### Chapter 4 Slides

```Chapter 4
Probability
4-1 Review and Preview
4-2 Basic Concepts of Probability
4-4 Multiplication Rule: Basics
4-5 Multiplication Rule: Complements and
Conditional Probability
4-6 Counting
4.1 - 1
Section 4-1
Review and Preview
4.1 - 2
Review
Necessity of sound sampling methods.
Common measures of characteristics of
data
Mean
Standard deviation
4.1 - 3
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 - 4
Section 4-2
Basic Concepts of
Probability
4.1 - 5
Part 1
Basics of Probability
4.1 - 6
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 - 7
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 - 8
Example

Which elements of the sample space correspond
to the event that the sum of each dice is 4?
3,1 2,2 1,3
4.1 - 9
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 - 10
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 - 11
Example
Problem 20 on page 149
F = event of a false negative on polygraph
test
9
P( F ) 
 0.0918
98
Thus this is not considered unusual since
it is more than 0.001 (see page 146). The
test is not highly accurate.
4.1 - 12
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 - 13
Example
Problem 21 on page 149
F = event of a selecting a female senator
16
16
P( F ) 

 0.160
84  16 100
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.
4.1 - 14
Example
Problem 28 on page 150
A = event that Delta airlines passenger is
involuntarily bumped from a flight
3
P( A) 
 0.000195
15378
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.
4.1 - 15
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 - 16
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:
3
P( A) 
 0.0833
36
4.1 - 17
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):
HHH, HHT, HTH , HTT , THH , THT , TTH , TTT
4.1 - 18
Example
Sample space has 8 simple events.
Event A corresponds to TTT only so that:
1
P ( A)   0.125
8
4.1 - 19
Example
Problem 36 on page 151
Let:
S = event that son inherits disease (xY or Yx)
D = event that daughter inherits disease (xx)
4.1 - 20
Example
Problem 36 on page 151
(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
P( S )   0
2
4.1 - 21
Example
Problem 36 on page 151
(b) Father: xY Mother: XX
Sample space for a daughter:
xX xX
Sample space has no simple
events that represent a daughter that
has the disease so:
0
P( D)   0
2
4.1 - 22
Example
Problem 36 on page 151
(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:
1
P( S )   0.5
2
4.1 - 23
Example
Problem 36 on page 151
(d) Father: XY Mother: xX
Sample space for a daughter:
Xx XX
Sample space has no simple
event that represents a daughter that
has the disease so:
0
P( D)   0
2
4.1 - 24
Example
Problem 18 on page 149
Table 4-1 on page 137 (polygraph data)
Did Not Lie
Did Lie
Positive Test Result
15
(false positive)
42
(true positive)
NegativeTest Result
32
(true negative)
9
(false negative)
4.1 - 25
Example
It is helpful to first total the data in the table:
TOTALS
Did Not Lie
Did Lie
Positive Test Result
15
(false positive)
42
(true positive)
57
NegativeTest Result
32
(true negative)
9
(false negative)
41
TOTALS
47
51
98
(a) How many responses were lies:
4.1 - 26
Example
(b) If one response is randomly selected, what
is the probability it is a lie?
L = event of selecting one of the lie responses
51
P ( L) 
98
(c)
51
 0.520
98
4.1 - 27
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 - 28
Example
Problem 4 on page 148
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)
4.1 - 29
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 - 30
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 - 31
Probability Limits
Always express a probability as a fraction or
decimal number between 0 and 1.
 The probability of an impossible event is 0.
 The probability of an event that is certain to
occur is 1.
 For any event A, the probability of A is
between 0 and 1 inclusive. That is:
0  P(A)  1
4.1 - 32
Possible Values
for Probabilities
4.1 - 33
Complementary Events
The complement of event A, denoted by
A, consists of all outcomes in which the
event A does not occur.
4.1 - 34
Example
If a fair coin is tossed three times and
A = event that exactly one heads occurs
Find the complement of A.
4.1 - 35
Example
Sample space:
HHH, HHT, HTH , HTT , THH , THT , TTH , TTT
Event A corresponds to HTT, THT, TTH
Therefore, the complement of A are the
simple events:
HHH, HHT, HTH, THH, TTT
4.1 - 36
Part 2
Beyond the
Basics of Probability: Odds
4.1 - 37
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 - 38
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 - 39
Example
Problem 38, page 149
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:
20
P (W ) 
38
(b) Odds against winning are
P(W ) 20 / 38 20


 10 / 9 or 10 : 9
P(W ) 18 / 38 18
4.1 - 40
Example
Problem 38, page 149
(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:
net profit 1

18
1
The casino returns \$18+\$18=\$36 to you.
