Partitions of the Sample Space, Axioms ofProbability and Properties of ProbabilityJuana [email protected] Department of StatisticsTextbook: Ross, Chapter 2, sections 1-4 and last lecture.June 25th, part 2 of lectureJ. Sanchez Partitions of the Sample Space, Axioms of Probability and Properties of ProbabilityAnnouncementsHomework 1 is due on Tuesday July 2nd at the beginning of lecture.Textbook reading for this next material: Ross (8th edition), Chapter2, Section 2.3-2.4J. Sanchez Partitions of the Sample Space, Axioms of Probability and Properties of ProbabilityOutlineI. Extension of Union and Intersection to Infinite Collection ofEvents.II. Partition of the sample spaceIII. Calculating the Probability of an Event when you know theprobability of all the outcomes in itIV. Axioms of ProbabilityV. Properties of Probabilities of eventsJ. Sanchez Partitions of the Sample Space, Axioms of Probability and Properties of ProbabilityWhile we get started can you represent the events on theleft hand side of the equal sign graphically on a piece ofpaper? Write the event symbolically on top of each figure.Your name and ID on the top right hand side, and theday,please. Use a 11x8. No small pieces of paper will becollected.ExampleFind the simplest expression for the following events and discuss why theprobability of the expression given and that of the simpler one you foundshould be the same(a) (E ∪ F)(E ∪ Fc)=(b) (E ∪ F)(Ec∪ F )(E ∪ Fc) =(c) (E ∪ F )(F ∪ G ) =J. Sanchez Partitions of the Sample Space, Axioms of Probability and Properties of ProbabilityI. Extension of Union and Intersection to an infinitecollection of eventsIf A1, A2, . . . is a collection of events, all defined on a sample space S,then∞[i=1Ai= {x ∈ S : x ∈ Aifor some i}where x is an outcome in S.∞\i=1Ai= {x ∈ S : x ∈ Aifor all i}ExampleLet S = (0, 1] and define Ai= ((1/i), 1], i = 1, 2, ...... Find the unionand the intersection of these events.∞[i=1Ai= (0, 1];∞\i=1Ai= 1J. Sanchez Partitions of the Sample Space, Axioms of Probability and Properties of ProbabilityI. Extension of Union and Intersection to an infinitecollection of eventsIf A1, A2, . . . is a collection of events, all defined on a sample space S,then∞[i=1Ai= {x ∈ S : x ∈ Aifor some i}where x is an outcome in S.∞\i=1Ai= {x ∈ S : x ∈ Aifor all i}ExampleLet S = (0, 1] and define Ai= ((1/i), 1], i = 1, 2, ...... Find the unionand the intersection of these events.∞[i=1Ai= (0, 1];∞\i=1Ai= 1J. Sanchez Partitions of the Sample Space, Axioms of Probability and Properties of ProbabilityII. Partition of the Sample SpaceMutually exclusive ( disjoint) eventsTwo events A and B are disjoint if they are mutually exclusive, ie, ifA ∩ B = ∅. The events Ai, Aj, are pairwise disjoint if Ai∩ Aj= ∅ for alli and j, i 6= j.Partition of SIf the union of pairwise disjoint events is S, then the collection of theseevents forms a partition of S.Partitions are very useful, allowing us to divide the sample space intosmall, non-overlapping pieces.J. Sanchez Partitions of the Sample Space, Axioms of Probability and Properties of ProbabilityExampleConsider the collection Ai= [i, i + 1) i = 0, 1, 2, · · · · · · . Does it consistof pairwise disjoint events? Do these events make a partition ofS = [0, ∞)?Ai∩ Aj= ∅Thus, yes, this collection consists of pairwise disjoint events because theintersection of every two events is the empty set.∞[i=0Ai= [0, ∞) = SThe union of these events is S.So yes, they form a partition of S because because the events are disjointas seen above, and their union is S.J. Sanchez Partitions of the Sample Space, Axioms of Probability and Properties of ProbabilityExampleIf we pull a card from a deck and consider the events A=spade, B=heart, are these events mutually exclusive? Why?Yes, mutually exclusive A ∩ B = ∅If we pull a card from a deck and consider the events A=spade andB= ace, are these events mutually exclusive? Why?J. Sanchez Partitions of the Sample Space, Axioms of Probability and Properties of ProbabilityExampleA consumer organization estimates that over a 1-year period 17% of carswill need to be repaired once, 7% will need repairs twice, and 4% willrequire three or more repairs.(a) What is the sample space ?S = { Repaired once, repaired twice, repaired three or more times,never repaired }(b) Do the events ”Repaired once”, ”repaired twice”, ”repaired three ormore times”, ”never repaired” form a partition of the sample space?Why ?Yes, mutually exclusive events, and the union is S.J. Sanchez Partitions of the Sample Space, Axioms of Probability and Properties of ProbabilityExampleA consumer organization estimates that over a 1-year period 17% of carswill need to be repaired once, 7% will need repairs twice, and 4% willrequire three or more repairs.(a) What is the sample space ?S = { Repaired once, repaired twice, repaired three or more times,never repaired }(b) Do the events ”Repaired once”, ”repaired twice”, ”repaired three ormore times”, ”never repaired” form a partition of the sample space?Why ?Yes, mutually exclusive events, and the union is S.J. Sanchez Partitions of the Sample Space, Axioms of Probability and Properties of ProbabilityExampleA consumer organization estimates that over a 1-year period 17% of carswill need to be repaired once, 7% will need repairs twice, and 4% willrequire three or more repairs.(a) What is the sample space ?S = { Repaired once, repaired twice, repaired three or more times,never repaired }(b) Do the events ”Repaired once”, ”repaired twice”, ”repaired three ormore times”, ”never repaired” form a partition of the sample space?Why ?Yes, mutually exclusive events, and the union is S.J. Sanchez Partitions of the Sample Space, Axioms of Probability and Properties of ProbabilityIII. Definition of ProbabilityIf siis an outcome in the sample space S, and A an event defined in S,then the probability of that event can always be found by adding theprobabilities of all the outcomes of S that are in that event.P(A) =nXi=1P(si∈ S | si∈ A)J. Sanchez Partitions of the Sample Space, Axioms of Probability and Properties of ProbabilityExampleAfter shipping 5 computers to users, a computer manufacturer realized thattwo out of the five computers were not configured properly (i.e., weredefective), without knowing specifically which ones. The manufacturerallocated resources to recall only two randomly chosen computers
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