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Lecture Slides

Lecture Slides. Elementary Statistics Twelfth Edition and the Triola Statistics Series by Mario F. Triola. Chapter 4 Probability. 4-1 Review and Preview 4-2 Basic Concepts of Probability 4-3 Addition Rule 4-4 Multiplication Rule: Basics

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Lecture Slides

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  1. Lecture Slides Elementary StatisticsTwelfth Edition and the Triola Statistics Series by Mario F. Triola

  2. Chapter 4Probability 4-1 Review and Preview 4-2 Basic Concepts of Probability 4-3 Addition Rule 4-4 Multiplication Rule: Basics 4-5 Multiplication Rule: Complements and Conditional Probability

  3. Experiments, Outcomes, and Sample Spaces • Example #3: • Draw a tree diagram for three tosses of a coin. List all outcomes for this experiment in a sample space S. • Solution: • Let “H” represent head and “T” represent tail. • Therefore, for each experiment, the outcome is either a “H” or “T”. 3rd Selection 2nd Selection 1st Selection HHH H HHT T H HTH H HTT S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} T THH T H H THT T H TTH H T TTT T 3 T

  4. Simple and Compound Events • Event is a collection of one or more of the outcomes of an experiment. An event could be the entire or portion of a sample space. Therefore, an event can be classified as: • Simple event • Compound event • A simple event, Ei, consists of one and only one of the final outcomes of an experiment. In Example #3, • A compound event consists of more than one outcome of an experiment. It is represented by • A, B, C, D,..., or A1, A2, A3,..., B1, B2, B3,,... • Reconsider Example #3, let A be the event that two of the three tosses will result in heads. Then, event A is given by E1=(HHH), E2=(HHT), E3=(HTH), E4=(HTT), E5=(THH), E6=(THT), E7=(TTH), and E8=(TTT) A = {HHH, HHT, HTH, THH} 4

  5. Simple and Compound Events • Solution: • Let, • D = a defective part • G = a good part • The experiment has the following outcomes: • DD = both parts are defective • DG = the 1st part is defective and the 2nd is good • GG = both parts are good • GD = the 1st part is good and the 2nd is defective • At least one part is good = {DG, GG, GD} compound event • Exactly one part is defective = {DG, GD} compound event • The 1st is good and 2nd defective = {GD} simple event • At most one part is good = {DD, DG, GD} compound event. • Example #4: • A box contains a certain number of computer parts, a few of which are defective. Two parts are selected at random from this box and inspected to determine if they are good or defective. List all the outcomes included in each of the following events. Indicate which are simple and which are compound events. • At least one part is good. • Exactly one part is defective. • The first part is good and the second is defective. • At most one part is good. 5

  6. MARGINAL AND CONDITIONAL PROBABILITIES The following table gives the responses of 2,000 randomly selected adults who were asked whether or not they have shopped on internet. Discussion • The table is a two-way classification of 2,000 adults. • The table is called contingency table and each box with a numeric entry is called a cell. • Each cell gives the frequency of two characteristics: • Gender (male or female) and • Opinion (have shopped or have never shopped).

  7. Marginal and Conditional Probabilities Discussion By adding the row totals and column totals, we obtain a new table. If only one characteristic, “have shopped”, “have never shopped”, “male”, or “female”, is being considered at a time, the probability of each event is called marginal probability or simple probability.

  8. Marginal and Conditional Probabilities The marginal probability or simple probability is a probability of a single event without consideration of any other event. From the table, the marginal probabilities of the characteristics are as follows: 8

  9. Marginal and Conditional Probabilities Now suppose we want to find the probability that the randomly selected adult has shopped on the internet, assuming that the adult is female. In other words, the event that the adult is female has already occurred. This probability is called conditional probability, and it is written, P(has shopped | female) and is read as The probability that the selected adult has shopped on the internet given that the event “female” has already occurred. 9

  10. Marginal and Conditional Probabilities General Statement Suppose A and B are two events, then the conditional probability of A given B is written as , P(A|B). Again from the contingent table,

  11. Marginal and Conditional Probabilities Tree Diagram HS | M M | HS HS M 500/1200 500/800 HNS HNS | M F M | HS 700/1200 300/800 M HS 1200/2000 HS | F HS 800/2000 M | HNS 300/800 M F HNS 700/1200 800/2000 HNS | F HNS 500/800 F 1200/2000 F | HNS 500/1200

  12. Definition Mutually exclusive events are events that do not have any outcome in common. Ex. Events for rolling a die: A = an even number is observed B = an odd number is observe C = a number less than five is observed Mutually Exclusive Mutually nonexclusive event Most importantly, the occurrence of one event prevents the occurrence of the other mutually exclusive events. MUTUALLY EXCLUSIVE EVENTS

