# sol2[1] - STAT 471 Assignment 2 solutions 1 Calculate the...

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STAT 471 - Assignment 2 solutions 1. Calculate the chance of needing a game 7 in a best-of-seven series when one team has a 3 / 5 chance of winning each individual game. Solution: Game seven is needed if each team wins exactly 3 of the ﬁrst 6 games. There are ( 6 3 ) = 20 ways that this can happen: WWWLLL, WLWLWL, LLWLWW, etc. By independence, each of these ways has the same probability of happening, for example, P ( WWWLLL ) = P ( W ) P ( W ) P ( W ) P ( L ) P ( L ) P ( L ) = (3 / 5) 3 (2 / 5) 3 = 216 / 15625. Therefore P ( need game 7) = (20)(216) 15625 = . 27648 . 2. Players A and B alternately throw an ordinary pair of dice in that order. Player A wins if he gets a sum of six, while Player B wins if he gets a sum of seven. They keep playing until somebody gets the sum he needs. If A starts the game, what is his probability of winning? Solution: There are two ways to solve this problem: a clever way using conditional probabilities and a brute force way that involves inﬁnite sums. Here are both solutions. a) Let A be the event that player A eventually wins, and B the event that player B eventually wins. Let A 1 be the event that player A wins (rolls a six) on the ﬁrst toss, and B 2 the event that player B wins (rolls a seven) on the second toss. Let C (for ‘continue’) be the event that nobody wins on the ﬁrst two tosses, i.e.,

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## This note was uploaded on 09/24/2009 for the course IE, KAIST IE+HSS taught by Professor Various. during the Spring '09 term at Korea Advanced Institute of Science and Technology.

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sol2[1] - STAT 471 Assignment 2 solutions 1 Calculate the...

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