

4.1 - 16
Basic Rules for
Computing Probability
Rule 2: Classical Approach to Probability
(Requires Equally Likely Outcomes)
Assume that for a given procedure each simple
event has an equal chance of occurring.
P(A) = number of ways A can occur
number of different simple events
in the sample space

4.1 - 17
Example
What is the probability of rolling two die
and getting a sum of 4?
A = event that sum of the dice is 4
Assume each number is equally likely to
be rolled on the die. Rolling a sum of 4
can happen in one of three ways (see
previous slide) with 36 simple events so:
0833
.
0
36
3
)
( 

A
P

4.1 - 18
Example
What is the probability of getting no heads
when a “fair” coin is tossed three times?
(A fair coin has an equal probability of
showing heads or tails when tossed.)
A = event that no heads occurs in three
tosses
Sample Space (in order of toss):
TTT
TTH
THT
THH
HTT
HTH
HHT
HHH ,
,
,
,
,
,
,

4.1 - 19
Example
Sample space has 8 simple events.
Event A corresponds to TTT only so that:
125
.
0
8
1
)
( 

A
P

4.1 - 20
Example
Let:
S = event that son inherits disease (xY or Yx)
D = event that daughter inherits disease (xx)

4.1 - 21
Example
(a) Father: xY Mother: XX
Sample space for a son:
YX YX
Sample space has no simple
events that represent a son that has
the disease so:
0
2
0
)
( 

S
P

4.1 - 22
Example
(b) Father: xY Mother: XX
Sample space for a daughter:
xX xX
events that represent a daughter that
has the disease so:
0
2
0
)
( 

D
P

4.1 - 23
Example
(c) Father: XY Mother: xX
Sample space for a son:
Yx YX
Sample space has one simple
event that represents a son that has
the disease so:
5
.
0
2
1
)
( 

S
P

4.1 - 24
Example
(d) Father: XY Mother: xX
Sample space for a daughter:
Xx XX
event that represents a daughter that
has the disease so:
0
2
0
)
( 

D
P

4.1 - 25
Table 4-1 on page 137 (polygraph data)
Example
Did Not Lie Did Lie
Positive Test Result 15
(false positive)
42
(true positive)
NegativeTest Result 32
(true negative)
9
(false negative)

4.1 - 26
It is helpful to first total the data in the table:
(a) How many responses were lies:
ANSWER: 51
Example
Did Not Lie Did Lie TOTALS
(false positive)
42
(true positive)
57
(true negative)
9
(false negative)
41
TOTALS 47 51 98

4.1 - 27
(b) If one response is randomly selected, what
is the probability it is a lie?
L = event of selecting one of the lie responses
(c)
Example
98
51
)
( 
L
P
520
.
0
98
51


4.1 - 28
Basic Rules for
Computing Probability - continued
Rule 3: Subjective Probabilities
P(A), the probability of event A, is
estimated by using knowledge of the
relevant circumstances.

4.1 - 29
Example
Probability should be high based on
experience (it is rare to be delayed
because of an accident).
Guess: P=99/100 (99 out of 100 times you
will not be delayed because of an
accident)
ANSWERS WILL VARY

4.1 - 30
Example:
Classical probability predicts the
probability of flipping a (non-biased)
coin and it coming up heads is ½=0.5
Ten coin flips will sometimes result in
exactly 5 heads and a frequency
probability of heads 5/10=0.5; but often
you will not get exactly 5 heads in ten
flips.

4.1 - 31
Law of
Large Numbers
As a procedure is repeated again and
again, the relative frequency probability
of an event tends to approach the actual
probability.
Example: If we flip a coin 1 million times
the frequency probability should be
approximately 0.5

4.1 - 32
Probability Limits
 The probability of an event that is certain to
occur is 1.
 The probability of an impossible event is 0.
 For any event A, the probability of A is
between 0 and 1 inclusive. That is:
0  P(A)  1
Always express a probability as a fraction or
decimal number between 0 and 1.

