This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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(Ø) = 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...
View
Full Document
 Spring '10
 Dunn
 Coin, exam ple, ple • Theprobability

Click to edit the document details