Introductory maths analysis chapter 08 official

INTRODUCTORY MATHEMATICALINTRODUCTORY MATHEMATICAL
ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
©2007 Pearson Education Asia
Chapter 8Chapter 8
Introduction to Probability and StatisticsIntroduction to Probability and Statistics

INTRODUCTORY MATHEMATICAL
ANALYSIS
0. Review of Algebra
1. Applications and More Algebra
2. Functions and Graphs
3. Lines, Parabolas, and Systems
4. Exponential and Logarithmic Functions
5. Mathematics of Finance
6. Matrix Algebra
7. Linear Programming
8. Introduction to Probability and
Statistics

9. Additional Topics in Probability
10. Limits and Continuity
11. Differentiation
12. Additional Differentiation Topics
13. Curve Sketching
14. Integration
15. Methods and Applications of Integration
16. Continuous Random Variables
17. Multivariable Calculus
INTRODUCTORY MATHEMATICAL
ANALYSIS

• To develop and apply a Basic Counting
Principle.
• Combinations and permutations.
• To determine a sample space.
• To define what is meant by the probability of an
event.
• To discuss conditional probability.
• To develop the notion of independent events.
• To develop Bayes’s formula.
Chapter 8: Introduction to Probability and Statistics
Chapter ObjectivesChapter Objectives

Basic Counting Principle and Permutations
Combinations and Other Counting Principles
Sample Spaces and Events
Probability
Conditional Probability and Stochastic Processes
Independent Events
Bayes’ Formula
8.1)
8.2)
8.3)
8.4)
Chapter OutlineChapter Outline
8.5)
8.6)
8.7)

8.1 Basic Counting Principle and Permutations8.1 Basic Counting Principle and Permutations
Example 1 – Travel Routes
Basic Counting Principle
• The total number of different ways a sequence can
occur is .knnn 21 ⋅
To drive from A, to B, to C, and then to city D, how
many different routes are possible?
Solution:
Total number of routes is 40542 =⋅⋅

8.1 Basic Counting Principle and Permutations
Example 3 – Answering a Quiz
In how many different ways can a quiz be answered
under each of the following conditions?
a. The quiz consists of three multiple-choice
questions with four choices for each.
Solution:
b. The quiz consists of three multiple-choice
questions (with four choices for each) and five
true–false questions.
Solution:
644444 3
==⋅⋅
( )( ) 20482422222444 53
=⋅=⋅⋅⋅⋅⋅⋅

Permutations
• An ordered selection of r objects, without
repetition, is a permutation of n objects, taken r at
a time.
• The number of permutations is denoted nPr .

Example 5 – Club Officers
A club has 20 members. The offices of president,
vice president, secretary, and treasurer are to be
filled. No member may serve in more than one
office. How many different slates of candidates are
possible?
Solution 1:
Solution 2:
( )( ) ( )121 +−−−= rnnnnPrn 
( )
280,11617181920
!16
!1617181920
!16
!20
!420
!20
420
=⋅⋅⋅=
⋅⋅⋅⋅
==
−
=P
280,11617181920420 =⋅⋅⋅=P

Example 7 – Name of a Legal Firm
• The number of permutations of n objects taken all
at a time is .
Lawyers Smith, Jones, Jacobs, and Bell want to form
a legal firm and will name it by using all four of their
last names. How many possible names are there?
Solution: Possible names for the firm,
( )
!
1
!
!0
!
!
!
n
nn
nn
n
Pnn ===
−
=
241234!4 =⋅⋅⋅=

8.2 Combinations and Other Counting Principles8.2 Combinations and Other Counting Principles
Example 1 – Comparing Combinations and Permutations
Combinations
• A combination of n objects taken r at a time is
denoted by
List all combinations and all permutations of the
four letters when they are taken three at a time.
Solution:
Combinations:
Permutations: 24
( )!!
!
rnr
n
Crn
−
=
434 =C

8.2 Combinations and Other Counting Principles
Example 3 – Poker Hand
Example 5 – A Basic Combinatorial Identity
A poker hand consists of 5 cards dealt from an
ordinary deck of 52 cards. How many different
poker hands are there?
Solution: Number of possible hands,
( )
960,598,2
!47!5
!52
!552!5
!52
552 ==
−
=C
Establish the identity
Solution:
111 +++ =+ rnrnrn CCC
( ) ( ) ( )
( )
( ) ( ) ( )( ) 11
1
!11!1
!1
!1!1
!
!!
!
++
+
=
+−++
+
=
−−+
+
−
=+
rn
rnrn
C
rnr
n
rnr
n
rnr
n
CC