4.1 - 33
Possible Values
for Probabilities

4.1 - 34
Complementary Events
The complement of event A, denoted by
A, consists of all outcomes in which the
event A does not occur.

4.1 - 35
Example
If a fair coin is tossed three times and
A = event that exactly one heads occurs
Find the complement of A.

4.1 - 36
Example
Sample space:
Event A corresponds to HTT, THT, TTH
Therefore, the complement of A are the
simple events:
HHH, HHT, HTH, THH, TTT
TTT
TTH
THT
THH
HTT
HTH
HHT
HHH ,
,
,
,
,
,
,

4.1 - 37
Part 2
Beyond the
Basics of Probability: Odds

4.1 - 38
Odds
The actual odds in favor of event A occurring are the
ratio P(A)/ P(A), usually expressed in the form of a:b (or
“a to b”), where a and b are integers having no common
factors.
The actual odds against event A occurring are the
ratio P(A)/P(A), which is the reciprocal of the actual
odds in favor of the event. If the odds in favor of A
are a:b, then the odds against A are b:a.
The payoff odds against event A occurring are the
ratio of the net profit (if you win) to the amount bet.
payoff odds against event A = (net profit) : (amount bet)

4.1 - 39
Example
Problem 38, page 149
W = simple event that you win due to an odd
number
Sample Space
00, 0,
1, 2, 3, 4, 5, 6, 7, 8, 9, 10,
11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25, 26, 27, 28, 29, 30,
31, 32, 33, 34, 35, 36
(a) There are 18 odd numbers so that
P(W) = 18/38

4.1 - 40
Example
There are 20 events that correspond to
winning from a number that is not odd (i.e.
you do not win due to an odd number) so:
(b) Odds against winning are
9
:
10
or
9
/
10
18
20
38
/
18
38
/
20
)
(
)
(



W
P
W
P
38
20
)
( 
W
P

4.1 - 41
Example
(c) Payoff odds against winning are 1:1
That is, $1 net profit for every $1 bet
Thus, if you bet $18 and win, your net profit
is $18 which can be found by solving the
proportion:
The casino returns $18+$18=$36 to you.
1
1
18
profit
net


4.1 - 42
Example
(d) Actual odds against winning are 10:9
That is, $10 net profit for every $9 bet
Thus, if you bet $18 and win, your net profit
is $20 which can be found by solving the
proportion:
The casino returns $18+$20=$38 to you.
9
10
18
profit
net


4.1 - 43
Recap
In this section we have discussed:
 Rare event rule for inferential statistics.
 Probability rules.
 Law of large numbers.
 Complementary events.
 Rounding off probabilities.
 Odds.

4.1 - 44
Section 4-3
Addition Rule

4.1 - 45
Key Concept
This section presents the addition rule as a
device for finding probabilities that can be
expressed as P(A or B), the probability that
either event A occurs or event B occurs (or
they both occur) as the single outcome of
the procedure.
The key word in this section is “or.” It is the
inclusive or, which means either one or the
other or both.

4.1 - 46
Compound Event
any event combining 2 or more simple events
Compound Event
Notation
P(A or B) = P (in a single trial, event A occurs
or event B occurs or they both occur)

4.1 - 47
When finding the probability that event
A occurs or event B occurs, find the
total number of ways A can occur and
the number of ways B can occur, but
find that total in such a way that no
outcome is counted more than once.
General Rule for a
Compound Event

4.1 - 48
A random survey of members of the class of
2005 finds the following:
What is the probability the student did not
graduate or was a man?
Example
Number of
students who
graduated
Number of students
who did not graduate
TOTALS
Women 672 22 694
Men 582 19 601
TOTALS 1254 41 1295

4.1 - 49
Method 1: directly add up those who did not
graduate and those who are men (without
counting men twice):
22 + 19 + 582 = 623
Method 2: add up the total number who did
not graduate and the total number of men,
then subtract the double count of men:
41 + 601 - 19 = 623
Example

4.1 - 50
The probability the student did not graduate
or was a man :
623/1295 = 0.481
Example

4.1 - 51
Compound Event
Formal Addition Rule
P(A or B) = P(A) + P(B) – P(A and B)
where P(A and B) denotes the probability
that A and B both occur at the same time as
an outcome in a trial of a procedure.

