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Unformatted text preview: If the coin and dice are both fair, then each of the 12 outcomes is equally likely. Additivity & normalization > P({})=1/12 Example Calculate probabilities of events by counting the number of outcomes and dividing by the total number of outcomes P(A) = 6/12 P(B) = 6/12 P(C) = 2/12 A = H 1, H 2, H 3, H 4, H 5, H 6, { } B = H 1, H 3, H 5, T 1, T 3, T 5 { } C = H 3, T 3 { } Does this conflict with the additivity axiom? More Probability Properties P() = 0 P(A c ) = 1  P(A) If A B then P(A) P(B) P(B) = P(A B) + P(A c B) P(A B) = P(A) + P(A c B) P(A B) = P(A) + P(B)  P(A B) P(A B C) = P(A) + P(A c B) + P(A c B c C) Exercises Exercise #1 Exercise #2 Exercise #3...
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This note was uploaded on 02/01/2011 for the course BME 335 taught by Professor Dunn during the Spring '10 term at University of Texas at Austin.
 Spring '10
 Dunn

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