Permutations with Repeated Objects
• The number of distinguishable permutations
such that n1 are of one type, n2 are of a second
type, …, and nk are of a kth type, where
n1 + n2 + … + nk = n, is !!...!
!
21 knnn
n

Example 7 – Name of a Legal Firm
A group of four lawyers, Smith, Jones, Smith, and
Bell (the Smiths are cousins), want to form a legal
firm and will name it by using all of their last names.
How many possible names exist?
Solution:
The number of distinguishable names is 12
!1!1!2
!4
=

Example 9 – Art Exhibit
An artist has created 20 original paintings, and she
will exhibit some of them in three galleries. Four
paintings will be sent to gallery A, four to gallery B,
and three to gallery C. In how many ways can this be
done?
Method 1:
Method 2:
Method 3:
000,938,939,1
!9!3!4!4
!20
=
00,938,939,1
!3!4!4
!11
!9!11
!20
!3!4!4
!11
1120 =⋅==C
!9!3!4!4
!20
!9!3
!12
!12!4
!16
!16!4
!20
312416420 =⋅⋅=⋅⋅ CCC

8.3 Sample Spaces and Events8.3 Sample Spaces and Events
Example 1 – Sample Space: Toss of Two Coins
Sample Spaces
• A sample space S is the set of all possible
outcomes.
• The number of sample points is denoted #(S).
Two different coins are tossed, and the result (H or
T) for each coin is observed. Determine a sample
space.
Solution: { }TTTH,HT,HH,=S

8.3 Sample Spaces and Events
Example 3 – Sample Space: Jelly Beans in a Bag
A bag contains four jelly beans: one red, one pink,
one black, and one white.
a.A jelly bean is withdrawn at random, its color is
noted, and it is put back in the bag. Then a jelly
bean is again randomly withdrawn and its color
noted. Describe a sample space and determine
the number of sample points.
Solution: Sample Space:
Sample Points:
{ }WWRB,PB,RW,=S
1644 =⋅

Example 3 – Sample Space: Jelly Beans in a Bag
b. Determine the number of sample points in the
sample space if two jelly beans are selected in
succession without replacement and the colors
are noted.
Solution: Sample Points: or1234 =⋅ 1224 =P

Example 5 – Sample Space: Roll of Two Dice
A pair of dice is rolled once, and for each die, the
number that turns up is observed. Determine the
number of sample points.
Solution:
Sample Points: 6 · 6 = 36
Events
• Event E is a subset of the sample space for the
experiment.

Example 7 – Complement, Union, Intersection
Given the usual sample space
for the rolling of a die, let E, F, and G be the events
Determine each of the following events.
Solution:
a. Complement, E’ 
b. Union: E ∪ F 
c. Intersect: E ∩ F 
d. Intersect: F ∩ G 
e. Union: E ∪ E’ 
f. Intersect: E ∩ E’ 
{ }65,4,3,2,1,=S
{ } { } { }165,4,3,53,1, === GFE
{ }64,2,'=E
{ }65,4,3,,1=∪ FE
{ }53,=∩ FE
φGF =∩
{ } { } { } SEE ==∪=∪ 65,4,3,2,1,64,2,53,1,'
{ } { } φEE =∩=∩ 64,2,53,1,'

Properties of Events

8.4 Probability8.4 Probability
Equiprobable Spaces
• S is called an equiprobable space if all events are
equally likely to occur.
• Probability of the simple event is
• If S is a finite equiprobable space, probability of E
is
( )
N
sP i
1
=
( ) ( ) ( ) ( )jsPsPsPEP +++= ...21
( ) ( )
( )S
E
EP
#
#
=

8.4 Probability
Example 1 – Coin Tossing
Two fair coins are tossed,
Determine the probability that
a. two heads occur
b. at least one head occurs
Solution:
a. E = {HH}, probability is
b. F = {at least one head} where
Thus probability is
( ) ( )
( ) 4
1
#
#
==
S
E
EP
{ }TTTH,HT,HH,=S
{ }THHT,HH,=F
( ) ( )
( ) 4
3
#
#
==
S
F
FP

8.4 Probability
Example 3 – Full House Poker Hand
Find the probability of being dealt a full house in a
poker game. A full house is three of one kind and
two of another, such as three queens and two 10’s.
Express your answer in terms of nCr .
Solution: ( ) ( )
( ) 553
2434 1213
#
#
housefull
C
CC
S
E
P
⋅⋅⋅
==