4.1 - 52
N = event that student did not graduate
M = event that student was a man
P(N) = 41/1295 = 0.0317
P(M) = 601/1295 = 0.464
P(N and M) = 19/1295 = 0.0147
P(N or M) = P(N) + P(M) - P(N and M)
= 0.0317 + 0.464 - 0.0147
= 0.481
Previous Example

4.1 - 53
What is the probability the subject had a
negative test result or lied?
Example
(false positive)
42
(true positive)
57
(true negative)
9
(false negative)
41
TOTALS 47 51 98

4.1 - 54
N = event that there is a negative test result
L = event that the subject lied
P(N) = 41/98 = 0.418
P(L) = 51/98 = 0.520
P(N and L) = 9/98 = 0.0918
P(N or L) = P(N) + P(L) - P(N and L)
= 0.418 + 0.520 - 0.0918
= 0.846
Example (cont.)

4.1 - 55
N=event that person refuses to respond
F=event that person’s age is 60 or older
Note: total number of people in the study is 1205
total number who refused to respond is 156 so:
P(N)=156/1205
total number who are 60 or older is 251 so:
P(F)=251/1205
total number who refuse to respond and 60 or older is 49 so:
P(N and F)=49/1205
P(N OR F) = 156/1205 + 251/1205 – 49/1205 = 358/1205 = 0.297
Example

4.1 - 56
Disjoint or Mutually Exclusive
Events A and B are disjoint (or mutually
exclusive) if they cannot occur at the same
time. (That is, disjoint events do not
overlap.)
Previous Example:
G = students who graduated
N = students who did not graduate
N and G are disjoint because no student
who did not graduate also graduated

4.1 - 57
Disjoint or Mutually Exclusive
Events A and B are not disjoint if they
overlap.
Previous Example:
M = male students
G = students who graduated
M and G are not disjoint because some
students who graduated are also male

4.1 - 58
Example
Problem 10 and 12 on page 157
10.These are disjoint since a subject
treated with Lipitor cannot be a subject
given no medication.
12. These are not disjoint since it is
possible for a homeless person to be a
college graduate.

4.1 - 59
Venn Diagram
A Venn diagram is a way to picture how
sets overlap.
Venn Diagram for Events That Are
Not Disjoint and overlap.
Venn Diagram for Disjoint Events
which do not overlap.

4.1 - 60
Complementary
Events
It is impossible for an event and its
complement to occur at the same time.
That is, for any event A,
are disjoint
A
A and

4.1 - 61
Probability Rule of
Complementary Events
P(A) + P(A) = 1
= 1 – P(A)
P(A) = 1 – P(A)
P(A)

4.1 - 62
Venn Diagram for the
Complement of Event A

4.1 - 63
Example
is the probability that a screened
driver is not intoxicated
)
(I
P
991
.
0
00888
.
0
1
)
(
1
)
( 



 I
P
I
P

4.1 - 64
Recap
 Compound events.
 Formal addition rule.
 Intuitive addition rule.
 Disjoint events.
 Complementary events.

4.1 - 65
Section 4-4
Multiplication Rule:
Basics

4.1 - 66
Tree Diagrams
A tree diagram is a picture of the
possible outcomes of a procedure,
shown as line segments emanating
from one starting point. These
diagrams are sometimes helpful in
determining the number of possible
outcomes in a sample space, if the
number of possibilities is not too
large.

4.1 - 67
Tree Diagrams
This figure
summarizes
the possible
outcomes
for a true/false
question followed
by a multiple choice
question.
Note that there are
10 possible
combinations.

