1. Chapter 3 Probability  The Concept of Probability  Sample Spaces and Events  Some Elementary Probability Rules  Conditional Probability and Independence

2. Chapter 3 Probability Section 3.1 The Concept of Probability  An experiment is any process of observation with an uncertain outcome. --- On any single trial of the experiment, one and only one of the possible outcomes will occur.  The possible outcomes for an experiment are called the experimental outcomes  Probability is a measure of the chance that an experimental outcome will occur when an experiment is carried out

3. Chapter 3 Probability Roll a die. The experimental outcomes are 1, 2, 3, 4, 5, and 6. Example 3.1 Example 3.1 Outcome An Outcome is the particular result of an experiment. An Event is the collection of one or more outcomes of an experiment. Possible outcomes: The numbers 1, 2, 3, 4, 5, 6 One possible event: The occurrence of an even number. That is, we collect the outcomes 2, 4, and 6.

4. Chapter 3 Probability  Regardless of the method used, probabilities must be assigned to the experimental outcomes so that two conditions are met: Conditions 1.0  P(E)  1 such that: If E can never occur, then P(E) = 0 If E is certain to occur, then P(E) = 1 2.The probabilities of all the experimental outcomes must sum to 1

5. Chapter 3 Probability Sample space (S): The sample space is defined as the set of all possible outcomes of an experiment. e.g. All 6 faces of a die: e.g. All 52 cards of a bridge deck: Section 3.2 Sample Spaces and Events

6. Chapter 3 Probability Example 3.2 Example 3.2 Genders of Two Children Let: B be the outcome that child is boy. G be the outcome that child is girl. G be the outcome that child is girl. Sample space S: S = {BB, BG, GB, GG} If B and G are equally likely, then P(B) = P(G) = ½ and P(BB) = P(BG) = P(GB) = P(GG) = ¼

7. Chapter 3 Probability An event is a set of sample space outcomes. Recall example 3.2: Genders of Two Children Events P(one boy and one girl) = P(BG) + P(GB) = ¼ + ¼ = ½. P(at least one girl) =P(BG) + P(GB) + P(GG) = ¼ + ¼ + ¼ = ¾. Note:Experimental Outcomes: BB, BG, GB, GG All outcomes equally likely: P(BB) = … = P(GG) = ¼

8. Chapter 3 Probability Example 3.3 Example 3.3 Answering Three True-False Questions A student takes a quiz that consists of three true-false questions. Let C and I denote answering a question correctly and incorrectly, respectively. The graph on the next slide shows the sample space outcomes for the experiment. The sample space consists of 8 outcomes: CCC CCI CIC CII ICC ICI IIC III Suppose the student is totally unprepared for the quiz and has to blindly guess the answers. That is, the student has a 50-50 chance of correctly answering each question.

9. Chapter 3 Probability So, each of the 8 outcomes is equally likely to occur. P(CCC)=P(CCI)=... = P(III)=1/8.

10. Chapter 3 Probability Probabilities: Equally Likely Outcomes If the sample space outcomes (or experimental outcomes) are all equally likely, then the probability that an event will occur is equal to the ratio:

11. Chapter 3 Probability The probability of an event is also equal the sum of the probabilities of the sample space outcomes that correspond to the event. The probability that the student will get exactly two questions correct is P(CCI) + P(CIC) + P(ICC) = 1/8 + 1/8 + 1/8 = 3/8. The probability that the student will get at least two questions correct is P(CCC) + P(CCI) + P(CIC) + P(ICC) = 1/8 + 1/8 + 1/8 + 1/8 = 1/2. Example 3.4 Example 3.4 Basic Computation of Probabilities

12. Chapter 3 Probability Relative Frequency Method Let E be an outcome of an experiment. If the experiment is performed many times, P(E) is the relative frequency of E. P(E) is the percentage of times E occurs in many repetitions of the experiment. Use sampled or historical data to calculate probabilities. Suppose that of 1000 randomly selected consumers, 140 preferred brand X. The probability of randomly picking a person who prefers brand X is 140/1000 = 0.14 or 14%. 140/1000 = 0.14 or 14%. Example 3.5 Example 3.5

13. Chapter 3 Probability Section 3.3 Some Elementary Probability Rules The complement of an event A is the set of all sample space outcomes not in A. Further, These figures are “Venn diagrams”.

14. Chapter 3 Probability Union of A and B, Is an event consisting of the outcomes that belong to either A or B (or both). Intersection of A and B, Is an event consisting of the outcomes that belong to both A and B.

