Basic Probability

Sample Spaces bridge cards Collection of all possible outcomes e.g.: All six faces of a die: e.g.: All 52 cards a deck of bridge cards

Events and Sample Spaces An event is a set of basic outcomes from the sample space, and it is said to occur if the random experiment gives rise to one of its constituent basic outcomes. Simple Event Outcome With 1 Characteristic Joint Event 2 Events Occurring Simultaneously Compound Event One or Another Event Occurring

Simple Event A: Male B: Over age 20 C: Has 3 credit cards D: Red card from a deck of bridge cards E: Ace card from a deck of bridge cards

Joint Event D and E, (DE): Red, ace card from a bridge deck A and B, (AB): Male, over age 20 among a group of survey respondents

Intersection Let A and B be two events in the sample space S. Their intersection, denoted AB, is the set of all basic outcomes in S that belong to both A and B. Hence, the intersection AB occurs if and only if both A and B occur. If the events A and B have no common basic outcomes, their intersection AB is said to be the empty set.

Compound Event D or E, (DE): Ace or Red card from bridge deck

Union Let A and B be two events in the sample space S. Their union, denoted AB, is the set of all basic outcomes in S that belong to at least one of these two events. Hence, the union AB occurs if and only if either A or B or both occurs

Event Properties Mutually Exclusive Collectively Exhaustive Two outcomes that cannot occur at the same time E.g. flip a coin, resulting in head and tail Collectively Exhaustive One outcome in sample space must occur E.g. Male or Female

 Special Events Null Event Complement of Event Club & Diamond on 1 Card Draw Complement of Event For Event A, All Events Not In A: A' or A 

What is Probability? 1. Numerical measure of likelihood that the event will occur Simple Event Joint Event Compound 2. Lies between 0 & 1 3. Sum of events is 1 1 Certain .5 Impossible

Concept of Probability A Priori classical probability, the probability of success is based on prior knowledge of the process involved. i.e. the chance of picking a black card from a deck of bridge cards Empirical classical probability, the outcomes are based on observed data, not on prior knowledge of a process. i.e. the chance that individual selected at random from the Kardan employee survey if satisfied with his or her job. (.89)

Concept of Probability Subjective probability, the chance of occurrence assigned to an event by a particular individual, based on his/her experience, personal opinion and analysis of a particular situation. i.e. The chance of a newly designed style of mobile phone will be successful in market.

Computing Probabilities The probability of an event E: Each of the outcomes in the sample space is equally likely to occur e.g. P( ) = 2/36 (There are 2 ways to get one 6 and the other 4)

Concept of Probability Experi- Number Number Relative menter of Trials of heads Frequency AITBAR 2048 1061 0.5181 ABBASI 4040 2048 0.5069 WAHIDI 12000 6019 0.5016 TARIQ 24000 12012 0.5005 What conclusion can be drawn from the observations?

Presenting Probability & Sample Space 1. Listing S = {Head, Tail} 2. Venn Diagram 3. Tree Diagram 4. Contingency Table

A Ā Venn Diagram S Example: Kardan Employee Survey Event: A = Satisfied, Ā = Dissatisfied Ā A Other compound events could be formed: Tail on the second toss {HT, TT} At least 1 Head {HH, HT, TH} S P(A) = 356/400 = .89, P(Ā) = 44/400 = .11

Tree Diagram Example: Kardan Employee Survey Advanced Satisfied Joint Probability Advanced .485 Satisfied Not Advanced .405 P(A)=.89 Kardan Employee Advanced .035 Not Satisfied Not Advanced .075 P(Ā)=.11

Joint Probability Using Contingency Table Event Event B B Total 1 2 A P(A  B ) P(A  B ) P(A ) 1 1 1 1 2 1 A P(A  B ) P(A  B ) P(A ) 2 2 1 2 2 2 Total P(B ) P(B ) 1 1 2 Joint Probability Marginal (Simple) Probability

Joint Probability Using Contingency Table Kardan Employee Survey Joint Probability Total Satisfied Not Satisfied Advanced .485 .035 .52 Not Advanced .405 .075 .48 Total .89 .11 1.00 Simple Probability

Use of Venn Diagram Fig. 3.1: AB, Intersection of events A & B, mutually exclusive Fig. 3.2: AB, Union of events A & B Fig. 3.3: Ā, Complement of event A Fig. 3.4 and 3.5: The events AB and ĀB are mutually exclusive, and their union is B. (A  B)  (Ā  B) = B

Use of Venn Diagram Let E1, E2,…, Ek be K mutually exclusive and collective exhaustive events, and let A be some other event. Then the K events E1  A, E2  A, …, Ek  A are mutually exclusive, and their union is A. (E1  A)  (E2  A)  …  (Ek  A) = A (See supplement Fig. 3.7)

Compound Probability Addition 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) 4. Probability of Complement P(A) + P(Ā) = 1. So, P(Ā) = 1  P(A)

Addition Rule: Example A hamburger chain found that 75% of all customers use mustard, 80% use ketchup, 65% use both. What is the probability that a particular customer will use at least one of these? A = Customers use mustard B = Customers use ketchup AB = a particular customer will use at least one of these Given P(A) = .75, P(B) = .80, and P(AB) = .65, P(AB) = P(A) + P(B)  P(AB) = .75 + .80  .65= .90

Conditional Probability 1. Event Probability Given that Another Event Occurred 2. Revise Original Sample Space to Account for New Information Eliminates Certain Outcomes 3. P(A | B) = P(A and B) , P(B)>0 P(B)

Example Recall the previous hamburger chain example, what is the probability that a ketchup user uses mustard? P(A|B) = P(AB)/P(B) = .65/.80 = .8125 Please pay attention to the difference from the joint event in wording of the question.

