Statistics for Business and Economics Chapter 3 Probability.

Statistics for Business and Economics

Chapter 3

Probability

Contents1. Events, Sample Spaces, and Probability2. Unions and Intersections3. Complementary Events4. The Additive Rule and Mutually Exclusive

Events

Learning Objectives

1. Develop probability as a measure of uncertainty

2. Introduce basic rules for finding probabilities

3. Use probability as a measure of reliability for an inference

4. Provide an advanced rule for finding probabilities

Thinking Challenge

• What’s the probability of getting a head on the toss of a single fair coin? Use a scale from 0 (no way) to 1 (sure thing).

• So toss a coin twice. Do it! Did you get one head & one tail? What’s it all mean?

Many Repetitions!*

Number of Tosses

Total Heads Number of Tosses

0.00

0.25

0.50

0.75

1.00

0 25 50 75 100 125

3.1

Events, Sample Spaces,and Probability

Experiments & Sample Spaces

1. Experiment• Process of observation that leads to a single

outcome that cannot be predicted with certainty

2. Sample point• Most basic outcome of an

experiment

3. Sample space (S) • Collection of all sample points

Sample Space Depends on Experimenter!

Visualizing Sample Space

1. Listing for the experiment of tossing a coin once and noting up face

S = {Head, Tail}

Sample point

2. A pictorial method for presenting the sample space Venn Diagram

HT

S

Example

• Experiment: Tossing two coins and recording up faces:

• Is sample space as below?

S={HH, HT, TT}

Tree Diagram

1st coin

H T

H T H T

2nd coin

Sample Space Examples

• Toss a Coin, Note Face {Head, Tail}• Toss 2 Coins, Note Faces {HH, HT, TH, TT}• Select 1 Card, Note Kind {2♥, 2♠, ..., A♦} (52)• Select 1 Card, Note Color {Red, Black} • Play a Football Game {Win, Lose, Tie}• Inspect a Part, Note Quality {Defective, Good}• Observe Gender {Male, Female}

Experiment Sample Space

Events

1. Specific collection of sample points

2. Simple Event

• Contains only one sample point

3. Compound Event

• Contains two or more sample points

S

HH

TT

THHT

Sample Space S = {HH, HT, TH, TT}

Venn Diagram

Outcome;

Sample point

Experiment: Toss 2 Coins. Note Faces.

Compound Event: At least one Tail

SHH

TTTH

HT


Venn Diagram


Compound Event: Exactly one head

SHH TT

TH

HT


Venn Diagram


Compound Event: Tail at the 2nd toss

SHH

TT

THHT


Venn Diagram


Simple Event: Tail for both tosses

Event Examples

• 1 Head & 1 Tail HT, TH

• Head on 1st Coin HH, HT

• At Least 1 Head HH, HT, TH

• Heads on Both HH


Sample Space: HH, HT, TH, TT

Event Outcomes in Event

Thinking challenge

• A fair coin is tossed till to get the first head or four tails in a row. Which one is the sample space for this experiment?

a. S={T, TH, TTH, TTTH, TTTT}

b. S={T, HT, TTH, TTTH, TTTT}

c. S={H, TH, TTH, TTTH, TTTT}

d. S={H, HT, HHT, HHHT, HHHH}

Probabilities

What is Probability?

1. Numerical measure of the likelihood that event will occur

• P(Event)• P(A)• Prob(A)

2. Lies between 0 & 1

3. Sum of probabilities for all sample points in the sample space is 1

11

.5 .5

00

CertainCertain

ImpossibleImpossible

Probability Rulesfor Sample Points

Let pi represent the probability of sample point i.

1. All sample point probabilities must lie between 0 and 1 (i.e., 0 ≤ pi ≤ 1).

2. The probabilities of all sample points within a sample space must sum to 1 (i.e., pi = 1).

Equally Likely Probability

P(Event) = X / T• X = Number of outcomes in the

event

• T = Total number of sample points in Sample Space

• Each of T sample points is equally likely

— P(sample point) = 1/T

© 1984-1994 T/Maker Co.

Steps for Calculating Probability

1. Define the experiment; describe the process used to make an observation and the type of observation that will be recorded

2. List the sample points

3. Assign probabilities to the sample points

4. Determine the collection of sample points contained in the event of interest

5. Sum the sample points probabilities to get the event probability

Thinking Challenge

• Consider the experiment of tossing two balanced dice. Which of the following are true for this experiment.

