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MA340 (2)
Lecture

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Department
Mathematics
Course
MA340
Professor
Joe Campolieti
Semester
Fall

Description
Lecture 2 Properties of Probability. Combinatorics Text: Sections 2.3, 2.4, 3.1, 3.2 Print version of the lecture in MA340 Probability Theory presented on September 13, 2012 2.1 Contents 1 Properties of Probability 1 2 Law of Partitions 3 3 Specifying Probabilities 5 3.1 Countable Probability Space . . . . . . . . . . . . . . . . . . 5 . . . . . . 3.2 Classical Probability . . . . . . . . . . . . . . . . . . . . . 5 . . . . . . . 3.3 Geometric Probability . . . . . . . . . . . . . . . . . . . . . 6 . . . . . . . 4 Buffons Needle Problem 6 5 Counting Methods 7 6 Combinatorial Coefcients 10 2.2 1 Properties of Probability Axioms of Probability The probability measure P dened on the set of events of W satises the following three axioms. Axiom 1 (Nonnegativity) P(E) 0 Axiom 2 (Certainty) P(W) = 1 Axiom 3 (Countable Additivity) For each countable collection of pairwise disjoint events E1;E2;::: s.n.mE E/ when n 6= m we have that [ ! P En = P(En) n1 n1 2.3 1 Properties of Probability (13) 1. P(0/) = 0 (Probability of an impossible event) Proof: 0/ = W ;0/ [W = W;0 / \W = 0 /. Hence, by exclusivity of the two sets/ and W, and by and axiom 3, we have: P(W) = P(0 / [W) = P(0/)+P(W) =) P(0 /) = 0: 2. P(A ) = 1P(A) (Complementation Rule) Proof:Write W as a disjoint union, W = A[A . Then, by axioms 2 and 3: 1 = P(W) = P(A)+P(A ): c 3. A B ) P(A) P(B) (Domination Principle) Proof: Since A[A = W, we can write B = B\W = (B\A)[(B\A ), i.e. as a c union of two mutually exclusive events. Hence, P(B) = P(B\A)+P(B\A ) = P(A)+P(B\A ) c where we used the fact that A B =) B\A = A. Then P(B) P(A) since the event E B\A W has probability P(E) 0 by axiom 1. 2.4 Properties of Probability (45) 4. 8A W P(A) 1 Proof: By property 3 above and axiom 2, P(A) P(W) = 1. 5. P(A[B) = P(A)+P(B)P(AB) (Addition Rule) Proof: Write A[B as union of 3 disjoint sets: A[B = (An(A\B))[(A\B)[(Bn(A\B)) and use P(E nF) = P(E)P(F) with axiom 3 to obtain P(A[B) = P(A)P(A\B)+P(A\B)+P(B)P(A\B) = P(A)+P(B)P(A\B) 2.5 2Properties of Probability (6) 6. 8A 1:::;A N W we have the Inclusion-Exclusion Identity: N ! N [ P A k = P(A k P(A k1)k2 k=1 k=1 k1
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