4.1 - 68
Example: Tree Diagrams
A bag contains three different
colored marbles: red, blue, and
green. Suppose two marbles are
drawn from the bag and after the
first marble is drawn, it is put back
into the bag before the second
marble is drawn. Construct a tree
diagram that depicts all possible
outcomes.

4.1 - 69

4.1 - 70
Example: Computing Probability
withTree Diagrams
Use the previous example of
drawing two marbles with
replacement to compute the
probability of drawing a red marble
on the first draw and a blue marble
on the second draw.

4.1 - 71
withTree Diagrams
P(Y) = number of ways Y can occur
in the sample space
9
1
)
( 
Y
P
Y = event of drawing red first then
blue

4.1 - 72
withTree Diagrams
Consider each event separately:
3
1
)
( 
R
P
R = event of drawing red on first draw
B = event of drawing blue on second
draw
3
1
)
( 
B
P

4.1 - 73
Notation
P(A and B) =
P(event A occurs in a first trial and
event B occurs in a second trial)

4.1 - 74
Key Concept
The basic multiplication rule is used for
finding P(A and B), the probability that
event A occurs in a first trial and event
B occurs in a second trial.

4.1 - 75
withTree Diagrams
The probability of drawing red on first
draw and drawing blue on second draw
is the product of the individual
probabilities:
9
1
3
1
3
1
)
and
( 


B
R
P

4.1 - 76
Key Concept
NEXT EXAMPLE SHOWS:
If the outcome of the first event A
somehow affects the probability of the
second event B, it is important to adjust
the probability of B to reflect the
occurrence of event A.
This is Conditional Probability

4.1 - 77
A bag contains three different
colored marbles: red, blue, and
green. Suppose two marbles are
drawn from the bag without
replacing the first marble after it is
drawn. Construct a tree diagram
that depicts all possible outcomes.

4.1 - 78

4.1 - 79
withTree Diagrams
Use the previous example to
compute the probability of drawing
a red marble on the first draw and
a blue marble on the second draw.

4.1 - 80
withTree Diagrams
P(Y) = number of ways Y can occur
in the sample space
6
1
)
( 
Y
P
Y = event of drawing red first then
blue

4.1 - 81
withTree Diagrams
Multiplication Method
3
1
)
( 
R
P
R = event of drawing red on first draw
B = event of drawing blue on second
draw (given that there are now only
two marbles in the bag)
2
1
)
( 
B
P

4.1 - 82
withTree Diagrams
The probability of drawing red on first
draw and drawing blue on second draw
is the product of the individual
probabilities:
6
1
2
1
3
1
)
and
( 


B
R
P

4.1 - 83
Key Point
Without replacement, we must
adjust the probability of the
second event to reflect the
outcome of the first event.

4.1 - 84
Important Principle
The probability for the second
event B should take into account
the fact that the first event A has
already occurred.

4.1 - 85
Notation for
P(B|A) represents the probability of
event B occurring after it is assumed
that event A has already occurred (read
B|A as “B given A.”)

4.1 - 86
Multiplication Rule and
P(A and B) = P(A) • P(B|A)

4.1 - 87
Intuitive
Multiplication Rule
When finding the probability that event
A occurs in one trial and event B occurs
in the next trial, multiply the probability
of event A by the probability of event B,
but be sure that the probability of event
B takes into account the previous

4.1 - 88
If three are selected without replacement,
what is probability they all had false positive
test results?
Example
(false positive)
42
(true positive)
57
(true negative)
9
(false negative)
41
TOTALS 47 51 98

4.1 - 89
On first selection there are 98 subjects,
15 of which are false positive:
Example
98
15
)
( 1 
F
P
selection
th
on
positive
false
a
is
there
event that n
Fn 

4.1 - 90
On second selection there are 97
subjects (without replacement)
If the first selection was false positive,
there are 14 false positives left:
Example
97
14
)
|
( 1
2 
F
F
P

4.1 - 91
On third selection there are 96 subjects
(without replacement)
If the first and second selections were
false positive, there are 13 false
positives left:
Example
96
13
)
and
|
( 2
1
3 
F
F
F
P

