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Unformatted text preview: EE 278 October 2, 2009 Statistical Signal Processing Handout #3 Homework #1 Solutions 1. Monty Hall. (Bonus) Gold is placed behind one of three curtains. A contestant chooses one of the curtains. Monty Hall, the game host, opens an unselected empty curtain. The contestant can choose either to switch his selection to the third curtain or not. a. What is the sample space for this random experiment? Hint: an outcome consists of the curtain with gold, the curtain chosen by the contestant, and the curtain chosen by Monty. b. Assume that placement of the gold behind the three curtains is random, that the contestant’s choice of curtains is random and independent of the gold placement, and that Monty Hall’s choice of an empty curtain is equally likely between the two alternatives (if the two unselected curtains are empty). Specify the probability measure for this random experiment and use it to compute the probability of winning the gold if the contestant decides to switch. Solution (15 points) a. The sample space consists of triplets of the form (curtain with gold behind it, curtain chosen by player, curtain that Monty opens). If we denote the curtains by A , B , and C , then the sample space is Ω = { ( A,A,B ) , ( A,A,C ) , ( A,B,C ) , ( A,C,B ) , ( B,B,A ) , ( B,B,C ) , ( B,A,C ) , ( B,C,A ) , ( C,C,A ) , ( C,C,B ) , ( C,B,A ) , ( C,A,B ) } . b. As discussed in lecture notes 1, for a discrete sample space, the probability measure is completely specified by the probabilities of the single outcome events. For this problem we can calculate the probabilities as follows: P( A,A,B ) = P { gold is behind A , player’s first choice is A } · P { Monty opens B  gold is behind A , player’s first choice is A } = 1 2 · P { gold is behind A } · P { player’s first choice is A } = 1 2 · 1 3 · 1 3 = 1 18 . Note that Monty Hall’s choice of curtain to open is not independent of where the gold is placed and what the player’s initial choice was. That is why we used conditional probability in the above derivation. Using the same approach we find P( A,A,C ) = P( B,B,A ) = P( B,B,C ) = P( C,C,A ) = P( C,C,B ) = 1 18 . This covers all the cases where the gold placement and the initial choice of contestant coincide. For the other cases, P( A,B,C ) = P { gold is behind A , player’s first choice is B } · P { Monty opens C  gold is behind A , player’s first choice is B } = 1 · P { gold is behind A } · P { player’s first choice is B } = 1 · 1 3 · 1 3 = 1 9 . Using the same argument, P( A,C,B ) = P( B,A,C ) = P( B,C,A ) = P( C,A,B ) = P( C,B,A ) = 1 9 . We now have a complete description of the probability space for this random experiment. To find the probability of winning if the player switches, we need to find the subset W of the sample space corresponding to this event, which is W = { ( A,B,C ) , ( A,C,B ) , ( B,A,C ) , ( B,C,A ) , ( C,A,B ) , ( C,B,A ) } ....
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 Fall '09
 BalajiPrabhakar
 Probability, Signal Processing, Probability theory

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