15. Chapter 3 Probability The probability that A or B (the union of A and B) will occur is where is the “joint” probability of A and B, i.e., both occurring. The Addition Rule A and B are mutually exclusive if they have no sample space outcomes in common, or equivalently, if If A and B are mutually exclusive, then

16. Chapter 3 Probability Newspaper Subscribers #1 Example 3.6 Example 3.6 Define events: Define events:  A = event that a randomly selected household subscribes to the Atlantic Journal.  B = event that a randomly selected household subscribes to the Beacon News. Given: Given:  total number in city, N = 1,000,000  number subscribing to A, N(A) = 650,000  number subscribing to B, N(B) = 500,000  number subscribing to both, N(A∩B) = 250,000

17. Chapter 3 Probability Newspaper Subscribers #2 Use the relative frequency method to assign probabilities

18. Chapter 3 Probability Table3.1 A Contingency Table Subscription Data for the Atlantic Journal and the Beacon News Events Subscribes to Beacon News, B Does Not Subscribe to Beacon News,Total Subscribes to Atlantic Journal, A250,000400,000650,000 Does not Subscribes to Atlantic Journal,250,000100,000350,000 Total500,000 1,000,000

19. Chapter 3 Probability Newspaper Subscribers #3 Refer to the contingency table in Table 3.1 for all probabilities  For example, the chance that a household does not subscribe to either newspaper Calculate, so from middle row and middle column of Table 3.1,

20. Chapter 3 Probability Newspaper Subscribers #4  The chance that a household subscribes to either newspaper: Note that if the joint probability was not subtracted, then we would have gotten 1.15, greater than 1, which is absurd. Note: The subtraction avoids double counting the joint probability.

21. Chapter 3 Probability A Mutually Exclusive Case If A and B are mutually exclusive, then Example 3.7: Consider randomly selecting a card from a standard deck of 52 playing cards, and define the events J= the randomly selected card is jack. Q=the randomly selected card is queen. P(J ∪ Q)=? Since there are four jacks, four queens, we have P(J)=4/52 and P(Q)=4/52. Furthermore, since there is no card that is both a jack And a queen, the events J and Q are mutually exclusive and thus P(J∩Q)=0. So we have P(J ∪ Q)=P(J)+P(Q)=

22. Chapter 3 Probability Interpretation: Restrict the sample space to just event B. The conditional probability is the chance of event A occurring in this new sample space. Section 3.4 Conditional Probability and Independence The probability of an event A, given that the event B has occurred, is called the “conditional probability of A given B” and is denoted as Further, Assume that P(B) is greater than 0 Assume that P(B) is greater than 0.

23. Chapter 3 Probability Similarly, if A occurred, then what is the chance of B occurring? To answer this question, we need to introduce the probability of event B, given that the event A has occurred, i.e., the conditional probability of B given A, denoted by P(B|A). Assume that P(A) is greater than 0.

24. Chapter 3 Probability Newspaper Subscribers Given that the households that subscribe to the Atlantic Journal, what is the chance that they also subscribe to the Beacon News? Calculate P(B|A), where

25. Example 7 Example: The Dean of the School of Business at Owens University collected the following information about undergraduate students in her college:

26. P(A|F)= If a student is selected at random, what is the probability that the student is a female (F) accounting major (A)? P(A and F) =. Given that the student is a female, what is the probability that she is an accounting major?

27. Chapter 3 Probability Independence of Events Two events A and B are said to be independent if and only if P(A|B) = P(A) or, equivalently, P(B|A) = P(B). That is, if the chance of event A occurring is not influenced by whether the event B occurs and vice versa; or if the occurrences of the events A and B have nothing to do with each other, then A and B are independent. In fact if one of the above two equations holds, so does the other, why? In fact if one of the above two equations holds, so does the other, why?

28. Chapter 3 Probability Newspaper Subscribers Given that the households that subscribe to the Atlantic Journal subscribers, what is the chance that they also subscribe to the Beacon News?  If independent, the P(B|A) = P(B). Is P(B|A) = P(B)?  Know that P(B) = 0.50.  Just calculated that P(B|A) = 0.3846.  0.50 ≠ 0.3846, so P(B|A) ≠ P(B). B is not independent of A.  A and B are said to be dependent.

29. Chapter 3 Probability The Multiplication Rule The joint probability that A and B (the intersection of A and B) will occur is If A and B are independent, then the probability that A and B (the intersection of A and B) will occur is

30. Chapter 3 Probability A Question Suppose in the following contingency table, where the numbers represent probabilities, some data are lost. 1. Can you recover the missing data? 2. Are events R and C independent?

31. Chapter 3 Probability Contingency Tables

32. Chapter 3 Probability P(A or B) = P(A) + P(B) - P(A and B) General Addition Rule: Special Rule of Addition: If A and B are mutually exclusive P(A or B) = P(A) + P(B) Complement Rule P(A) + P(~A) = 1 or P(A) = 1 - P(~A). General Rule of Multiplication P(A and B) = P(A)P(B|A) or P(A and B) = P(B)P(A|B) Special Rule of Multiplication: If A and B are independent P(A and B) = P(A)P(B)