4.1 - 92
ANSWER:
This is considered unusual since
probability is less than 0.05
Example
)
and
|
(
)
|
(
)
(
)
and
and
( 2
1
3
1
2
1
3
2
1 F
F
F
P
F
F
P
F
P
F
F
F
P 


00299
.
0
96
13
97
14
98
15





4.1 - 93
Dependent and Independent
Two events A and B are independent if
the occurrence of one does not affect
the probability of the occurrence of the
other. (Several events are similarly
independent if the occurrence of any
does not affect the probabilities of the
occurrence of the others.) If A and B
are not independent, they are said to be
dependent.

4.1 - 94
Finding that your calculator works
Finding that your computer works
Assuming your calculator and your
computer are not both running off the
same source of power, these are
independent events.
Examples

4.1 - 95
Previous example of drawing two
marbles with replacement
Since the probability of drawing the
second marble is not affected by
drawing the first marble, these are
independent events.
Examples

4.1 - 96
Dependent Events
Two events are dependent if the
occurrence of one of them affects the
probability of the occurrence of the
other, but this does not necessarily
mean that one of the events is a cause
of the other.

4.1 - 97
Previous example of drawing two
marbles without replacement
Since the probability of drawing the
second marble is affected by drawing
the first marble, these are dependent
events.
Examples

4.1 - 98
Multiplication Rule for
Independent Events
 Note that if A and B are independent
events, then P(B|A)=P(B) and the
multiplication rule is then:
P(A and B) = P(A) • P(B)

4.1 - 99
Applying the
Multiplication Rule

4.1 -
Applying the
Multiplication Rule

4.1 -
Caution
When applying the multiplication rule,
always consider whether the events
are independent or dependent, and
adjust the calculations accordingly.

4.1 -
Multiplication Rule for
Several Events
In general, the probability of any
sequence of independent events is
simply the product of their
corresponding probabilities.

4.1 -
Example
Page 169, problem 26
These events are independent since the
probability of getting a girl on any try
is not affected by the occurence of
getting a girl on a previous try.
th try
on
birth
girl
a
is
there
event that n
Gn 
2
1
)
( 
n
G
P

4.1 -
Example
ten factors of 1/2
10
2
1
2
1
...
2
1
2
1











)
(
...
)
(
)
(
)
and
...
and
and
(
10
2
1
10
2
1
G
P
G
P
G
P
G
G
G
P





4.1 -
Example
Calculator: use “hat key” to evaluate
powers:
  000977
.
0
5
.
0
2
1 10
10








10
^
5
.
0
)
5
.
0
( 10


4.1 -
Example
Since:
We see that getting 10 girls by
chance alone is unusual and
conclude that the gender selection
method is effective.
05
.
0
000977
.
0 

4.1 -
Treating Dependent Events
as Independent
Some calculations are cumbersome,
but they can be made manageable by
using the common practice of treating
events as independent when small
samples are drawn from large
populations. In such cases, it is rare to
select the same item twice (sample
with replacement).

4.1 -
The 5% Guideline for
Cumbersome Calculations
If a sample size is no more than 5% of
the size of the population, treat the
selections as being independent (even
if the selections are made without
replacement, so they are technically
dependent).

4.1 -
Example
a) If we select without replacement,
then randomly selecting an ignition
system are not independent. But
since 3/200 = 0.015 =1.5%, we could
use the 5% guideline and regard
these events as independent.

4.1 -
Example
b) If these events are not independent
(dependent) then:
)
and
|
(
)
|
(
)
(
)
and
and
( 2
1
3
1
2
1
3
2
1 G
G
G
P
G
G
P
G
P
G
G
G
P 


926
.
0
198
193
199
194
200
195





4.1 -
Example
c) If these events are independent then:
)
(
)
(
)
(
)
and
and
( 3
2
1
3
2
1 G
P
G
P
G
P
G
G
G
P 


3
200
195
200
195
200
195
200
195










  927
.
0
975
.
0
3



4.1 -
Example
d) The answer from part (b) is exact so
it is better.

