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Chapter 3 Probability The Concept of Probability Sample Spaces and Events Some Elementary Probability Rules Conditional Probability and Independence 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 Chapter 3 Probability Example 3.1 Roll a die. The experimental outcomes are 1, 2, 3, 4, 5, and 6. 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. 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 Chapter 3 Probability Section 3.2 Sample Spaces and Events 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: Chapter 3 Probability Example 3.2 Genders of Two Children Let: B be the outcome that child is boy. 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 Chapter 3 Probability Recall example 3.2: Genders of Two Children An event is a set of sample space outcomes. 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 Chapter 3 Probability 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. Chapter 3 Probability So, each of the 8 outcomes is equally likely to occur. P(CCC)=P(CCI)= ... = P(III)=1/8. 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: the number of outcomes that correspond to the event The total number of outcomes Chapter 3 Probability Basic Computation of Probabilities The probability of an event is also equal the sum of the probabilities of the sample space outcomes that correspond to the event. Example 3.4 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. 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. Example 3.5 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%. Chapter 3 Probability Section 3.3 Some Elementary Probability Rules The complement A of an event A is the set of all sample space outcomes not in A. Further, P(A) = 1 - P(A). These figures are “Venn diagrams”. Chapter 3 Probability Union of A and B, A B Is an event consisting of the outcomes that belong to either A or B (or both). Intersection of A and B, A B Is an event consisting of the outcomes that belong to both A and B. Chapter 3 Probability The Addition Rule The probability that A or B (the union of A and B) will occur is P(A B) = P(A) + P(B) - P(A B) where P(A B) is the “joint” probability of A and B, i.e., both occurring. A and B are mutually exclusive if they have no sample space outcomes in common, or P(A B) = 0. equivalently, if If A and B are mutually exclusive, then P(A B)=P(A)+P(B). Chapter 3 Probability Example 3.6 Newspaper Subscribers #1 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: 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 Chapter 3 Probability Newspaper Subscribers #2 Use the relative frequency method to assign probabilities 650,000 P A 0.65 1,000,000 500,000 P B 0.50 1,000,000 250,000 P A B 0.25 1,000,000 Chapter 3 Probability Table3.1 A Contingency Table Subscription Data for the Atlantic Journal and the Beacon News Events Subscribes to Does Not Beacon News, Subscribe to B Beacon News, Total Subscribes to Atlantic Journal, A 250,000 400,000 650,000 Does not Subscribes to Atlantic Journal, 250,000 100,000 350,000 Total 500,000 500,000 1,000,000 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 PA B , so from middle row and middle column of Table 3.1, 100,000 P A B 0.10. 1,000,000 Chapter 3 Probability Newspaper Subscribers #4 The chance that a household subscribes to either newspaper: P(A B)=P(A)+P ( B) P( A B) 0.65 0.50 0.25 0.90. 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. Chapter 3 Probability A Mutually Exclusive Case If A and B are mutually exclusive, then P(A B)=P(A)+P(B). 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 8 2 P(J ∪Q)=P(J)+P(Q)= 52 13 Chapter 3 Probability 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, P(A B) P(A|B) = P(B) Assume that P(B) is greater than 0. Interpretation: Restrict the sample space to just event B. The conditional probability P(A|B) is the chance of event A occurring in this new sample space. 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). P(A B) P(B | A) = P(A) Assume that P(A) is greater than 0. 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 P A B P B | A P A 0.25 0.3846. 0.65 Chapter 3 Probability Example: The Dean of the School of Business at Owens University collected the following information about undergraduate students in her college: Major Accounting Male 170 Female 110 Total 280 Finance 120 100 220 Marketing 160 70 230 Management 150 120 270 Total 600 400 1000 Example 7 If a student is selected at random, what is the probability that the student is a female (F) accounting major (A)? 110 P(A and F) =. 1000 Given that the student is a female, what is the probability that she is an accounting major? 110 P ( A and F ) 1000 0.275 P(A|F)= 400 P( F ) 1000 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? 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. Chapter 3 Probability The Multiplication Rule The joint probability that A and B (the intersection of A and B) will occur is P(A B) = P(A) P(B|A) = P(B) P(A|B). If A and B are independent, then the probability that A and B (the intersection of A and B) will occur is P(A B) = P(A) P(B) P(B) P(A). 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? R C .4 C .3 .5 R Total Chapter 3 Probability Total .6 1.00 Contingency Tables P(R ) P(R C) R R Total C .4 .1 .5 P(R C) As P( R C ) 0.4 C .2 .3 .5 Total .6 .4 1.00 P(C) P ( R) P (C ) 0.6 0.5 0.3 P ( R C ) P ( R ) P(C ) the events R and C are dependent. Chapter 3 Probability General Addition Rule: P(A or B) = P(A) + P(B) - P(A and B) 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) Chapter 3 Probability