4 - 18.05 Lecture 4 February 11, 2005 Union of Events P(A1...

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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 stu± 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 ² n ³ ( n - 2)! ² n ³ ( n - 3)! - ... + ( - 1) n +1 ² n ³ ( n - n )! P ( A 1 ... A n ) = n + × n - 2 n !3 n ! nn ! general term: ² n ³ ( n - k )! n !( n - k )! 1 = = kn ! k !( n - k )! n ! k ! 11 SUM = 1 - + 3! - ... + ( - 1) n +1 1 2! n ! 23 Recall: Taylor series for e x = 1 + x + x + x + ... 2! 3! 1 for x= -1, e - 1 = 1 - 1 + 1 + ... 3! 2 - therefore, SUM = 1 -l im it of series as n →∞ When n is large, the probability converges to 1 - e - 1 = 0 . 63 § 2.1 - Conditional Probability Given that B “happened,” what is the probability that A also happened?
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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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4 - 18.05 Lecture 4 February 11, 2005 Union of Events P(A1...

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