Conditional Probability Draw a card, what is the probability of black ace? What is the probability of black ace when black happens? Black Event (Ace and Black) Black (S) Ace S Black ‘Happens’: Eliminates All Other Outcomes and Thus Increase the Conditional Probability

Conditional Probability Using Contingency Table Conditional Event: Draw 1 Card. Note Black Ace Color Type Red Black Total Revised Sample Space Ace 2 2 4 Try other examples using this table. Non-Ace 24 24 48 Total 26 26 52 P(Ace and Black) P(Black) 2/52 26/52 P(Ace|Black) = = = 2/26

Game Show Imagine you have been selected for a game show that offers the chance to win an expensive new car. The car sits behind one of three doors, while monkeys reside behind the other two. You choose a door, and the host opens one of the remaining two, revealing a monkey. The host then offers you a choice: stick with your initial choice or switch to the other, still-unopened door. Do you think that switch to the other door would increase your chance of winning the car?

Statistical Independence 1. Event Occurrence Does Not Affect Probability of Another Event e.g. Toss 1 Coin Twice, Throw 3 Dice 2. Causality Not Implied 3. Tests For Independence P(A | B) = P(A), or P(B | A) = P(B), or P(A and B) = P(A)P(B)

Statistical Independence Kardan Employee Survey Note: (.52)(.89) = .4628  .485 Total Satisfied Not Satisfied Advanced .485 .035 .52 Not Advanced .405 .075 .48 Total .89 .11 1.00 P(A1and B1) P(A1) P(B1)

Multiplication Rule 1. Used to Get Joint Probabilities for Intersection of Events (Joint Events) 2. P(A and B) = P(A  B) P(A  B) = P(A)P(B|A) = P(B)P(A|B) 3. For Independent Events: P(A and B) = P(AB) = P(A)P(B)

Randomized Response An approach for solicitation of honest answers to sensitive questions in surveys. A survey was carried out in spring 1997 to about 150 undergraduate students taking Statistics for Business and Economics at Peking University. Each student was faced with two questions. Students were asked first to flip a coin and then to answer question a) if the result was the national emblem and b) otherwise.

Practice of Randomized Response Question Is the second last digit of your office phone number odd? Recall your undergraduate course work, have you ever cheated in midterm or final exams? Respondents, please do as follows: 1.  Flip a coin. 2.  If the result is the head, answer question a); otherwise answer question b). Please circle “Yes” or “No” below as your answer.   Yes No

Bayes’ Theorem 1. Permits Revising Old Probabilities Based on New Information 2. Application of Conditional Probability 3. Mutually Exclusive Events Prior Probability New Information Apply Bayes' Theorem Revised Probability

Permutation and Combination Counting Rule 1 Example：In TV Series: Kangxi Empire, Shilang tossed 50 coins, the number of outcomes is 2·2 ·… ·2 ＝ 250. What is the probability of all coins with heads up? If any one of n different mutually exclusive and collectively exhaustive events can occur on each of r trials, the number of possible outcomes is equal to：n·n ·… ·n ＝ nr

Permutation and Combination Application: Lottery Post Card (Post of China) Six digits in each group (2000 version) Winning the first prize: No.035718 Winning the 5th prize: ending with number 3 If there are k1 events on the first trial, k2 events on the second trial, and kr events on the rth trial, then the number of possible outcomes is: k1· k2 ·… · kr

Permutation and Combination Applicatin: If a license plate consists of 3 letters followed by 3 digits, the total number of outcomes would be? (most states in the US) Application: China License Plates How many licenses can be issued? Style 1992: one letter or digit plus 4 digits. Style 2002: 1) three letters + three digits 2) three digits + three letters 3) three digits + three digits

Permutation and Combination Counting Rule 2 Example: The number of ways that 5 books could be arranged on a shelf is: (5)(4)(3)(2)(1) = 120 The number of ways that all n objects can be arranged in order is： = n(n -1)(n -2)(2)(1) = n! Where n! is called factorial and 0! is defined as 1.

Permutation and Combination Counting Rule 3: Permutation Example：What is the number of ways of arranging 3 books selected from 5 books in order? (5)(4)(3) = 60 The number of ways of arranging r objects selected from n objects in order is：

Permutation and Combination Counting Rule 4: Combination Example：The number of combinations of 3 books selected from 5 books is (5)(4)(3)/[(3)(2)(1)] = 10 Note: 3! possible arrangements in order are irrelevant The number of ways that arranging r objects selected from n objects, irrespective of the order, is equal to