Stat 100A. Introduction to ProbabilityProbability defined in Sample Spaces with equally likelyoutcomes.Juana [email protected] Department of StatisticsReference: Ross, chapter 2.5, Chapter 1.Lecture, Part I.July 2ndJ. Sanchez Stat 100A. Introduction to ProbabilityAnnouncementsHomework 1 due today.Quiz 1 today. Key will be posted after all TA sessions havefinished. Please, wait to ask questions about quiz until youhave received your quiz back.J. Sanchez Stat 100A. Introduction to ProbabilityTodayReference reading for this lecture’s material: Section 5 of chapter 2(8th edition); chapter 1.I. More on Fundamental Theorem of Counting.II. Probability theorems apply to sample spaces with equallylikely outcomes.J. Sanchez Stat 100A. Introduction to ProbabilityI. More on fundamental theorem of countingTheoremIf a job consists of k separate tasks, the ith of which can be donein niways, i = 1, · · · , k, then the entire job can be done inn1× n2× · · · × nkways.The lecture posted for last day contains examples on how to applythis theorem to count under different scenarios. Today, we focuson some examples of probabilities in sample spaces with equallylikely outcomes, where those counting techniques are needed.J. Sanchez Stat 100A. Introduction to ProbabilityAllo cating objects to groupsExampleIf 12 people are to be divided into 3 committees of respective sizes3, 4 and 5, how many divisions are possible?Apply Counting theorem:1239455=12!3!4!5!= 27, 720. But this result is more efficiently expressed as123,4,5, which is themultinomial coefficient.nk,w ,m=n!k!w!m!is the multinomial coefficient. It is the result ofapplying the Fundamental Theorem to this problem of allocationto groups.J. Sanchez Stat 100A. Introduction to ProbabilityAllo cating objects to groupsExampleIf 12 people are to be divided into 3 committees of respective sizes3, 4 and 5, how many divisions are possible?Apply Counting theorem:1239455=12!3!4!5!= 27, 720. But this result is more efficiently expressed as123,4,5, which is themultinomial coefficient.nk,w ,m=n!k!w!m!is the multinomial coefficient. It is the result ofapplying the Fundamental Theorem to this problem of allocationto groups.J. Sanchez Stat 100A. Introduction to ProbabilityAllo cating objects to groupsExampleIf 12 people are to be divided into 3 committees of respective sizes3, 4 and 5, how many divisions are possible?Apply Counting theorem:1239455=12!3!4!5!= 27, 720. But this result is more efficiently expressed as123,4,5, which is themultinomial coefficient.nk,w ,m=n!k!w!m!is the multinomial coefficient. It is the result ofapplying the Fundamental Theorem to this problem of allocationto groups.J. Sanchez Stat 100A. Introduction to ProbabilityProbability in the special case of equally likely outcomes i nSP(A) =number of outcomes of S in Anumber of outcomes in SJ. Sanchez Stat 100A. Introduction to ProbabilityThe importance of constraintsWhen applying the fundamental theorem, we should pay attentionto constraints, which often require special interpretation of thefundamental theorem of counting.ExampleA random sample of 45 instructors from different state universitieswere selected randomly and asked whether they are happy withtheir teaching loads. The responses of 32 were negative. If Drs.Smith, Brown, and Jones were among those questioned, what isthe probability that all three of them gave negative responses?J. Sanchez Stat 100A. Introduction to ProbabilityExampleA random sample of 45 instructors from different state universitieswere selected randomly and asked whether they are happy withtheir teaching loads. The responses of 32 were negative. If Drs.Smith, Brown, and Jones were among those questioned, what isthe probability that all three of them gave negative responses?Number of possible different groups of 32 teachers with negativeresponses is4532= 73006209045If three of the 32 are Drs. Smith, Brown, and Jones the other 29are from the remaining 42 faculty members questioned. SoNumber of different groups that include Drs. Smith, Brown, andJones is4229= 25518731280 Hence the desired probability is42294532≈ 0.35J. Sanchez Stat 100A. Introduction to ProbabilityExampleA random sample of 45 instructors from different state universitieswere selected randomly and asked whether they are happy withtheir teaching loads. The responses of 32 were negative. If Drs.Smith, Brown, and Jones were among those questioned, what isthe probability that all three of them gave negative responses?Number of possible different groups of 32 teachers with negativeresponses is4532= 73006209045If three of the 32 are Drs. Smith, Brown, and Jones the other 29are from the remaining 42 faculty members questioned. SoNumber of different groups that include Drs. Smith, Brown, andJones is4229= 25518731280 Hence the desired probability is42294532≈ 0.35J. Sanchez Stat 100A. Introduction to ProbabilityExampleA random sample of 45 instructors from different state universitieswere selected randomly and asked whether they are happy withtheir teaching loads. The responses of 32 were negative. If Drs.Smith, Brown, and Jones were among those questioned, what isthe probability that all three of them gave negative responses?Number of possible different groups of 32 teachers with negativeresponses is4532= 73006209045If three of the 32 are Drs. Smith, Brown, and Jones the other 29are from the remaining 42 faculty members questioned. SoNumber of different groups that include Drs. Smith, Brown, andJones is4229= 25518731280 Hence the desired probability is42294532≈ 0.35J. Sanchez Stat 100A. Introduction to ProbabilityExampleA random sample of 45 instructors from different state universitieswere selected randomly and asked whether they are happy withtheir teaching loads. The responses of 32 were negative. If Drs.Smith, Brown, and Jones were among those questioned, what isthe probability that all three of them gave negative responses?Number of possible different groups of 32 teachers with negativeresponses is4532= 73006209045If three of the 32 are Drs. Smith, Brown, and Jones the other 29are from the remaining 42 faculty members questioned. SoNumber of different groups that include Drs. Smith, Brown, andJones is4229= 25518731280Hence the desired probability is42294532≈ 0.35J. Sanchez Stat 100A.
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