ProbabilitySample Spaces and EventsRandom ExperimentDiscrete Sample SpacesContinuous Sample SpacesSlide 6Tree DiagramEventsSlide 9Set Notation / Venn DiagramsSlide 11Slide 12Set Rules for EventsSlide 14Venn Diagrams: ExampleProbability TheorySlide 17Probability AxiomsProbability: Additive RulesSlide 20Slide 21Slide 22Conditional Probability: ExampleSlide 24Multiplicative Rules in ProbabilityIndependence in ProbabilityIndependent Events: ExampleSlide 28Slide 29Slide 30Slide 31Bayes’ Rule DevelopmentSlide 33Slide 34Bayes’ Rule: General FormulaBayes’ Rule ExampleSlide 37Counting Sample PointsSlide 39Slide 40Slide 41Counting PermutationsSlide 43Slide 44Slide 45Counting Ways to PartitionSlide 47Counting CombinationsSlide 49Probability Calculation ExampleSlide 51Slide 521ProbabilitySample Spaces and EventsProbability Theory•Definition and Axioms•Conditional Probability•Independence•Bayes’ RuleCounting Sample Points2Sample Spaces and EventsRandom ExperimentSample Space (S)•Discrete•Continuous•Tree diagramEvent (A, B)Set Notation3Random ExperimentThe task that results in a random outcome.•Toss a coin / Roll a die•See how long a light bulb lasts•Count how many people enter the post office between 2:00 and 3:00 pm tomorrow. S = the set of all possible outcomes.An event consists of one or more outcomes•A = {odd number on a die}•A = {light bulb lasts < 200 hours}•A = {at least 3 people enter the post office between 2:00 and 3:00 pm tomorrow}4Discrete Sample SpacesTossing a coin: S = {H, T}Tossing a die: S = {1, 2, 3, 4, 5, 6}Tossing 3 coins: S = { HHH, HHT, HTH, THH, HTT, THT, TTH, TTT }Count # people that enter the post office: S = {0, 1, 2, 3, 4, 5,…}5Continuous Sample SpacesValues between 0 and 130: S = {x | 0 < x < 130}Lifetime of a light bulb: S = {x | x ≥ 0 }Points on a circle of radius 2: S = {(x, y) | x 2 + y 2 = 4}Points inside a circle of radius 2:}4|),{(22 yxyxS6An engineer in charge of the maintenance of a particular machine notices that its breakdowns can be characterized as due to an electrical failure within the machine, a mechanical failure of some component of the machine, or operator misuse. When the machine is running, the engineer is uncertain what will be the cause of the next breakdown. The problem can be thought of as an experiment with the sample space:Sample SpacesS = {electrical, mechanical, misuse}7Tree DiagramExample 2.2 Experiment•Toss a coin once. If heads, then toss the coin again; if tails, then roll a six-sided die.Tree diagram for example 2.28EventsAn event is a subset of a sample space. •Toss 3 coins: A = {exactly two heads} = {HHT, HTH, THH}•Light bulb: A = {lasts < 200 hours} = {x | 0  x < 200}•Post office: A = { 3 people enter} = {3, 4, 5, ….}•Example 2.2: A = {odd number on die} = {T1, T3, T5}9The control of errors in a computer software product is obviously of great importance. The number of separate errors in a particular piece of software can be viewed as having a sample space.S = {0 errors, 1 error, 2 errors, 3 errors, …}Consider the event A that there are no more than two errors in a software product. EventsA = {0 errors, 1 error, 2 errors}  S10Set Notation / Venn DiagramsA is the complement of A = S  A•Light bulb: A = {x | 0  x < 200} A = {x | 200  x < } = {x | x  200}•Toss 3 coins: A = {HHT, HTH, THH} A = {HHH, HTT, THT, TTH, TTT}•Venn Diagram: S = entire rectangle11Set Notation / Venn DiagramsA  B is the intersection of A and B•Events for which both A and B occur•Light bulb: A = {x | 0  x < 200}, B = {x | x > 50}  A  B = {x | 50 < x < 200} •Toss 3 coins: A = {HHT, HTH, THH}, B = {HHH, THH, THT}  A  B = {THH}12Set Notation / Venn DiagramsA  B is the union of A and B•Events for which either A or B (or both) occur•Light bulb: A = {x | 0  x < 200}, B = {x | 5  x < 500}  A  B = {x | 0  x < 500} •Toss 3 coins: A = {HHT, THH}, B = {HHH, THH}  A  B = {HHH, HHT, THH}13Set Rules for EventsMutually exclusive (or disjoint) events •Events that have nothing in common•A  B =  (empty set)14A  B = B  AA  A = AA  S = AA   = A  A = A  (B  C) = (A  B )  C (A  B) = A  BA  B = B  AA  A = AA  S = SA   = AA  A = SA  (B  C) = (A  B )  C(A  B) = A  B Set Rules for Events15Venn Diagrams: ExampleFind A  B, B  C, A  C, B   A, A  B  C, (A  B)  C1237654CBA16Probability TheoryWhat is Probability?•Axioms•Additive RulesConditional Probability•Definition•Multiplicative RulesIndependent EventsBayes’ Rule17Probability TheoryInterpretations of probabilities•Subjective: Based on people’s experience and general knowledge•Equally likely: Outcomes have equal probability Roll a six-sided die: P(1) = … = P(6) = 1/6•Limiting Relative Frequency:as n  )( trialsof#totaloccurseventtimesof #APAnnASANnAPAin outcomes of#totaleventin outcomesof #)( 18Probability Axioms0  P(A)  1P() = 0P(S) = 1If events A1, A2, … , Ak are disjoint, then P(A1  A2  …  Ak) = P(A1) + P(A2) +…+ P(Ak)• Roll a six-sided die: P(even) = P(2) + P(4) + P(6) = 3/6 P({2, 3, 4, 6}) = P(even) + P(3) = 3/6 + 1/6 = 4/6 P({1, 2, 3, 4, 5, 6}) = P(S) = 119Probability: Additive RulesFor event A: P(A) = 1  P(A ) • Roll a six-sided die: P({2, 3, 4, 6}) =4/6  P({1, 5}) = 1  4/6 = 2/6For events A and B:P(A  B ) = P(A ) + P(B )  P(A  B )• Roll a six-sided die: P(even) = 3/6, P({2, 3}) = 2/6 P(even  {2, 3}) = P(even) + P({2, 3})  P({2}) = 4/620Probability: Additive RulesFor 3 events A, B, and C:P(A  B  C ) = P(A ) + P(B ) + P(C )  P(A  B )  P(A  C )  P(B  C ) + P(A  B  C )Note: This is a generalization of the last axiom.21The conditional probability of A, given B, is the probability that event A occurs when it is known that event B occurs: where P(B) > 0( )( )P A BP B=I# of outcomes in events and ( | )# of outcomes in event A BP A BB=Conditional Probability22( )( | )( )( | ) ( | ) ( | ) ( | )( | ) 1 ( | )P A B CP C A BP A BP A B C P A C P B C P A B CP A B P A B== + -�= -I IIIU IExtensions