Stat 100A. Introduction to ProbabilitySyllabus questions, a couple of review exercises, and Introductionto Sample Spaces with equally likely outcomes.Juana [email protected] Department of StatisticsReference: Ross, chapter 2.5, Chapter 1.Lecture, part IJune 27, 2013J. Sanchez Stat 100A. Introduction to ProbabilityAnnouncementsHomework 1 is due on July 2nd at the beginning of lecture.Homework turned in after 11:05 gets points deducted.Quiz 1 will be done at 11:05 July 2nd. All homework musthave been turned in by then. I will post a practice quiz thisFriday for you to get idea of what it will look like.Enrollment questions: contact [email protected] students must be enrolled to attend this class at all times.J. Sanchez Stat 100A. Introduction to ProbabilitySyllabusAny questions on the course syllabus handed out the first day?If you did not get a copy of the course syllabus in lecture lastday, there are a few extra copies in the front today. OR get acopy from CCLECourse web site ccle.ucla.edu (find our course). Required.Now let’s quickly go over the syllabus. Please, review on yourown as well. You do not need to ask a question that isanswered already in the syllabus or for which the policy is setin the syllabus.J. Sanchez Stat 100A. Introduction to ProbabilitySyllabusAny questions on the course syllabus handed out the first day?If you did not get a copy of the course syllabus in lecture lastday, there are a few extra copies in the front today. OR get acopy from CCLECourse web site ccle.ucla.edu (find our course). Required.Now let’s quickly go over the syllabus. Please, review on yourown as well. You do not need to ask a question that isanswered already in the syllabus or for which the policy is setin the syllabus.J. Sanchez Stat 100A. Introduction to ProbabilityNotesNotesNotesNotesSyllabusAny questions on the course syllabus handed out the first day?If you did not get a copy of the course syllabus in lecture lastday, there are a few extra copies in the front today. OR get acopy from CCLECourse web site ccle.ucla.edu (find our course). Required.Now let’s quickly go over the syllabus. Please, review on yourown as well. You do not need to ask a question that isanswered already in the syllabus or for which the policy is setin the syllabus.J. Sanchez Stat 100A. Introduction to ProbabilityTodayReference reading for today’s material: Ross (8th edition), Chapter2, section 2.3, 2.4, 2.5 and beginning of Chapter 1(sections 1-5).(Section 2.5 will make more sense after we have discussed Chapter1).I. Quick review of some of a couple of the results seen in thelast lesson and some applications of those results.II. Probabilities of events in a Sample Space with equally likelyoutcomes. Small sample spaces.III. Introduction to counting.J. Sanchez Stat 100A. Introduction to ProbabilityI. Some applications of results seen last day.ExampleA consumer organization estimates that over a 1-year period 17%of cars will need to be repaired once, 7% will need repairs twice,and 4% will require three or more repairs.(a) What is the sample space and the probability of its outcomes?S = { Repaired once, repaired twice, repaired three or moretimes, never repaired }(b) What is the probability that a car chosen at random will needno more than one repair? P( once or never ) = 0.17 + (1-0.07 - 0.04 -0.17) =0.17 + 0.72 = 0.89(c) What is the probability that the car will need some repairs?P(some repairs) = P(at least one repair) =1-P(no repairs) =1- 0.72= 0.28.J. Sanchez Stat 100A. Introduction to ProbabilityI. Some applications of results seen last day.ExampleA consumer organization estimates that over a 1-year period 17%of cars will need to be repaired once, 7% will need repairs twice,and 4% will require three or more repairs.(a) What is the sample space and the probability of its outcomes?S = { Repaired once, repaired twice, repaired three or moretimes, never repaired }(b) What is the probability that a car chosen at random will needno more than one repair? P( once or never ) = 0.17 + (1-0.07 - 0.04 -0.17) =0.17 + 0.72 = 0.89(c) What is the probability that the car will need some repairs?P(some repairs) = P(at least one repair) =1-P(no repairs) =1- 0.72= 0.28.J. Sanchez Stat 100A. Introduction to ProbabilityNotesNotesNotesNotesI. Some applications of results seen last day.ExampleA consumer organization estimates that over a 1-year period 17%of cars will need to be repaired once, 7% will need repairs twice,and 4% will require three or more repairs.(a) What is the sample space and the probability of its outcomes?S = { Repaired once, repaired twice, repaired three or moretimes, never repaired }(b) What is the probability that a car chosen at random will needno more than one repair?P( once or never ) = 0.17 + (1-0.07 - 0.04 -0.17) =0.17 + 0.72 = 0.89(c) What is the probability that the car will need some repairs?P(some repairs) = P(at least one repair) =1-P(no repairs) =1- 0.72= 0.28.J. Sanchez Stat 100A. Introduction to ProbabilityI. Some applications of results seen last day.ExampleA consumer organization estimates that over a 1-year period 17%of cars will need to be repaired once, 7% will need repairs twice,and 4% will require three or more repairs.(a) What is the sample space and the probability of its outcomes?S = { Repaired once, repaired twice, repaired three or moretimes, never repaired }(b) What is the probability that a car chosen at random will needno more than one repair? P( once or never ) = 0.17 + (1-0.07 - 0.04 -0.17) =0.17 + 0.72 = 0.89(c) What is the probability that the car will need some repairs?P(some repairs) = P(at least one repair) =1-P(no repairs) =1- 0.72= 0.28.J. Sanchez Stat 100A. Introduction to ProbabilityI. Some applications of results seen last day.ExampleA consumer organization estimates that over a 1-year period 17%of cars will need to be repaired once, 7% will need repairs twice,and 4% will require three or more repairs.(a) What is the sample space and the probability of its outcomes?S = { Repaired once, repaired twice, repaired three or moretimes, never repaired }(b) What is the probability that a car chosen at random will needno more than one repair? P( once or never ) = 0.17 + (1-0.07 - 0.04 -0.17) =0.17 + 0.72 = 0.89(c) What is the probability that the car will need some repairs?P(some repairs) = P(at least one repair) =1-P(no repairs) =1- 0.72= 0.28.J. Sanchez Stat 100A. Introduction to ProbabilityI. Some applications of results seen last day.ExampleA
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