4.1 -
Summary of Fundamentals
 In the addition rule, the word “or” in
P(A or B) suggests addition. Add P(A)
and P(B), being careful to add in such a
way that every outcome is counted only
once.
 In the multiplication rule, the word
“and” in P(A and B) suggests
multiplication. Multiply P(A) and P(B),
but be sure that the probability of event
B takes into account the previous

4.1 -
Recap
 Notation for P(A and B).
 Notation for conditional probability.
 Independent events.
 Formal and intuitive multiplication rules.
 Tree diagrams.

4.1 -
Section 4-5
Multiplication Rule:
Complements and

4.1 -
Key Concepts
Probability of “at least one”:
Find the probability that among several
trials, we get at least one of some
specified event.
Conditional probability:
Find the probability of an event when we
have additional information that some
other event has already occurred.

4.1 -
Complements: The Probability
of “At Least One”
 The complement of getting at least one item
of a particular type is that you get no items
of that type.
 “At least one” is equivalent to “one or
more.”

4.1 -
Example
Page 175, problems 6
If not all 6 are free from defects, that
means that at least one of them is
defective.

4.1 -
Example
Page 175, problems 8
If it is not true that at least one of the
five accepts an invitation, then all five
did not accept the invitation.

4.1 -
Finding the Probability
of “At Least One”
To find the probability of at least one of
something, calculate the probability of
none, then subtract that result from 1.
That is,
P(at least one) = 1 – P(none).

4.1 -
Example
P(at least one girl) = 1 – P(all boys)
)
(
...
)
(
)
(
1
)
and
...
and
and
(
1
)
boys
all
(
1
8
2
1
8
2
1
B
P
B
P
B
P
B
B
B
P
P








th try
on
birth
boy
a
is
there
event that n
Bn 
996
.
0
00391
.
0
1
2
1
1
8












4.1 -
Example
The probability of having 8 children and
none of them are girls (all boys) is
0.00391 which means this is a rare
event.

4.1 -
Example
P(at least one working calculator)
= 1 – P(all calculators fail)
Note: these are independent events and
fails
calculator
th
event that n
Fn 
04
.
0
96
.
0
1
)
( 


n
F
P
P(a calculator fails) = 1 – P(a calculator does not fail)

4.1 -
Example
 
998
.
0
0016
.
0
1
04
.
0
1
)
(
)
(
1
)
and
(
1
)
fail
s
calculator
all
(
1
2
2
1
2
1











F
P
F
P
F
F
P
P

4.1 -
Example
With one calculator,
P(working calculator) = 0.96
With two calculators,
P(at least one working calculator)
= 0.998
The increase in chance of a working
calculator might be worth the effort.

4.1 -
A conditional probability of an event is a
probability obtained with the additional
information that some other event has
already occurred.
P(B|A) denotes the conditional probability
of event B occurring, given that event A
has already occurred.

4.1 -
Intuitive Approach to
The conditional probability of B given A
can be found by assuming that event A
has occurred, and then calculating the
probability that event B will occur.

4.1 -
Table 4-1 on page 137 (polygraph data)
Example
Did Not Lie Did Lie
(false positive)
42
(true positive)
(true negative)
9
(false negative)

4.1 -
Example
(a)There are 47 subjects who did not lie,
32 of which had a negative test
result.
681
.
0
47
32
lie)
not
did
subject
|
result
test
negative
( 

P

4.1 -
Conditional Probability Formula
Conditional probability of event B
occurring, given that event A has already
occurred
P(B A) =
P(A and B)
P(A)

4.1 -
Example
(a)Using the formula
681
.
0
98
/
47
98
/
32
lie)
not
did
subject
(
result)
test
negative
had
and
lie
not
did
subject
(
lie)
not
did
subject
|
result
test
negative
(



P
P
P

4.1 -
Confusion of the Inverse
To incorrectly believe that P(A|B) and
P(B|A) are the same, or to incorrectly use
one value for the other, is often called
confusion of the inverse.

