Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
96.37% of statistics are made up. Professor: David Sumner Office: LeConte 400H Phone: 777-7529 Email: [email protected] Office Hours: Tuesday - Wednesday - 2:00 - 3:30PM And by Appointment. I also respond promptly to email. Textbook: Mathematical Statistics, Wackerly, et al 7th Edition Room: LeConte 310 – Tu - Th 11:00AM - 12:15PM WEB Site A great deal of material will be available on the class web site at: http://www.math.sc.edu/~sumner/math511spring2013 This will include problem sets, problem set solutions, review materials, practice exams, quiz solutions, handouts, software and much more. Problem Sets and other assignments will be posted on the WEB site by 6:00PM each Tuesday and Thursday. Grading Three exams 100 points each 300 points Twelve quizzes 10 points each 100 points Final Exam 150 points 90 - 100 A 88 - 89 B+ 80 - 87 B 78 - 79 C+ 70 - 77 C 68 - 69 D+ 60 - 67 D Total: 550 points Homework will be assigned but not collected and not graded. Exam Schedule (tentative) Exam 1: Thursday, February 7 Exam 2: Thursday, March 7 Exam 3: Thursday, April 11 Final Exam: Friday, May 3 9:00AM These dates (other than the final exam) are tentative. But you will have a week’s notice if they should change. What is probability? When the weather anchor says there is a 30% chance of rain, what does that mean? If we roll a pair of dice what is the probability of getting a sum of 7? Possible Sums are: 2 3 4 5 6 7 8 9 10 11 12 That’s 11 possible sums. So, is the probability of getting 7 equal to ? There are actually 36 equally likely things that can happen when tossing a pair of dice. Which of these yield a sum of seven? There are 6 ways to get a sum of 7. The probability of a sum of 7 is So suppose that S is the set of all outcomes of some experiment. We refer to S as the sample space of outcomes. If S is finite or countable, then it is said to be a discrete sample space. Suppose that S is finite and that all the outcomes in S are equally likely to occur. An event in this context simply refers to a subset A of S i.e., to a set of possible outcomes. We usually describe an event in ordinary English. So suppose that S is the set of all outcomes of rolling a pair of fair dice. Then there are 36 possible equally likely outcomes that we would typically consider. S = {(1, 1), (1, 2), ..., (3, 4), ..., (5, 5),..., (6,6) } Subsets of the sample space are called events. Event: Roll doubles – {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)} 1 6 The probability of this event is = 36 6 Event: Roll a sum of 10 – {(4, 6), (5, 5), (6, 4)} 1 3 The probability of this event is = 36 12 Event: Sum is 8 but no doubles – {(2, 6), (3, 5), (5, 3), (6, 2)} 1 4 The probability of this event is = 36 9 So, although there is still much to say about sample spaces of experiments in which not all the sample points are equally likely, it is evident that many problems dealing with discrete probability can be settled by basic counting (enumeration). If S is a sample space of an experiment and all the elements of S are equally likely to occur, then for any event A in the sample space, the probability P(A) of the event is Number of ways A can occur A P(A) = Number of equally S likely outcomes Suppose that we toss three coins. What is the probability that exactly two heads? S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT} Each of these 8 outcomes is as likely as any of the others. Three of these outcomes consist of two heads and one tail. The probability is 3/8. More generally, we may have sample spaces S where the outcomes are not equally likely. In such a case we will endeavor to find a way to assign probabilities properly to the events of the sample space. So if S is a finite sample space, a function P defined on the events of S is a probability function on S if (i). 0 ≤ P(A) ≤ 1 for each event A. (ii). P(S) = 1 (iii). If A and B are disjoint events, then P ( A ∪ B ) = P(A) + P(B) (iii). If S is countably infinite and A1 , A2 , A3 ,… An ,… ⎛ ∞ ⎞ ∞ are disjoint events in S, then P ⎜ Ai ⎟ = ∑ P ( Ai ) ⎝ i=1 ⎠ i=1 Disjoint events are said to be mutually exclusive. So if A and B are mutually exclusive then P ( A ∩ B ) = 0, and P ( A ∪ B ) = P ( A ) + P ( B ) For now we will concentrate on discrete probability where the sample space S is finite (or occasionally countable). In order to do this effectively, we need to learn how to count well. Set Notation Suppose that A and B are both subsets of the Sample Space set S. We express this by, A ⊆ S and B ⊆ S A ∩ B = { x : x ∈A and x ∈B} The intersection of A and B. A ∪ B = { x : x ∈A or x ∈B} The union of A and B. Key conjunctions! Set Notation Suppose that A and B are both subsets of the Sample Space set S. We express this by, A ⊆ S and B ⊆ S A − B = { x ∈A and x ∉B} = A ∩ B The difference A minus B. A = {x ∈S : x ∉A} = S − A The complement of A. “Not A” Also, A = A′ and A − B = A − A ∩ B S = {a, b, c, d, e, f , g} A = {a, c, f } , B = {a, c, d} A ∩ B = {a, c} A ∪ B = {a, c, d, f } A− B ={f} A = {b, d, e, g} A =3 A∪ B = 4 A Fundamental Law of Counting One of the more effective ways to determine the number of objects of some type is to devise a procedure that will produce each object exactly once and then count the number of ways to carry out that procedure. A × B = {(a,b) : a ∈A, b ∈B} The Cartesian Product of A and B {a, 1, 2} × {a, b} = {(a, a), (a, b), (1, a), (1, b), (2, a), (2, b)} A× B = A ⋅ B Rule of Product If one action can be done in n ways, and then after that a second action can be done in m ways, then the total number of ways to perform the first followed by the second is the product nm. This example will be useful someday when you’re in an institution different from this one. How many license plates are there that consist of 3 letters followed by 3 digits? 26 x 26 x 26 x 10 x 10 x 10 This example will be useful someday when you’re in an institution different from this one. What if the letters and digits must all be different? How many license plates are there that consist of 3 letters followed by 3 digits? 26 x 25 x 24 x 10 x 9 x 8