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1.4.1 Bayes

# 1.4.1 Bayes - open box B therefore • • If the key is in...

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9/13/2010 1 The Law of Total Probability Suppose that the events A 1 , A 2 , …, A n are a partition of the sample space . That is, A 1 , A 2 , …, A n are mutually exclusive and A i = . Then for any event B, P(B) = P(B|A 1 )P(A 1 ) + … + P(B|A n )P(A n ) Thomas Bayes (1702–1761) Bayes Rule If P ( B ) > 0, then ) ( ) | ( ) | ( c c A P A B P A P A B P A P A B P B A P If P ( B ) > 0 and A 1 , A 2 , …, A n are a partition of the sample space , then . ) ( ) | ( ) ( ) | ( ) | ( 1 n j j j i i i A P A B P A P A B P B A P ) ( ) | ( ) ( ) | ( Monty Hall Let A , B , C be the event that the keys are in box A, B, C respectively. Let M be the event that Monty Hall opens box B. Suppose A is chosen. If the key is in box A, then both B and C are empty and Monty can randomly open any one of them at his own choice, therefore If the key is in box B, then Monty surely won’t

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Unformatted text preview: open box B, therefore • • If the key is in box C, then Monty can only open box B because surely he won’t open A or C, therefore Bayes 9/13/2010 2 Misconception • Many people would think that since we know that B is empty after Monty had opened box B, the chance that the key is in either boxes will become half and it would make no difference to switch or not. • However, this kind of reasoning actually misinterpreted the information we obtained from Monty’s action and lead to incorrect calculation. • What we really want?...
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1.4.1 Bayes - open box B therefore • • If the key is in...

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