4.1 -
Example
(b) Using the formula
780
.
0
98
/
41
98
/
32
result)
test
negative
(
lie)
not
did
subject
and
result
test
negative
(
result)
test
negative
|
lie
not
did
subject
(



P
P
P

4.1 -
Example
(c) The results from parts (a) and (b) are
not equal.

4.1 -
Recap
 Concept of “at least one.”
 Conditional probability.
 Intuitive approach to conditional
probability.

4.1 -
Section 4-6
Counting

4.1 -
Key Concept
In many probability problems, the big obstacle
is finding the total number of outcomes, and
this section presents several methods for
finding such numbers without directly listing
and counting the possibilities.

4.1 -
Fundamental Counting Rule
For a sequence of two events in which
the first event can occur m ways and
the second event can occur n ways,
the events together can occur a total of
m n ways.

4.1 -
Example
A coin is flipped and then a die is
rolled. What are the total number of
outcomes?

4.1 -
Example
A coin is flipped and then a die is
rolled. What are the total number of
outcomes?
ANSWER: There are 2 outcomes for
the coin flip and 6 outcomes for the die
roll. Total number of outcomes:
12
6
2 


4.1 -
Example
Page 185, Problem 32 (a)
If 20 newborn babies are randomly
selected, how many different gender
sequences are possible?

4.1 -
Example
ANSWER:
This is a sequence of 20 events each of
which has two possible outcomes (boy or
girl). By the fundamental counting rule, total
number of gender sequences will be:
576
,
048
,
1
2
2
2
2
2 20








4.1 -
Example
Page 185, Problem 28
A safe combination consists of four
numbers between 0 and 99. If four
numbers are randomly selected, what
is the probability of getting the correct
combination on the first attempt?

4.1 -
Example
ANSWER:
Total number of possible combinations is
Since there is only one correct combination:
It is not feasible to try opening the safe this
way.
000
,
000
,
100
100
100
100
100
100 4





00000001
.
0
000
,
000
,
100
1
n)
combinatio
correct
( 

P

4.1 -
Notation
The factorial symbol ! denotes the product of
decreasing positive whole numbers.
For example,
4! 4 3 2 1 24.
    
By special definition, 0! = 1.

4.1 -
A collection of n different items can be
arranged in order n! different ways.
(This factorial rule reflects the fact that
the first item may be selected in n
different ways, the second item may be
selected in n – 1 ways, and so on.)
Factorial Rule

4.1 -
Example
Page 183, Problem 6
Find the number of different ways that
the nine players on a baseball team
can line up for the National Anthem by
evaluating 9 factorial.
880
,
362
1
2
3
4
5
6
7
8
9
!
9 










4.1 -
Factorial on Calculator
Calculator
9 MATH PRB 4:!
To get:
Then Enter gives the result
!
9

4.1 -
Permutations Rule
(when items are all different)
(n - r)!
n r
P =
n!
If the preceding requirements are satisfied, the number of
permutations (or sequences) of r items selected from n
available items (without replacement) is
Requirements:
1. There are n different items available. (This rule does not
apply if some of the items are identical to others.)
2. We select r of the n items (without replacement).
3. We consider rearrangements of the same items to be
different sequences. (The permutation of ABC is different
from CBA and is counted separately.)

4.1 -
Example
A bag contains 4 colored marbles: red,
blue, green, yellow. If we select 3 of
the four marbles from the bag without
replacement, how many different color
orders are there?

