18_05_lec4

18_05_lec4 - 18.05 Lecture 4 Union of Events P(A1 An = i P(Ai i

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18.05 Lecture 4 February 11, 2005 Union of Events P ( A 1 ... A n ) = P ( A i ) P ( A i A j ) + P ( A i A j A k ) + ... ⇒ ⇒ i i<j i<j<k It is often easier to calculate P (intersections) than P (unions) Matching Problem : You have n letters and n envelopes, randomly stuff the letters into the envelopes. What is the probability that at least one letter will match its intended envelope? P ( A 1 ... A n ) , A i = { ith position will match } ( n 1)! P ( A i ) = 1 = n ! n (permute everyone else if just A i is in the right place.) P ( A i A j ) = ( n 2)! ( A i and A j are in the right place) n ! P ( A i 1 A i 2 ...A ik ) = ( n k )! n ! 1 ± n ² ( n 2)! ± n ² ( n 3)! ... + ( 1) n +1 ± n ² ( n n )! P ( A 1 ... A n ) = n + ⇒ ⇒ × n 2 n ! 3 n ! n n ! general term: ± n ² ( n k )! n !( n k )! 1 = = k n ! k !( n k )! n ! k ! 1 1 SUM = 1 + 3! ... + ( 1) n +1 1 2! n ! 2
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This note was uploaded on 01/24/2011 for the course MAT 111 taught by Professor Drstag during the Summer '06 term at MIT.

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18_05_lec4 - 18.05 Lecture 4 Union of Events P(A1 An = i P(Ai i

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