4.1 - 41
Example
Problem 38, page 149
(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:
net profit 10

18
9
The casino returns \$18+\$20=\$38 to you.
4.1 - 42
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 - 43
Section 4-3
4.1 - 44
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 - 45
Compound Event
Compound Event
any event combining 2 or more simple events
Notation
P(A or B) = P (in a single trial, event A occurs
or event B occurs or they both occur)
4.1 - 46
General Rule for a
Compound Event
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.
4.1 - 47
Example
A random survey of members of the class of
2005 finds the following:
Number of
students who
Number of students
TOTALS
Women
672
22
694
Men
582
19
601
TOTALS
1254
41
1295
What is the probability the student did not
4.1 - 48
Example
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
4.1 - 49
Example
The probability the student did not graduate
or was a man :
623/1295 = 0.481
4.1 - 50
Compound Event
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 - 51
Previous Example
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
4.1 - 52
Example
Problem 20 on page 157
TOTALS
Did Not Lie
Did Lie
Positive Test Result
15
(false positive)
42
(true positive)
57
NegativeTest Result
32
(true negative)
9
(false negative)
41
TOTALS
47
51
98
What is the probability the subject had a
negative test result or lied?
4.1 - 53
Example (cont.)
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
4.1 - 54
Example
Problem 30 on page 158
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
4.1 - 55
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:
N = students who did not graduate
N and G are disjoint because no student
4.1 - 56
Disjoint or Mutually Exclusive
Events A and B are not disjoint if they
overlap.
Previous Example:
M = male students
M and G are not disjoint because some
students who graduated are also male
4.1 - 57
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
4.1 - 58
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 - 59
Complementary
Events
It is impossible for an event and its
complement to occur at the same time.
That is, for any event A,
A and A
are disjoint
4.1 - 60
Probability Rule of
Complementary Events
P(A) + P(A) = 1
P(A) = 1 – P(A)
P(A) = 1 – P(A)
4.1 - 61
Venn Diagram for the
Complement of Event A
4.1 - 62
Example
Problem 16 on page 157
P(I ) is the probability that a screened
driver is not intoxicated
P( I )  1  P( I )  1  0.00888 0.991
4.1 - 63
Recap
In this section we have discussed:
 Compound events.
 Disjoint events.
 Complementary events.
4.1 - 64
Section 4-4
Multiplication Rule:
Basics
4.1 - 65
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
determining the number of possible
outcomes in a sample space, if the
number of possibilities is not too
large.
4.1 - 66
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 - 67
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 - 68
Example: Tree Diagrams
4.1 - 69
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 - 70
Example: Computing Probability
withTree Diagrams
Y = event of drawing red first then
blue
P(Y) =
number of ways Y can occur
number of different simple events
in the sample space
1
P(Y ) 
9
4.1 - 71
Example: Computing Probability
withTree Diagrams
Consider each event separately:
R = event of drawing red on first draw
1
P( R) 
3
B = event of drawing blue on second
draw
1
P( B) 
3
4.1 - 72
Notation
P(A and B) =
P(event A occurs in a first trial and
event B occurs in a second trial)
4.1 - 73
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 - 74
Example: Computing Probability
withTree Diagrams
The probability of drawing red on first
draw and drawing blue on second draw
is the product of the individual
probabilities:
1 1 1
P( R and B)   
3 3 9
4.1 - 75
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 - 76
Example: Tree Diagrams
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 - 77
Example: Tree Diagrams
4.1 - 78
Example: Computing Probability
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 - 79
Example: Computing Probability
withTree Diagrams
Y = event of drawing red first then
blue
P(Y) =
number of ways Y can occur
number of different simple events
in the sample space
1
P(Y ) 
6
4.1 - 80
Example: Computing Probability
withTree Diagrams
Multiplication Method
R = event of drawing red on first draw
1
P( R) 
3
B = event of drawing blue on second
draw (given that there are now only
two marbles in the bag)
1
P( B) 
2
4.1 - 81
Example: Computing Probability
withTree Diagrams
The probability of drawing red on first
draw and drawing blue on second draw
is the product of the individual
probabilities:
1 1 1
P ( R and B )   
3 2 6
4.1 - 82
Conditional Probability
Key Point
Without replacement, we must
second event to reflect the
outcome of the first event.
4.1 - 83
Conditional Probability
Important Principle
The probability for the second
event B should take into account
the fact that the first event A has
4.1 - 84
Notation for
Conditional Probability
P(B|A) represents the probability of
event B occurring after it is assumed
B|A as “B given A.”)
4.1 - 85
Multiplication Rule and
Conditional Probability
P(A and B) = P(A) • P(B|A)
4.1 - 86
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
occurrence of event A.
