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Matematika Bisnis
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INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
2007 Pearson Education Asia
Chapter 8 Chapter 8 Introduction to Probability and StatisticsIntroduction to Probability and Statistics
2007 Pearson Education Asia
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
2007 Pearson Education Asia
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
2007 Pearson Education Asia
• 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
2007 Pearson Education Asia
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 8: Introduction to Probability and Statistics
Chapter OutlineChapter Outline
8.5)
8.6)
8.7)
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.1 Basic Counting Principle and Permutations
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 .
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.1 Basic Counting Principle and Permutations
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
!20420
P
280,11617181920420 P
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.1 Basic Counting Principle and Permutations
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
nPnn
241234!4
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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
nCrn
434 C
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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
!52552
C
Establish the identity
Solution:
111 rnrnrn CCC
11
1
!11!1
!1
!1!1
!
!!
!
rn
rnrn
Crnr
n
rnr
n
rnr
nCC
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.2 Combinations and Other Counting Principles
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
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.2 Combinations and Other Counting Principles
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
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.2 Combinations and Other Counting Principles
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
!111120 C
!9!3!4!4
!20
!9!3
!12
!12!4
!16
!16!4
!20312416420 CCC
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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: TT TH, HT, HH,S
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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,S1644
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.3 Sample Spaces and Events
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
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.3 Sample Spaces and Events
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.
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.3 Sample Spaces and Events
Example 7 – Complement, Union, IntersectionGiven the usual sample spacefor 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’
6 5, 4, 3, 2, 1,S
1 6 5, 4, 3, 5 3, 1, GFE
6 4, 2,'E
6 5, 4, 3, ,1FE
5 3,FE
φGF
SEE 6 5, 4, 3, 2, 1,6 4, 2,5 3, 1,'
φEE 6 4, 2,5 3, 1,'
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.3 Sample Spaces and Events
Properties of Events
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.4 Probability8.4 ProbabilityEquiprobable 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
EEP
#
#
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.4 Probability
Example 1 – Coin Tossing
Two fair coins are tossed,
Determine the probability thata. two heads occurb. at least one head occurs
Solution:
a. E = {HH}, probability is
b. F = {at least one head} where
Thus probability is
4
1
#
#
S
EEP
TT TH, HT, HH,S
TH HT, HH,F
4
3
#
#
S
FFP
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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
#
#house full
C
CC
S
EP
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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
EEP
02.05000
1100
EP
98.002.011' EPEP
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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.
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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
0110110 ......
s
kkss EPEPEPEPEEEP
1
0 2
s
nn
knC
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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
aEP
• 0 ≤ P(si) ≤ 1 for i = 1 to N
• P(s1) + P(s2) + ·· ·+ P(sN) = 1
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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
FEFEP
#
#
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.
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.5 Conditional Probability and Stochastic Processes
Example 1 – Jelly Beans in a Bag
Solution:
Event W ∩ B consists of the outcomes in B for which the second jelly bean is white:
3
2
6
4BWP
Conditional probability of an event E is given as
0 and
FPFP
FEPFEP
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.5 Conditional Probability and Stochastic Processes
Example 3 – Quality Control
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
ADPADP
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.5 Conditional Probability and Stochastic Processes
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
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.5 Conditional Probability and Stochastic Processes
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
2612121 RRPRPRRP
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.5 Conditional Probability and Stochastic Processes
Example 9 – Jelly Beans in a Bag
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.
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.5 Conditional Probability and Stochastic Processes
Example 9 – Jelly Beans in a Bag
Solution: This is a compound experiment with three trials:a. Select a bagb. Taking a jelly bean outc. Putting it in the other bag and then taking a jelly bean from that bag
8
34
11
2
1
2
1
3
1
2
1
2
1
3
2
2
1draw 2nd on beanjelly pink
P
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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 eventsE = {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
EEP
2
1
#
#
#
#
F
HH
F
FEFEP
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
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
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.6 Independent Events
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
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.6 Independent Events
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 ,1,6F
12
1
36
3FEP
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.6 Independent Events
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
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.7 Bayes’ Formula8.7 Bayes’ Formula
• The conditional probability of Fi given that event E has occurred is expressed by
nn
iii FEPFPFEPFPFEPFP
FEPFPEFP
...2211
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.7 Bayes’ Formula
Example 1 – Quality Control
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.
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.7 Bayes’ Formula
Example 1 – Quality Control
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?
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.7 Bayes’ Formula
Example 1 – Quality Control
Solution: We define the following events,
a.
microchip defective
C supplier
B supplier
Asupplier
3
2
1
D
S
S
S
35
12
01.045.002.035.003.02.0
03.02.0
to paths all ofy probabilit
and through path ofy probabilit 11
D
DSDSP
2007 Pearson Education Asia
Chapter 8: Introduction to Probability and Statistics
8.7 Bayes’ Formula
Example 1 – Quality Control
b.
c.
35
140175.0
02.035.0
to paths all ofy probabilit
and through path ofy probabilit 22
D
DSDSP
35
90175.0
01.045.0
to paths all ofy probabilit
and through path ofy probabilit 33
D
DSDSP