4.1 -
Example
ANSWER:
24
1
1
2
3
4
!
1
!
4
)!
3
4
(
!
4
3
4








P

4.1 -
Example
In horse racing, a trifecta is a bet that
the first three finishers in a race are
selected, and they are selected in the
correct order. Find the number of
different possible trifecta bets in a race
with ten horses by evaluating
3
10 P

4.1 -
Example
ANSWER:
720
1
8
9
10
1
2
3
4
5
6
7
1
2
3
4
5
6
7
8
9
10
!
7
!
10
)!
3
10
(
!
10
3
10























P

4.1 -
Permutations on Calculator
Calculator
10 MATH PRB 2:nPr 3
To get:
3
10 P

4.1 -
Permutations Rule
(when some items are identical to others)
n1! . n2! .. . . . . . . nk!
n!
If the preceding requirements are satisfied, and if there are n1
alike, n2 alike, . . . nk alike, the number of permutations (or
sequences) of all items selected without replacement is
Requirements:
1. There are n items available, and some items are identical to
others.
2. We select all of the n items (without replacement).
3. We consider rearrangements of distinct items to be different
sequences.

4.1 -
Example
Find the number of different ways the
letters AGGYB can be arranged.

4.1 -
Example
ANSWER: there are two letters of the
five that are alike. Then the total
number of arrangements will be
60
1
2
1
2
3
4
5
!
1
!
1
!
1
!
2
!
5











4.1 -
Example
Page 185, Problem 32 (b)
How many different ways can 10 girls
and 10 boys be arranged in a
sequence?

4.1 -
Example
ANSWER: there are twenty total children in
the arrangement. There are two groups of
10 that are alike. This gives a total number
of arrangements:
756
,
184
!
10
!
10
!
20



4.1 -
Example
Page 185, Problem 32 (c)
What is the probability of getting 10
girls and 10 boys when twenty babies
are born?

4.1 -
Example
ANSWER: as found on previous slides
The total number of boy/girl arrangements of
20 newborns is:
The total number of ways 10 girls and 10
boys can be arranged in a sequence is
576
,
048
,
1
220

756
,
184
!
10
!
10
!
20



4.1 -
Example
ANSWER:
176
.
0
576
,
048
,
1
756
,
184
born)
are
babies
20
when
boys
10
and
girls
10
(


P

4.1 -
(n - r )! r!
n!
nCr =
Combinations Rule
If the preceding requirements are satisfied, the number of
combinations of r items selected from n different items is
Requirements:
1. There are n different items available.
2. We select r of the n items (without replacement).
3. We consider rearrangements of the same items to be the
same. (The combination of ABC is the same as CBA.)

4.1 -
Example
Page 183, Problem 8
Find the number of different possible 5
card poker hands by evaluating
5
52C

4.1 -
Example
Page 183, Problem 8
ANSWER:
960
,
598
,
2
!
5
48
49
50
51
52
!
5
!
47
!
52
!
5
)!
5
52
(
!
52
5
52











C

4.1 -
Combinations on Calculator
Calculator
52 MATH PRB 3:nCr 5
To get:
5
52C

4.1 -
Example
What is the probability of winning a
lottery with one ticket if you select five
winning numbers from 1,2,…,31.
Note: numbers selected are different
and order does not matter.

4.1 -
Example
ANSWER:
Total number of possible combinations is:
Since only one combination wins:
911
,
169
!
5
!
26
!
31
5
31 


C
00000589
.
0
911
,
169
1
winning)
( 

P

4.1 -
Example
What is the probability of winning a
lottery with one ticket if you select five
winning numbers from 1,2,…,31.
Note: assume now that you must
select the numbers in the same order
they were drawn so that order does
matter.

4.1 -
Example
ANSWER:
Total number of possible combinations is:
Since only one permutation wins:
320
,
389
,
20
!
26
!
31
5
31 

P
0000000490
.
0
320
,
389
,
20
1
winning)
( 

P

4.1 -
When different orderings of the same
items are to be counted separately, we
have a permutation problem, but when
different orderings are not to be counted
separately, we have a combination
problem.
Permutations versus
Combinations

4.1 -
Recap
 The fundamental counting rule.
 The permutations rule (when items are all
different).
 The permutations rule (when some items
are identical to others).
 The combinations rule.
 The factorial rule.

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