4.1 - 87
Example
Problem 14 on page 168
TOTALS
Did Not Lie
Did Lie
Positive Test Result
15
(false positive)
42
(true positive)
57
NegativeTest Result
32
(true negative)
9
(false negative)
41
TOTALS
47
51
98
If three are selected without replacement,
what is probability they all had false positive
test results?
4.1 - 88
Example
Problem 14 on page 168
Fn  event thatthereis a false positiveon nth selection
On first selection there are 98 subjects,
15 of which are false positive:
15
P( F1 ) 
98
4.1 - 89
Example
Problem 14 on page 168
On second selection there are 97
subjects (without replacement)
If the first selection was false positive,
there are 14 false positives left:
14
P( F2 | F1 ) 
97
4.1 - 90
Example
Problem 14 on page 168
On third selection there are 96 subjects
(without replacement)
If the first and second selections were
false positive, there are 13 false
positives left:
13
P( F3 | F1 and F2 ) 
96
4.1 - 91
Example
Problem 14 on page 168
P( F1 and F2 and F3 )  P( F1 )  P( F2 | F1 )  P( F3 | F1 and F2 )
15 14 13
    0.00299
98 97 96
This is considered unusual since
probability is less than 0.05
4.1 - 92
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 - 93
Examples
Problem 10 on page 168
computer are not both running off the
same source of power, these are
independent events.
4.1 - 94
Examples
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.
4.1 - 95
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 - 96
Examples
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.
4.1 - 97
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 - 98
Applying the
Multiplication Rule
4.1 - 99
Applying the
Multiplication Rule
4.1 -
Caution
When applying the multiplication rule,
always consider whether the events
are independent or dependent, and
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.
Gn  event thatthereis a girl birth on nth try
1
P (G n ) 
2
4.1 -
Example
Page 169, problem 26
P(G1 and G2 and ... and G10 )
 P(G1 )  P(G2 )  ... P(G10 )
10
1 1
1 1
   ...   
2 2
2 2
ten factors of 1/2
4.1 -
Example
Page 169, problem 26
10
1
10
   0.5  0.000977
2
Calculator: use “hat key” to evaluate
powers:
(0.5)  0.5 ^ 10
10
4.1 -
Example
Page 169, problem 26
Since:
0.000977  0.05
We see that getting 10 girls by
chance alone is unusual and
conclude that the gender selection
method is effective.
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
Page 169, problem 30
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
Page 169, problem 30
b) If these events are not independent
(dependent) then:
P(G1 and G2 and G3 )  P(G1 )  P(G2 | G1 )  P(G3 | G1 and G2 )
195 194 193



 0.926
200 199 198
4.1 -
Example
Page 169, problem 30
c) If these events are independent then:
P(G1 and G2 and G3 )  P(G1 )  P(G2 )  P(G3 )
195 195 195  195 





200 200 200  200
3
 0.975  0.927
3
4.1 -
Example
Page 169, problem 30
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
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
occurrence of event A.
4.1 -
Recap
In this section we have discussed:
 Notation for P(A and B).
 Tree diagrams.
 Notation for conditional probability.
 Independent events.
 Formal and intuitive multiplication rules.
4.1 -
Section 4-5
Multiplication Rule:
Complements and
Conditional Probability
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
4.1 -
Complements: The Probability
of “At Least One”
 “At least one” is equivalent to “one or
more.”
 The complement of getting at least one item
of a particular type is that you get no items
of that type.
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
Page 175, problem 10
P(at least one girl) = 1 – P(all boys)
Bn  event thatthereis a boy birth on nth try
1  P(all boys)  1  P( B1 and B2 and ... and B8 )
 1  P( B1 )  P( B2 )  ... P( B8 )
8
1
 1     1  0.00391
2
 0.996
4.1 -
Example
Page 175, problem 10
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
Page 175, problem 12
P(at least one working calculator)
= 1 – P(all calculators fail)
Fn  event thatnth calculatorfails
Note: these are independent events and
P(a calculator fails) = 1 – P(a calculator does not fail)
P( Fn )  1  0.96  0.04
4.1 -
Example
Page 175, problem 12
1  P(all calculators fail)  1  P( F1 and F2 )
 1  P( F1 )  P( F2 )
 1  0.04  1  0.0016
2
 0.998
4.1 -
Example
Page 175, problem 12
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 -
Conditional Probability
A conditional probability of an event is a
information that some other event has
P(B|A) denotes the conditional probability
of event B occurring, given that event A
4.1 -
Intuitive Approach to
Conditional Probability
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 -
Example
Problem 22 on page 175
Table 4-1 on page 137 (polygraph data)
Did Not Lie
Did Lie
Positive Test Result
15
(false positive)
42
(true positive)
NegativeTest Result
32
(true negative)
9
(false negative)
4.1 -
Example
Page 175, problem 22
(a)There are 47 subjects who did not lie,
32 of which had a negative test
result.
