16-322-stochastic-estimation-and-control-fall-2004

16-322-stochastic-estimation-and-control-fall-2004 - 16.322...

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16.322 Stochastic Estimation and Control Professor Vander Velde Lecture 2 Lecture 2 Last time: Given a set of events with are mutually exclusive and equally likely , () nE PE N = . Example: card games Number of different 5-card poker hands 52 2,598,960 5 ⎛⎞ == ⎜⎟ ⎝⎠ Number of different 13-card bridge hands 52 635,013,559,600 13 Poker probabilities: 1. 4444 13 12 11 10 2111 (one pair) 0.423 52 3! 5 P ⎛⎞ ⎛⎞ ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎝⎠ 2. 444 13 12 11 221 (two pair) 0.0476 52 2! 5 P ⎛⎞ ⎛⎞ ⎛⎞ ⎜⎟ ⎜⎟ ⎜⎟ ⎝⎠ ⎝⎠ ⎝⎠ Independence ( ) ( ) PAB PAPB = when A,B independent ( ) ( | ) PAB PAPB A = when A,B dependent Definition of independence of events: 123 1 2 ( ... ) ( ) ( )... ( ) mm PAAA A PAPA PA = for all m (pairwise, threewise, etc.) Definition of conditional probability : (|) PAB PB A PA = restricts our attention to the situation where A has already occurred. If A,B are independent ()() PAPB PB Two useful results:
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16.322 Stochastic Estimation and Control Professor Vander Velde Lecture 2 1. ( ...) ( ) ( | ) ( | ) ( | )... P ABCD P A P B A P C AB P D ABC = Derive this by letting A=CD. Then () ( ) ( | ) ( ) ( | ) ( | ) PBCD PCDPBCD PCPDCPDCD == 2. If 1 A , 2 A ,… is a set of mutually exclusive and collectively exhaustive events, then 12 1 1 2 2 ( ) ( ) ( ) ... ( ) ( ) ( | ) ( ) ( | ) ... ( ) ( | ) nn n PE PEA PAPE A PA PE A = + ++ = + ( ) ( ) ( ) PA B PA PB PAB += + must subtract off P(AB) because it is counted twice by the first two terms of the RHS. ( ) ( ) ( ) ( ) ( ) ( ) ( ) PA B C PA PB PC PAB PAC PBC PABC ++ = + + + (for three events). If four events, PA B C D +++ would be the sum of probabilities of the single events, minus the probability of the joint events taken two at a time, plus the
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This note was uploaded on 11/07/2011 for the course AERO 16.322 taught by Professor Wallacevandervelde during the Fall '04 term at MIT.

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16-322-stochastic-estimation-and-control-fall-2004 - 16.322...

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