I. The probability of having 4 or less for the sum of the dots on the up faces is 1/6.

II. The probability of having sum of the dots on the upfaces larger than 4 is 5/6

III.The probability of having 6 or less for the sum of the dots on the up faces is 7/12

a. I and II b. I and III c. II and III d. all

Thinking Challenge (sol.)• Sample space for tossing two dice

S={(1,1),(1,2),…(1,6),(2,1),…,(2,6),…,(6,6)}

with 36 sample points

• Let event A=Having 4 or less for the sum of upfaces. So,

• A={(1,1),(1,2),(1,3),(2,1),(2,2),(3,1)}

• Prob:{1/36,1/36,1/36,1/36,1/36,1/36}

• P(A)=6/36=1/6

Thinking Challenge (sol.)• Let event B=Having the sum of upfaces

larger than 4. So,

• B={(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,2),(3,3),…(3,6),(4,1),…(4,6),(5,1),…(5,6),(6,1),…(6,6)}

With 30 sample points each with prob.1/36

• P(B)=30/36=5/6

Thinking Challenge (sol.)• Let event C=Having the sum of upfaces 6 or

less. So,

• C={ (1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2),(3,3), (4,1), (4,2), (5,1)}

each with prob.1/36

• P(C)=15/36=5/12

So the correct choice is a. I and II

Combinations RuleA sample of n elements is to be drawn from a set of N elements. The, the number of different samples possible

is denoted byN

n

and is equal to

N

n

N!

n! N n !where the factorial symbol (!) means that

n!=n*(n-1)*…*3*2*1

5!54 321For example, 0! is defined to be 1.

Thinking Challenge

• The price of a european tour includes four stopovers to be selected from among 10 cities. In how many different ways can one plan such a tour if the order of the stopovers does not matter?

P(A)=0.3, P(B)=0.2

P(A)=0.25, P(B)=0.3

Ex.3.2 from text book (p.139)

Ex.3.9 from text book

a. S={Brown, yellow, red, blue, orange,green}

b. P={0.13, 0.14, 0.13, 0.24, 0.20, 0.16}

c. Let event A=selecting brown candy

P(A)=P(Brown)=0.13

d. Let event B=selecting red, green or yellow candy

P(B)= 0.13+0.16+0.14=0.43

e. Let event C= selecing a candy other than blue

P(C) = 0.13 +0.14+ 0.13+ 0.20+ 0.16=0.76

or P(C) = 1-0.24=0.76

Ex.3.9 from text book: solution

Ex.3.25 from text book

• Odds in favor of E= P(E) / [1-P (E)]

• Odds against E = [1-P (E)] / P(E)

a. Odds in favor of Oxford Shoes winning=(1/3)/(2/3)=1/2 meaning 1 to 2

b.1/1= P(E) / [1-P (E)] P(E)=1/2

c. Odds against Oxford Shoes winning = 3/2

3/2 = [1-P (E)] / P(E) P(E)=2/5

Ex.3.25 from text book: solution

3.2

Unions and Intersections

Compound Events

Compound events:

Composition of two or more other events.

Can be formed in two different ways.

Unions & Intersections

1. Union• Outcomes in either events A or B or both• ‘OR’ statement• Denoted by symbol (i.e., A B)

2. Intersection• Outcomes in both events A and B• ‘AND’ statement• Denoted by symbol (i.e., A B)

S

BlackAce

Event Union: Venn Diagram

Event Ace Black:

A, ..., A, 2, ..., K

Event Black:

2,

2,...,

A

Sample Space:

2,2,

2, ..., A

Event Ace:

A, A, A, A

Experiment: Draw 1 Card. Note Kind, Color & Suit.

EventAce Black:

A,..., A, 2, ..., K

Event Union: Two–Way Table

Sample Space (S):

2, 2,

2, ..., A

Simple Event Ace:

A,

A,

A,

A

Simple Event Black:

2, ..., A

Experiment: Draw 1 Card. Note Kind, Color & Suit. Color

Type Red Black TotalAce Ace &

RedAce &Black

Ace

Non &Red

Non &Black

Non-Ace

Total Red Black S

Non-Ace

S

BlackAce

Event Intersection: Venn Diagram

Event Ace Black:

A, A

Event Black:

2,...,A

Sample Space:

2, 2,

2, ..., A


Event Ace:

A, A, A, A

Sample Space (S):

2, 2,

2, ..., A

Event Intersection: Two–Way Table


Event

Ace Black:

A, A

Simple Event Ace:

A, A,

A, A

Simple Event Black: 2, ..., A

ColorType Red Black Total

Ace Ace &Red

Ace &Black

Ace

Non &Red

Non &Black

Non-Ace

Total Red Black S

Non-Ace

Compound Event Probability

1. Numerical measure of likelihood that compound event will occur

2. Can often use two–way table• Two variables only

EventEvent B1 B2 Total

A1 P(A 1 B1) P(A1 B2) P(A1)

A2 P(A 2 B1) P(A2 B2) P(A2)

P(B1) P(B2) 1

Event Probability Using Two–Way Table

Joint Probability Marginal (Simple) Probability

Total


Ace 2/52 2/52 4/52

Non-Ace 24/52 24/52 48/52

Total 26/52 26/52 52/52

Two–Way Table Example

Experiment: Draw 1 Card. Note Kind & Color.