8.4 Probability
Example 5 – Quality Control
From a production run of 5000 light bulbs, 2% of
which are defective, 1 bulb is selected at random.
What is the probability that the bulb is defective?
What is the probability that it is not defective?
Solution:
The number of outcomes in E is 0.02 · 5000 = 100.
Alternatively, probability (defective) is
Probability (not defective) is
( ) ( )
( )
02.0
5000
100
#
#
===
S
E
EP
( ) 02.0
5000
1
100 =





=EP
( ) ( ) 98.002.011' =−=−= EPEP

8.4 Probability
Example 7 – Interrupted Gambling
Obtain Pascal and Fermat’s solution to the problem
of dividing the pot between two gamblers in an
interrupted game of chance, as described in the
introduction to this chapter. Recall that when the
game was interrupted, Player 1 needed r more
“rounds” to win the pot outright and that Player 2
needed s more rounds to win. It is agreed that the
pot should be divided so that each player gets the
value of the pot multiplied by the probability that he
or she would have won an uninterrupted game.

8.4 Probability
Example 7 – Interrupted Gambling
Solution:
Probability that Player 1 will win is given by
Number of these outcomes which consist of k T’s is
the number of ways of choosing k from among n.
( ) ( ) ( ) ( ) ( )∑
−
=
−− =+++=∪∪∪
1
0
110110 ......
s
k
kss EPEPEPEPEEEP
∑
−
=
1
0 2
s
n
n
knC

8.4 Probability
P is a probability function, if both of the following
are true:
Odds
• The odds in favor of event E occurring are the
ratio
Finding Probability from Odds
• If the odds that event E occurs are a : b, then
( )
( )'EP
EP
( )
ba
a
EP
+
=
• 0 ≤ P(si) ≤ 1 for i = 1 to N
• P(s1) + P(s2) + ·· ·+ P(sN) = 1

8.4 Probability
Example 9 – Odds for an A in an Exam
A student believes that the probability of getting an
A on the next mathematics exam is 0.2. What are
the odds (in favor) of this occurring?
Solution:
The odds of getting an A are
( )
( )
4:1
4
1
8.0
2.0
'
===
EP
EP

8.5 Conditional Probability and Stochastic Processes8.5 Conditional Probability and Stochastic Processes
Example 1 – Jelly Beans in a Bag
Conditional Probability
• If E and F are events associated with an
equiprobable sample space and F = , then∅
( ) ( )
( )F
FE
FEP
#
# ∩
=
A bag contains two blue jelly beans (say, B1 and B2)
and two white jelly beans (W1 and W2). If two jelly
beans are randomly taken from the bag, without
replacement, find the probability that the second jelly
bean taken is white, given that the first one is blue.

8.5 Conditional Probability and Stochastic Processes
Solution:
Event W ∩ B consists of the outcomes in B for which
the second jelly bean is white:
( )
3
2
6
4
==BWP
Conditional probability of an event E is given as
( ) ( )
( )
( ) 0and ≠
∩
= FP
FP
FEP
FEP

After the initial production run of a new style of steel
desk, a quality control technician found that 40% of
the desks had an alignment problem and 10% had
both a defective paint job and an alignment problem.
If a desk is randomly selected from this run, and it
has an alignment problem, what is the probability that
it also has a defective paint job?
Solution:
Let A and D be the events
We have P(A) = 0.4 and P(D ∩ A) = 0.1, thus
( ) ( )
( ) 4
1
4.0
1.0
==
∩
=
AP
ADP
ADP

Example 5 – Advertising
A computer hardware company placed an ad for its
new modem in a popular computer magazine. The
company believes that the ad will be read by 32% of
the magazine’s readers and that 2% of those who
read the ad will buy the modem. Assume that this is
true, and find the probability that a reader of the
magazine will read the ad and buy the modem.
General Multiplication Law
( ) ( ) ( ) ( ) ( )FEPFPEFPEPFEP ==∩

Example 5 – Advertising
Example 7 – Cards
Solution:
R is “read ad” and B is “buy modem”, thus
( ) ( ) ( ) ( )( ) 0064.002.032.0 ===∩ RBPRPBRP
Two cards are drawn without replacement from a
standard deck of cards. Find the probability that both
cards are red.
Solution:
The desired probability is
( ) ( ) ( )
102
25
51
25
52
26
12121 =⋅==∩ RRPRPRRP

Bag I contains one black and two red jelly beans,
and Bag II contains one pink jelly bean. A bag is
selected at random. Then a jelly bean is randomly
taken from it and placed in the other bag. A jelly
bean is then randomly taken from that bag. Find the
probability that this jelly bean is pink.

