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slides8 - Lecture Stat 302 Introduction to Probability...

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Lecture Stat 302 Introduction to Probability - Slides 8 AD Feb. 2010 AD () Feb. 2010 1 / 17
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Expected Value The expectation of X , denoted E ( X ) E ( X ) = i = 1 x i p ( x i ) = x : p ( x ) > 0 x p ( x ) It corresponds to a weighted average of the possible values that X experiments then the average will be E ( X ) . Example : Assume you have a fair die and X is the number rolled then p ( i ) = 1 6 for i = 1 , 2 , ..., 6 and E ( X ) = 1 1 6 + 2 1 6 + ± ± ± + 6 1 6 = 1 + 2 + 3 + 4 + 5 + 6 6 = 21 6 = 7 2 AD () Feb. 2010 2 / 17
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Example: Expected Payout of Roulette Wheel Red/Black : We have P ( X 1 = 1 ) = 18 38 and P ( X 1 = 1 ) = 20 38 so E ( X 1 ) = 1 ± 18 38 + ( 1 ) ± 20 38 = 1 19 ² & 0 . 053 Straight up : We have P ( X 2 = 35 ) = 1 38 and P ( X 2 = 1 ) = 37 38 so E ( X 2 ) = 35 ± 1 38 + ( 1 ) ± 37 38 = 1 19 . Split bet : We have P ( X 3 = 17 ) = 2 38 and P ( X 3 = 1 ) = 36 38 so E ( X 3 ) = 17 ± 2 38 + ( 1 ) ± 36 38 = 1 19 . Street bet : We have P ( X 4 = 11 ) = 3 38 and P ( X 4 = 1 ) = 35 38 so E ( X 4 ) = 11 ± 3 38 + ( 1 ) ± 35 38 = 1 19 . AD () Feb. 2010 3 / 17
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Example: Tramways in Melbourne In Melbourne (Australia), no one collects fare when you board a tram and no one checks systematically whether you have a ticket. Only meeting inspectors 5%. Assume you are interested in minimizing the average cost, should you buy tram tickets or not? Let C being the cost. If you buy a ticket each time you board a tram then the cost is C = 2 $ with proba. 1 (i.e. there is no uncertainty, i.e. P ( C = 2 ) = 1) so the average cost is E ( C ) = 2 1 = 2 $ . If you never buy tram tickets then the cost is
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