Chapter 8 - Introduction to Probability and Statistics

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


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 Probability10. Limits and Continuity11. Differentiation12. Additional Differentiation Topics13. Curve Sketching14. Integration15. Methods and Applications of Integration16. Continuous Random Variables17. 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


Chapter 8: Introduction to Probability and Statistics8.1 Basic Counting Principle and Permutations

Example 3 – Answering a QuizIn 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!

!! nnnnn

nPnn

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

!!!

rnrnCrn

434 C


Chapter 8: Introduction to Probability and Statistics8.2 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 identitySolution:

111 rnrnrn CCC

11

1

!11!1!1

!1!1!

!!!

rn

rnrn

Crnr

nrnr

nrnr

nCC



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 knnnn



Example 7 – Name of a Legal FirmA 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 ExhibitAn 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: TT TH, HT, HH,S


Chapter 8: Introduction to Probability and Statistics8.3 Sample Spaces and Events

Example 3 – Sample Space: Jelly Beans in a BagA 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


Chapter 8: Introduction to Probability and Statistics8.3 Sample Spaces and EventsExample 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, 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,'



Properties of Events



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 i1

jsPsPsPEP ...21

SEEP

##


Chapter 8: Introduction to Probability and Statistics8.4 Probability

Example 1 – Coin TossingTwo 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##

SEEP

TT TH, HT, HH,S

TH HT, HH,F

4

3##

SFFP



Example 3 – Full House Poker HandFind 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

CCC

SEP



Example 5 – Quality ControlFrom 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

5000100

##

SEEP

02.05000

1100

EP

98.002.011' EPEP



Example 7 – Interrupted GamblingObtain 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.


Chapter 8: Introduction to Probability and Statistics8.4 ProbabilityExample 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



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

'EPEP

ba

aEP

• 0 ≤ P(si) ≤ 1 for i = 1 to N

• P(s1) + P(s2) + ·· ·+ P(sN) = 1



Example 9 – Odds for an A in an ExamA 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

41

8.02.0

'

EPEP



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.


Chapter 8: Introduction to Probability and Statistics8.5 Conditional Probability and Stochastic ProcessesExample 1 – Jelly Beans in a Bag

Solution:Event W ∩ B consists of the outcomes in B for which the second jelly bean is white:

32

64

BWP

Conditional probability of an event E is given as

0 and

FPFP

FEPFEP


Chapter 8: Introduction to Probability and Statistics8.5 Conditional Probability and Stochastic Processes

Example 3 – Quality ControlAfter 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 eventsWe have P(A) = 0.4 and P(D ∩ A) = 0.1, thus

4

14.01.0

APADPADP



Example 5 – AdvertisingA 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


Chapter 8: Introduction to Probability and Statistics8.5 Conditional Probability and Stochastic ProcessesExample 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

10225

5125

5226

12121 RRPRPRRP



Example 9 – Jelly Beans in a BagBag 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.


Chapter 8: Introduction to Probability and Statistics8.5 Conditional Probability and Stochastic ProcessesExample 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

83

411

21

21

31

21

21

32

21draw 2nd on beanjelly pink

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 eventsE = {head on first toss}F = {head on second toss}Determine whether or not E and F are independent events.

Solution:

FPEFPEPFEP or

2

142

##

SEEP

2

1#

##

#

FHH

FFEFEP


Chapter 8: Introduction to Probability and Statistics8.6 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


Chapter 8: Introduction to Probability and Statistics8.6 Independent EventsExample 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 – DiceTwo 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

121

363

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

87881

5213

524

524

524

HPJPQPKPHJQKP



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


Chapter 8: Introduction to Probability and Statistics 8.7 Bayes’ Formula

Example 1 – Quality ControlMicrochips 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.


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?


Chapter 8: Introduction to Probability and Statistics8.7 Bayes’ FormulaExample 1 – Quality Control

Solution: We define the following events,

a.

microchip defective

C supplierB supplier

Asupplier

3

2

1

DSSS

3512

01.045.002.035.003.02.003.02.0

to paths all ofy probabilit and through path ofy probabilit 1

1

D

DSDSP


Chapter 8: Introduction to Probability and Statistics8.7 Bayes’ FormulaExample 1 – Quality Control

b.

c.

3514

0175.002.035.0


2

D

DSDSP

359

0175.001.045.0


3

D

DSDSP