P(Ace)

P(Ace Red)P(Red)

Example from text book (p.145)

c. 0.16+0.03=0.19

0.10+0.16+0.03+0.10+0.08+0.22+0.14=0.83

P(A)=0.10+0.16+0.03=0.29

P(B)= 0.16+0.03+0.10+0.08+0.22+0.14=0.73

Thinking Challenge

1. P(A) =

2. P(D) =

3. P(C B) =

4. P(A D) =

5. P(B D) =

EventEvent C D Total

A 4 2 6

B 1 3 4

Total 5 5 10

What’s the Probability?

Solution*

The Probabilities Are:

1. P(A) = 6/10

2. P(D) = 5/10

3. P(C B) = 1/10

4. P(A D) = 9/10

5. P(B D) = 3/10


A 4 2 6

B 1 3 4

Total 5 5 10

3.3

Complementary Events

Complementary Events

Complement of Event A• The event that A does not occur• All events not in A• Denote complement of A by AC

S

AC

A

Rule of Complements

The sum of the probabilities of complementary events equals 1:

P(A) + P(AC) = 1

S

AC

A

S

Black

Complement of Event Example

Event Black:

2, 2, ..., A

Complement of Event Black,

BlackC: 2, 2, ..., A, A

Sample Space:

2, 2,

2, ..., A

Experiment: Draw 1 Card. Note Color.

Back to M&M example

a. S={Brown, yellow, red, blue, orange,green}

b. P={0.13, 0.14, 0.13, 0.24, 0.20, 0.16}

c. Let event A=selecting brown candy

P(A)=P(Brown)=0.13

d. Let event B=selecting red, green or yellow candy

P(B)= 0.13+0.16+0.14=0.43

e. Let event C= selecing a candy other than blue

P(C) = 0.13 +0.14+ 0.13+ 0.20+ 0.16=0.76

or P(C) = 1-0.24=0.76 P(Cc) = 0.24

3.4

The Additive Rule and Mutually Exclusive Events

Mutually Exclusive Events

• Events do not occur simultaneously

• A does not contain any sample points

Mutually Exclusive Events

S

Mutually Exclusive Events Example

Events and are Mutually Exclusive

Experiment: Draw 1 Card. Note Kind & Suit.

Outcomes in Event Heart:

2, 3, 4,

..., A

Sample Space:

2, 2,

2, ..., A

Event Spade:

2, 3, 4, ..., A

Additive Rule

1. Used to get compound probabilities for union of events

2. P(A OR B) = P(A B) = P(A) + P(B) – P(A B)

3. For mutually exclusive events:P(A OR B) = P(A B) = P(A) + P(B)

Additive Rule Example

Experiment: Draw 1 Card. Note Kind & Color.

P(Ace Black) = P(Ace) + P(Black) – P(Ace Black)


Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

52 52 52 52 4 26 2 28

= + – =

Thinking Challenge

1. P(A D) =

2. P(B C) =


A 4 2 6

B 1 3 4

Total 5 5 10

Using the additive rule, what is the probability?

10 10 10 10 6 5 2 9

Solution*

Using the additive rule, the probabilities are:

P(A D) = P(A) + P(D) – P(A D)1.

2. P(B C) = P(B) + P(C) – P(B C)

10 10 10 10

4 5 1 8

= + – =

= + – =

a. AB

b. Ac

c. BC

d. Ac Bc

Exercise from text book (p.152)

Example from text book (p.153)

Let events

• F= being a fully compensated worker

• R= being a partially compensated worker

• N=being a noncompasated volunteer

• L= leaving because of retirement

a. P(F)=127/244

b. P(FL)= 7/244

c. P(Fc)=1-(127/244)=117/244

d. P(FL)=P(F)+P(L)-P(FL)=(127/244)+(28/244)-(7/244)=148/244

Example from text book (p.153): solution

Statistics for Business and Economics Chapter 3 Probability.

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