18_05_lec2

18_05_lec2 - 18.05 Lecture 2 February 4, 2005 1.5...

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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 by #2., therefore, P ( ) = 0. 5. P ( A c ) = 1 P ( A ) because A, A c are disjoint, P ( A A 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 ) by 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
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18_05_lec2 - 18.05 Lecture 2 February 4, 2005 1.5...

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