Statistics for Business and Economics Chapter 3 Probability

Contents 1.Events, Sample Spaces, and Probability 2.Unions and Intersections 3.Complementary Events 4.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

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 H T S

Example Experiment: Tossing two coins and recording up faces: Is sample space as below? S={HH, HT, TT}

Tree Diagram 1 st coin H T H TH T 2 nd 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 TH HT Sample Space S = {HH, HT, TH, TT} Venn Diagram Outcome; Sample point Experiment: Toss 2 Coins. Note Faces. Compound Event: At least one Tail

S HH TT TH HT Sample Space S = {HH, HT, TH, TT} Venn Diagram Experiment: Toss 2 Coins. Note Faces. Compound Event: Exactly one head

S HH TT TH HT Sample Space S = {HH, HT, TH, TT} Venn Diagram Experiment: Toss 2 Coins. Note Faces. Compound Event: Tail at the 2 nd toss

S HH TT TH HT Sample Space S = {HH, HT, TH, TT} Venn Diagram Experiment: Toss 2 Coins. Note Faces. 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 Experiment: Toss 2 Coins. Note Faces. 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 Certain Impossible

Probability Rules for Sample Points Let p i represent the probability of sample point i. 1.All sample point probabilities must lie between 0 and 1 (i.e., 0 ≤ p i ≤ 1). 2.The probabilities of all sample points within a sample space must sum to 1 (i.e.,  p i = 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 © 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 IIb. I and IIIc. 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 Rule A sample of n elements is to be drawn from a set of N elements. The, the number of different samples possible is denoted byand is equal to where the factorial symbol (!) means that n!=n*(n-1)*…*3*2*1 For 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)= =0.43 e. Let event C= selecing a candy other than blue P(C) = =0.76 or P(C) = =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.

Event Ace  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 RedBlack Total AceAce & Red Ace & Black Ace Non & Red Non & Black Non- Ace TotalRedBlack S Non-Ace

S BlackAce Event Intersection: Venn Diagram Event Ace  Black: A , A  Event Black: 2 ,..., A  Sample Space: 2, 2 , 2 ,..., A  Experiment: Draw 1 Card. Note Kind, Color & Suit. Event Ace: A, A , A , A 

Sample Space (S): 2, 2 , 2 ,..., A  Event Intersection: Two–Way Table Experiment: Draw 1 Card. Note Kind, Color & Suit. Event Ace  Black: A , A  Simple Event Ace: A, A , A , A  Simple Event Black: 2 ,..., A  Color Type RedBlack Total AceAce & Red Ace & Black Ace Non & Red Non & Black Non- Ace TotalRedBlack 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

Event B 1 B 2 Total A 1 P(AP(A 1  B 1 ) P(AP(A 1  B 2 ) P(AP(A 1 ) A 2 P(AP(A 2  B 1 ) P(AP(A 2  B 2 ) P(AP(A 2 ) P(BP(B 1 ) P(BP(B 2 )1 Event Probability Using Two–Way Table Joint ProbabilityMarginal (Simple) Probability Total

Color Type RedBlack Total Ace 2/52 4/52 Non-Ace 24/52 48/52 Total 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 = =0.83 P(A)= =0.29 P(B)= =0.73

Thinking Challenge 1. P(A) = 2. P(D) = 3. P(C  B) = 4. P(A  D) = 5. P(B  D) = Event CDTotal A 426 B 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 Event CDTotal A 426 B

3.3 Complementary Events

Complement of Event A The event that A does not occur All events not in A Denote complement of A by A C S ACAC A

Rule of Complements The sum of the probabilities of complementary events equals 1: P(A) + P(A C ) = 1 S ACAC A

S Black Complement of Event Example Event Black: 2 , 2 ,..., A  Complement of Event Black, Black C : 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)= =0.43 e. Let event C= selecing a candy other than blue P(C) = =0.76 or P(C) = =0.76 P(C c ) = 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)  Color Type RedBlack Total Ace 224 Non-Ace Total = + – =

Thinking Challenge 1. P(A  D) = 2. P(B  C) = Event CDTotal A 426 B Using the additive rule, what is the probability?

Solution* Using the additive rule, the probabilities are: P(A  D) = P(A) + P(D) – P(A  D) P(B  C) = P(B) + P(C) – P(B  C) = + – =

a.A  B b.A c c.B  C d.A c  B c 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(F  L)= 7/244 c.P(F c )=1-(127/244)=117/244 d.P(F  L)=P(F)+P(L)-P(F  L)=(127/244)+(28/244)- (7/244)=148/244 Example from text book (p.153): solution