Partitions of the Sample Space, Axioms ofProbabilityJuana [email protected] Department of StatisticsJ. Sanchez Partitions of the Sample Space, Axioms of ProbabilityOutlineThe syllabus again.I. 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 it. Definition of probability.IV. Axioms of ProbabilityJ. Sanchez Partitions of the Sample Space, Axioms 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.J. Sanchez Partitions of the Sample Space, Axioms 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 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 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 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. Suppose the experiment consists ofdrawing a car at random.(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 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 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 in successionout of the five for examination.1What is the sample space for the status of the two recalled computers?S = {dd , dn, nd, nn}2What is the event A that the second recalled computer is a defectivecomputer?A = {dd , nd }3We know that the probabilities of all the outcomes of the sample spacedefined in part (a) are, respectively, 1/10, 3/10, 3/10, 3/10. What is theprobability that the second recalled computer is a defective computer?That would be the probability of AP(A) = 1/10 + 3/10 = 4/10J. Sanchez Partitions of the Sample Space, Axioms of ProbabilityIV. Axiomatic approach to probabilityProbability: a function defined on events in S such that it satisfiesthe Kolmogorov Axioms. It is not concerned with the interpretationof this probability.0 ≤ P(A) ≤ 1 for all AP(S ) = 1Axiom of countable additivity: If A1, A2, ....., are pairwise disjoint,thenP ∞[i=1Ai!=∞Xi=1P(Ai)J. Sanchez Partitions of the Sample Space, Axioms of ProbabilityWhat function?Modern approach to probabilityThe assumption of the existence of a set function P, defined on theevents of a sample space S, and satisfying Axioms 1,2,3, constitutes themodern mathematical approach to probability theory.P must satisfy the axioms. For any sample space S, many differentprobability functions can be defined.J. Sanchez Partitions of the Sample Space, Axioms of ProbabilityExampleLet S = [0, 1] 0 ≤ a ≤ b ≤ 1. DefineP([a , b]) = b − a P(a ) = 0, P(b) = 0.This function satisfies the three axioms of probability.(a) 0 < P([a, b]) = b − a < 1(b) P(S) = P ([0, 1]) = 1 − 0 = 1(c) If [ai, bi]are disjoint intervals,P n[i=1[ai, bi]!=nXi=1[bi− ai]J. Sanchez Partitions of the Sample Space, Axioms of ProbabilityAssumptionWe assume in this class that if Eii = 1, 2, .... is an event in S, P(Ei) isdefined for all the events Eiof the sample space. Actually, when thesample space is an uncountably infinite set, P(Ei) is defined only for aclass of events called measurable. However, this restriction needs notconcern us as all events of any practical interest are measurable.J. Sanchez Partitions of the Sample Space, Axioms of ProbabilitySummarySome review.I. 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 ProbabilityJ. Sanchez Partitions of the Sample Space, Axioms of ProbabilityPractice problemsTossing a coin. Let’s find a couple of probability functions and see howwe do it.J. Sanchez Partitions of the Sample Space, Axioms of
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