Solution:
This is a compound experiment with three trials:
a. Select a bag
b. Taking a jelly bean out
c. Putting it in the other bag and then taking a jelly
bean from that bag
( )
8
3
4
1
1
2
1
2
1
3
1
2
1
2
1
3
2
2
1
draw2ndonbeanjellypink
=
⋅⋅+⋅⋅+⋅⋅=P

8.6 Independent Events8.6 Independent Events
Example 1 – Showing That Two Events Are Independent
• E and F are said to be independent events if either
A fair coin is tossed twice. Let E and F be the events
E = {head on first toss}
F = {head on second toss}
Determine whether or not E and F are independent
events.
Solution:
( ) ( ) ( ) ( )FPEFPEPFEP == or
( ) ( )
( ) 2
1
4
2
#
#
===
S
E
EP
( ) ( )
( )
{ }( )
( ) 2
1
#
#
#
#
==
∩
=
F
HH
F
FE
FEP

8.6 Independent Events
Example 3 – Survival Rates
Special Multiplication Law
• If E and F are independent events, then
Suppose the probability of the event “Bob lives 20
more years” (B) is 0.8 and the probability of the
event “Doris lives 20 more years” (D) is 0.85.
Assume that B and D are independent events.
a. Find the probability that both Bob and Doris live
20 more years.
Solution:
( ) ( ) ( )FPEPFEP =∩
( ) ( ) ( ) ( )( ) 68.085.08.0 ===∩ DPBPDBP

Example 3 – Survival Rates
b. Find the probability that at least one of them lives
20 more years.
Solution:
c. Find the probability that exactly one of them lives
20 more years.
Solution:
( ) 97.068.085.08.0 =−+=∪DBP
( ) ( ) ( ) ( )( ) 17.085.02.0'' ===∩ DPBPDBP
( ) 29.017.012.0 =+=EP
( ) ( ) ( ) ( )( ) 12.015.08.0'' ===∩ DPBPDBP

Example 5 – Dice
Two fair dice, one red and the other green, are rolled,
and the numbers on the top faces are noted. Test
whether P(E ∩ F ) = P(E)P(F ) to determine whether
E and F are independent.
Solution: Event F has 6 outcomes which is
Thus the probability is
( ) ( ) ( ) ( ) ( ) ( ){ }6,1,5,2,4,3,3,4,2,5,1,6=F
( )
12
1
36
3
==∩FEP

Example 7 – Cards
Four cards are randomly drawn, with replacement,
from a deck of 52 cards. Find the probability that the
cards chosen, in order, are a king (K), a queen (Q), a
jack (J), and a heart (H).
Solution:
We obtain ( ) ( ) ( ) ( ) ( )
8788
1
52
13
52
4
52
4
52
4
=⋅⋅⋅=
=∩∩∩ HPJPQPKPHJQKP

8.7 Bayes’ Formula8.7 Bayes’ Formula
• The conditional probability of Fi given that event E
has occurred is expressed by
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )nn
ii
i
FEPFPFEPFPFEPFP
FEPFP
EFP
+++
=
...2211

8.7 Bayes’ Formula
Microchips are purchased from A, B, and C and are
randomly picked for assembling each camcorder.
20% of the microchips come from A, 35% from B,
and rest from C. The probabilities that A is defective
is 0.03, and the corresponding probabilities for B
and C are 0.02 and 0.01, respectively. A camcorder
is selected at random from a day’s production, and
its microchip is found to be defective.

Find the probability that it was supplied
(a) from A,
(b) from B, and
(c) from C.
(d) From what supplier was the microchip most
likely purchased?

Solution: We define the following events,
a.
{ }
{ }
{ }
{ }microchipdefective
Csupplier
Bsupplier
Asupplier
3
2
1
=
=
=
=
D
S
S
S
( )
( )( )
( )( ) ( )( ) ( )( )
35
12
01.045.002.035.003.02.0
03.02.0
topathsallofyprobabilit
andthroughpathofyprobabilit 1
1
=
++
=
=
D
DS
DSP

b.
c.
( )
( )( )
35
14
0175.0
02.035.0
2
=
=
=
D
DS
DSP
( )
( )( )
35
9
0175.0
01.045.0
3
=
=
=
D
DS
DSP

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Introductory maths analysis chapter 08 official