  13. Mutually Exclusive Events • For example, the outcomes of tossing a coin are mutually exclusive because both Head and Tail outcomes could not occur at the same time. The occurrence of Head prevents occurrence of Tail to occur. HH H T HT H TH H T T TT

  14. Mutually Exclusive Events • Solution • Example #12 There are 160 practicing physicians in a city. Of them, 75 are female and 25 are pediatricians. Of the 75 female, 20 are pediatricians. Are the events “female” and “pediatrician” mutually exclusive? Explain why or why not. The events “female” and “pediatrician” are not mutually exclusive because a physician could be a female and a pediatrician as shown above. • Example #13 Define the following two events for two tosses of a coin: A = at least one head B = both tails are obtained Are A and B mutually exclusive? Explain why or why not. • Solution:The experiment involves tossing a coin twice. The sample space • S = {HH, HT, TT, TH} where H = Head and T = Tail. • The events are: A = {HH, HT, TH} & B = {TT} • A and B are mutually exclusive. They do not have any outcome in common. 14

  15. INDEPENDENT VERSUS DEPENDENT EVENTS Definition • Two events are said to be independent if the occurrence of one does not affect the probability of the occurrence of the other. In other words, A and B are independenteventsif either P(A | B) = P(A) or P(B | A) = P(B). • If P(A | B) = P(A) is true, then P(B | A) = P(B) is also true. • If P(A | B) = P(A) is false, then P(B | A) = P(B) is also false. • If the occurrence of one event affects the probability of the other, then we say that the events are dependent. In other words, two events are dependentifeitherP(A | B) ≠ P(A) or P(B | A) ≠ P(B).

  16. Independent Versus Dependent Events • General Statement • Two events are either mutually exclusive or independent. • Mutually exclusive events are dependent • Independent events are never mutually exclusive • Dependent events may or may not be mutually exclusive. • Solution • Example #14 There are 160 practicing physicians in a city. Of them, 75 are female and 25 are pediatricians. Of the 75 female, 20 are pediatricians. Are the events “female” and “pediatrician” independent? Explain why or why not. Since P(female | Pediatrician) ≠ P(female), then the events are not independent. 16

  17. Independent Versus Dependent Events • Solution • Example #15: Two donut bakers baked 1000 donut holes. Baker A baked 600 donuts, of which 450 were sold and the remaining were discarded. Baker B baked the remaining donuts, of which 100 were discarded. The events are “Baker A”, “Baker B”, “sold donuts”, and “discarded donuts”. Prepare a contingent table for this experiment. Are the events “Baker A” and “sold donuts” independent? Explain why or why not. S | A A 450/750 B S | B 300/750 S 750/1000 D | A A 150/250 Since P(Baker A | sold donut) = P(Baker A), then the events are independent because the occurrence of event “sold donuts” does affect the probability of event “Baker A”. D B 250/1000 D | B 100/250 17

  18. Independent Versus Dependent Events • Example #15.1: A statistical experiment has 10 equally likely outcomes that are denoted by 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. Let event A = { 10, 12, 14, 16} and event B = {11, 13, 15} a. Are events A and B mutually exclusive? yes b. Are event A and B independent events? Because the two probabilities are not the same, the two events are not independent. Also, we know that mutually exclusive events are always dependent.

  19. Definition The complement of event A, denoted by Ā and is read as “A bar” or “A complement,” is the event that includes all the outcomes for an experiment that are not in A. Therefore, complementary events are always mutually exclusive. Two complementary events, combined together, includes all the outcomes of the experiment. COMPLEMENTARY EVENTS

  20. Complementary Events • Example #15 – Problem Let A be the event that a number less than 3 is obtained if we roll a die once. What is the probability of A? What is the complementary event of A, and what is its probability? • Solution 20

  21. Complementary Events • Example #15.1: A statistical experiment has 10 equally likely outcomes that are denoted by 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. Let event A = { 10, 12, 14, 16} and event B = {11, 13, 15} What are the complements of event A and B, respectively, and their probabilities? • Solution

  22. Suppose an experiment resulted in a sample space described as, S = {1, 2, 3, 4, 5, 6, 7, 8} Also, three events from the experiment are define as consisting of the following outcomes: A = {1, 2, 3, 4} B = {3, 4, 5, 6} C = {5, 6, 7, 8} INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE A and B A  B, or simply AB • You can see that the Events A and B are not mutually exclusive because they have two common outcomes, 3 and 4. • Likewise, Events B and C are not mutually exclusive because they have two outcomes, 5 and 6, in common. • Given two Events, A and B, we can say that the intersection of Events A and B is the collection of all outcomes that are common to both A and B. It can be written as,

  23. Multiplication Rule The probability of the intersection of Events A and B is called the joint probability and is define as the product of the marginal and conditional probabilities. Joint probability is written as, P(A  B) = P(A)  P(B|A) or P(A  B) = P(B)  P(A|B) From the two formulas, we can calculate the conditional probabilities as,

