2 - 18.05 Lecture 2 February 4, 2005 1.5 Properties of...

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18.05 Lecture 2 February 4, 2005 § 1.5 Properties of Probability. 1. P ( A ) [0 , 1] 2. P ( S ) = 1 3. P ( A i ) = P ( A i ) if disjoint A i A j = , i = j ± The probability of a union of disjoint events is the sum of their probabilities. 4. P ( ) , P ( S ) = P ( S ∪∅ ) = P ( S ) + P ( ) = 1 where S and are disjoint by de±nition, P (S) = 1 #2., therefore, P ( ) = 0. 5. P ( A c ) = 1 - P ( A ) because A, A c are disjoint, P ( AA c ) = P ( S ) = 1 = P ( A ) + P ( A c ) the sum of the probabilities of an event and its complement is 1. 6. If A B, P ( A ) P ( B ) de±nition, B = A ( B \ A ) , two disjoint sets. P ( B ) = P ( A ) + P ( B \ A ) P ( A ) 7. P ( A B ) = P ( A ) + P ( B ) - P ( AB ) must subtract out intersection because it would be counted twice, as shown: write in terms of disjoint pieces to prove it: P ( A ) = P ( A \ B ) + P ( AB ) P ( B ) = P ( B \ A ) + P ( AB ) P ( A B ) = P ( A \ B ) + P ( B \ A ) + P ( AB ) Example: A doctor knows that P (bacterial infection) = 0.7 and P (viral infection) = 0.4 What is P (both) if P (bacterial viral) = 1? P (both) = P (B V) 1 = 0.7 + 0.4 - P (BV) P (BV) = 0.1 Finite Sample Spaces There are a ±nite # of outcomes S = s 1 , ..., s n } { De±ne p i = P ( s i ) as the probability function.
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This note was uploaded on 05/02/2011 for the course DYNAM 101 taught by Professor Matuka during the Spring '11 term at MIT.

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2 - 18.05 Lecture 2 February 4, 2005 1.5 Properties of...

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