32
P(negative test result | subject did not lie) 
 0.681
47
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
Page 175, problem 22
(a)Using the formula
P (negat ivetest result | subject did not lie)
P (subject did not lie and had negat ivetest result )

P (subject did not lie)
32 / 98

47 / 98
 0.681
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
Page 175, problem 22
(b) Using the formula
P(subject did not lie | negat ivetest result )
P(negat ivetest result and subject did not lie)

P(negat ivetest result )
32 / 98

41/ 98
 0.780
4.1 -
Example
Page 175, problem 22
(c) The results from parts (a) and (b) are
not equal.
4.1 -
Recap
In this section we have discussed:
 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?
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:
2  6  12
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
Page 185, Problem 32 (a)
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:
2  2  2    2  2  1,048,576
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
Page 185, Problem 28
Total number of possible combinations is
100100100100  100  100,000,000
4
Since there is only one correct combination:
1
P(correctcombination) 
 0.00000001
100,000,000
It is not feasible to try opening the safe this
way.
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 -
Factorial Rule
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.)
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.
9! 9  8  7  6  5  4  3  2 1  362,880
4.1 -
Factorial on Calculator
Calculator
9
MATH
PRB
4:!
To get: 9!
Then Enter gives the result
4.1 -
Permutations Rule
(when items are all different)
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.)
If the preceding requirements are satisfied, the number of
permutations (or sequences) of r items selected from n
available items (without replacement) is
nPr =
n!
(n - r)!
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
4!
4 P3 
(4  3)!
4! 4  3  2 1
 
 24
1!
1
4.1 -
Example
Page 183, Problem 12
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
10
P3
4.1 -
Example
Page 183, Problem 12
10!
10 P3 
(10  3)!
10! 10  9  8  7  6  5  4  3  2 1


7!
7  6  5  4  3  2 1
10  9  8

 720
1
4.1 -
Permutations on Calculator
Calculator
10
MATH
To get:
10
PRB
2:nPr
3
P3
Then Enter gives the result
4.1 -
Permutations Rule
(when some items are identical to others)
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.
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
n!
n1! . n2! .. . . . . . . nk!
4.1 -
Example
Page 185, Problem 26
Find the number of different ways the
letters AGGYB can be arranged.
4.1 -
Example
Page 185, Problem 26
ANSWER: there are two letters of the
five that are alike. Then the total
number of arrangements will be
5!
5  4  3  2 1

 60
2!1!1!1!
2 1
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
Page 185, Problem 32 (b)
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:
20!
 184 ,756
10!10!
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
Page 185, Problem 32 (c)
ANSWER: as found on previous slides
The total number of boy/girl arrangements of
20 newborns is:
2  1,048,576
20
The total number of ways 10 girls and 10
boys can be arranged in a sequence is
20!
 184 ,756
10!10!
4.1 -
Example
Page 185, Problem 32 (c)
P(10 girls and10 boys when 20 babies are born)
184,756

 0.176
1,048,576
4.1 -
Combinations Rule
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.)
If the preceding requirements are satisfied, the number of
combinations of r items selected from n different items is
n!
nCr = (n - r )! r!
4.1 -
Example
Page 183, Problem 8
Find the number of different possible 5
card poker hands by evaluating
52
C5
4.1 -
Example
Page 183, Problem 8
52!
52 C5 
(52  5)!5!
52! 52  51 50  49  48


47!5!
5!
 2,598,960
4.1 -
Combinations on Calculator
Calculator
52
MATH
To get:
52
PRB
3:nCr
5
C5
Then Enter gives the result
4.1 -
Example
Page 184, Problem 16 (a)
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
Page 184, Problem 16 (a)
Total number of possible combinations is:
31!
 169 ,911
31 C5 
26!5!
Since only one combination wins:
1
P( winning) 
 0.00000589
169,911
4.1 -
Example
Page 184, Problem 16 (b)
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
Page 184, Problem 16 (b)
Total number of possible combinations is:
31!
 20,389 ,320
31 P5 
26!
Since only one permutation wins:
1
P( winning) 
 0.0000000490
20,389,320
4.1 -
Permutations versus
Combinations
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.
4.1 -
Recap
In this section we have discussed:
 The fundamental counting rule.
 The factorial rule.
 The permutations rule (when items are all
different).
 The permutations rule (when some items
are identical to others).
 The combinations rule.