  24. Multiplication Rule for Independent Events Recall, P(A  B) = P(A)  P(B|A) or P(A  B) = P(B)  P(A|B) This is true for dependent events if the probability of one is affected by the occurrence of the other event. In other words, P(A|B) ≠ P(A) and P(B|A)≠ P(B). For independent events, the occurrence of one event does not affect the probability of the other. Therefore, P(A|B) = P(A) and P(B|A) = P(B). Hence, we can rewrite the formula for calculating probability of intersection of two independent events as, P(A  B) = P(A)  P(B) Note: You can extend the multiplication rule to calculate the joint probability of as many events as you want. 24

  25. Joint Probability of Mutually Exclusive Events We know from discussion of mutually exclusive events that mutually exclusive events have no common outcomes. Therefore, they do not have an intersection. In this case, we write the intersection of two or more mutually exclusive events as P(AB) = 0 • Solution • Example #17 a. P(AB)= P(B) P(A|B) = (0.59)(0.77) = 0.4543 b. P(AB)= P(A) P(B|A) = (0.28)(0.35) = 0.098 Find the joint probability of A and B for the following: a. P(B) = 0.59 and P(A|B) = 0.77 b. P(A) = 0.28 and P(B|A) = 0.35 • Solution • Example #18 a. P(ABC)= P(A)P(B)P(C) = (.49)(.67)(.75) = 0.2462 b. P(ABC)= P(A)P(B)P(C) = (.71)(.34)(.45) = 0.1086 Find the joint probability for the following three independent events: a. P(A) = 0.49, P(B) = 0.67, P(C) = .75 b. P(A) = 0.71, P(B) = 0.34, P(C) = 0.45 25

  26. Intersection of Events and Multiplication Rule • Solution • Example #19 The following table gives two way classification of all basketball players at a state university who began their college careers between 2001 and 2005, based on gender and whether or not they graduate. • If one of these players is selected at random, find the following probabilities: • P(female and graduate) • P(male and did not graduate) • Find P(graduate and did not graduate). Is this probability zero? If yes, why? P(graduate and did not graduate) = 0, because these events are mutually exclusive and you could not have someone that is both “a graduate” and “a non graduate”. 26

  27. Intersection of Events and Multiplication Rule • Example #20 The following table gives two way classification of the responses based on the education levels of the persons included in the survey and whether they are financially better off, the same as, or worse off than their parents. • Suppose one adult is selected at random from these 2000 adults. Find the following probabilities: • P(better off and high school) • P(more than high school and worse off) • Find the joint probability of the events “worse off” and “better off.” Is this probability zero? Explain why or why not. 27

  28. Intersection of Events and Multiplication Rule • Solution 28

  29. Definition Let a sample space, S, consist of all outcomes in Events A and B. Then the union of the two events is the collection of all outcomes that belong to either A or B or to both A and B. This is denoted by A  B or just A or B UNION OF EVENTS AND THE ADDITION RULE S B A S Addition Rule Addition rule is the method for calculating the probability of the union of events. It is defined as, • P(A  B) = P(A)+P(B)-P(A  B) B A Essentially, we calculate the probability of union of events by: 1. Adding the probability of each event and 2. Subtract the probability of the intersection of the events from result in (1).

  30. Let re-examine the formula for calculating the probability of union of events. P(A  B) = P(A)+P(B)-P(A  B) However, we have said that for mutually exclusive events, P(A  B) = 0 Then, for mutually exclusive events, the union of two events is, P(A  B) = P(A)+P(B) Addition Rule for Mutually Exclusive Events

  31. Union of Events and the Addition Rule • Example #24 - Solution • Example #24 The following table gives two way classification of all basketball players at a state university who began their college careers between 2001 and 2005, based on gender and whether or not they graduate. If one of these players is selected at random, find the following probabilities: a. P(female or did not graduate) b. P(graduate or male) 31

  32. Union of Events and the Addition Rule • Example #25 The following table gives two way classification of the responses based on the education levels of the persons included in the survey and whether they are financially better off, the same as, or worse off than their parents. • Suppose one adult is selected at random from these 2000 adults. Find the following probabilities: • P(better off or high school) • P(more than high school or worse off) • P(better off or worse off) 32

  33. Union of Events and the Addition Rule • Solution 33

  34. Union of Events and the Addition Rule • Example #26 The probability of a student getting an A grade in an economics class is 0.24 and that of getting a B grade is 0.28. What is the probability that a randomly selected student from this class will get an A or a B in this class? Explain why the probability is not equal to 1.0. • Solution P(A) = 0.24 P(B) = 0.28 Then, the probability of getting an A or B is, P(A or B) = P(A) + P(B) = 0.24 + 0.28 = 0.52 The probability is not equal to 1.0 because the student can get a grade of C, D, or F. 34

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