Events A and B are independent if:knowing whether A occured does not change the probability of B.Mathematically, can say in two equivalent ways:P(B|A)=P(B)P(A and B)=P(B ∩ A)=P(B) × P(A).Important to distinguish independenc e from m utual ly exclusive whichwould say B ∩ A is empty (cannot happen) .Example. Deal 2 cards from deckAfirstcardisAceC second card is AceP(C |A)=351P(C )=452(last class).So A and C are dependent.Example. Throw 2 diceAfirstdielands1B second die shows larger number than first dieC both dice show same numberP(B|A)=56P(B)=?=1536by countingso A and B dependent.P(C |A)=16P(C )=636=16so A and C independent.Note 1: here B and C are mutually exclusive.Note 2: writing B�= ”second die shows smaller number than first die ”we haveP(B�)=P(B)bysymmetryP(B ∪ B�)=P(Cc)=1− P(C )=56giving a “non-counting” argument that P(B)=512.Example. Deal 1 card from deckAcardisAceScardisSpadeP(A)=452P(S)=1352P(A ∩ S)=152.Here P(A ∩ S)=P(A) × P(S)soindependent.Conceptual point.(a) In a fully-specified math model, two events are either dependent orindependent; can be checked by calc ul a ti on.(b) Often we use independence as an as s um pti on in making a model.For instance we assume that different die throws give indepen d en tresults. Most probability models one encounters in enginee ri n g or sciencehave some assumption of “bottom level” independence; bu t one n eed s tobe careful about which other events within the model are independent.(silly) Example.Throw 2 dice. If s um is at least 7 I show you the dice; if not, Idon’t.A: I show you first die lands 1B: I show you second die lands 1P(A)=136, P(B)=136, P(A ∩ B)=0so A and B dependent.Conceptual point. This illustrates a subtle poin t: being told by atruthful person that “A happened” is not (for probability/statisticspurposes) exactly the same as “knowing A happened ”.[car accident example]Systems of componentsWill show logic diagrams: system works if there is some pathleft-to-right which passes only though worki n g compon en ts.Assume components work/fail indepen de nt ly,P(Ciworks ) = pi, P(Cifails ) = 1 − pi.Note in practice the independence assumpt ion i s u su all y u n re al i s ti c.Math question: calculate P( system works ) in terms of the numbers piand the network structure.Example: “in series”.[picture on board]P(systemworks)=p1p2p3.Example: “in parallel”.[picture on board]P(systemfails)=(1− p1)(1 − p2)(1 − p3).P(systemworks)=1− (1 − p1)(1 − p2)(1 − p3).More complicated exampl e:[picture on board]We could write out all 16 combinations; instead l et’ s cond i ti on onwhether or not C1works.P(system works) = P(system works|C1works)P(C1works)+ P(system works|C1fails)P(C1fails)[continue on board]Example: Deal 4 cards. What is chance we get exactly one Spade?event 1st 2nd 3rd 4thF1SNNNF2NSNNF3F4NNNS[board: repeated conditioning]P(F1)=1352×3951×3850×3749P(F1)=P(F2)=P(F3)=P(F4)P(exactly one Sp a de ) = P(F1or F2or F3or F4))= P(F1)+P(F2)+P(F3)+P(F4)=4× P(F1) ≈ 44%.Example: Deal 4 cards. What is chance we get one card of each suit?event 1st 2nd 3rd 4thA1CDHSA2CDSH..........P(A1)=1352×1351×1350×1349P(A1)=P(A2)=...Number of possible orders = 4 × 3 × 2 × 1 = 24 = 4!P(one card of each suit) = 24 × P(A1) ≈ 10.5%.Bayes rule: updating probabilities as new information is acq u i re d.(silly) Example There are 2 coi n s:one is fair: P(Heads) = 1/2; one is biased: P(Heads) = 9/10Pick one coin at random. Toss 3 times. Suppose we get 3 Heads. Whatthen is the chance that the coin we picked is the bia sed coin?Abstract set-up: Partition (B1, B2,...) of “alternate possibilities”.Know prior probabi l i ti es P(Bi).Then observe some event A happens (the “new information”) for whichwe know P(A|Bi). We want to calculate the posterior probabilitiesP(Bi|A).Bayes formula:P(Bi|A)=P(A|Bi)P(Bi)P(A|B1)P(B1)+P(A|B2)P(B2)+....[example above on