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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## The probability of winning game A is X and the probability ##### This topic has 4 expert replies and 0 member replies ### Top Member ## The probability of winning game A is X and the probability ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once? A. X-YX B. XY C. X+Y D. X+Y-2XY E. X^2Y^2-XY^2-X^2Y OA D Source: Economist GMAT ### GMAT/MBA Expert Elite Legendary Member Joined 23 Jun 2013 Posted: 10130 messages Followed by: 494 members Upvotes: 2867 GMAT Score: 800 Top Reply Hi All, We're told that the probability of winning Game A is X and the probability of winning Game B is Y. We're asked when playing a SINGLE round of EACH of the two games, what is the probability of winning EXACTLY ONCE. This question can be solved in a couple of different ways, including by TESTing VALUES. IF... X = .4 AND Y = .5 then there is a 40% chance to win Game A and a 60% chance to lose Game A and a 50% chance to win Game B and a 50% chance to lose Game B. The probability of winning A and losing B = (.4)(.5) = .2 The probability of winning B and losing A = (.5)(.6) = .3 Total probability of winning just one Game = .2 + .3 = .5 Thus, we're looking for an answer that equals .5 when X = .4 and Y = .5 There's only one answer that matches.... Final Answer: D GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 Top Reply BTGmoderatorDC wrote: The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once? A. X-YX B. XY C. X+Y D. X+Y-2XY E. X^2Y^2-XY^2-X^2Y Source: Economist GMAT $$? = P\left( {{\text{win}}\,\,{\text{just}}\,\,{\text{A}}\,\,{\text{or}}\,\,{\text{win}}\,\,{\text{just}}\,\,{\text{B}}} \right)\,\,\,\, = \,\,\,\,P\left( {{\text{win}}\,\,{\text{A}}\,{\text{or}}\,\,{\text{B}}} \right) - P\left( {{\text{win}}\,\,{\text{both}}} \right)$$ $$P\left( {{\rm{win}}\,\,{\rm{both}}} \right) = xy\,\,\,\,\,\,\,\,\left[ {\,x,y\,\,\,{\rm{constants}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{independency}}\,} \right]$$ $$P\left( {{\text{win}}\,\,{\text{A}}\,{\text{or}}\,\,{\text{B}}} \right) = x + y - xy\,\,\,\,\,\left[ {\,{\text{simplifier}}\,} \right]$$ $$? = x + y - 2xy$$ This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. "Kolmogorov - PoincarÃ© - Gauss - Euler - Newton, are only five lives separating us from the source of our science". (Vladimir Arnold) _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br Last edited by fskilnik@GMATH on Sat Oct 13, 2018 1:30 pm; edited 2 times in total ### GMAT/MBA Expert GMAT Instructor Joined 22 Aug 2016 Posted: 1898 messages Followed by: 30 members Upvotes: 470 BTGmoderatorDC wrote: The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once? A. X-YX B. XY C. X+Y D. X+Y-2XY E. X^2Y^2-XY^2-X^2Y OA D Source: Economist GMAT Given: 1. The probability of winning game A is X, and 2. The probability of winning game B is Y Thus, The probability of NOT winning game A = 1 - X The probability of NOT winning game B = 1 - Y Thus, the probability of winning exactly once = the probability of winning either A or B but not both the games = [(Probability of winning game A)*(Probability of NOT winning game B)] + [(Probability of winning game B)*(Probability of NOT winning game A)] = X*(1 - Y) + Y*(1 - X) = X - XY + Y - XY = X + Y - 2XY The correct answer: D Hope this helps! -Jay _________________ Manhattan Review GMAT Prep Locations: New York | Bangkok | Abu Dhabi | Rome | and many more... Schedule your free consultation with an experienced GMAT Prep Advisor! Click here. ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2444 messages Followed by: 18 members Upvotes: 43 BTGmoderatorDC wrote: The probability of winning game A is X and the probability of winning game B is Y. When playing a single round of each of the two games, what is the probability of winning exactly once? A. X-YX B. XY C. X+Y D. X+Y-2XY E. X^2Y^2-XY^2-X^2Y P(winning exactly once) = 1 - P(winning zero times) - P(winning both times) P(wWinning exactly once) = 1 - (1 - X)(1 - Y) - XY = 1 - (1 - X - Y + XY) - XY = X + Y - 2XY Alternate Solution: There are two ways of winning exactly once, either (Win A, Lose B) OR (Lose A, Win B). The probability of winning A is X, so the probability of losing A is (1 - X). Similarly, the probability of winning B is Y, and so the probability of losing B is (1 - Y). P(Win A, Lose B) = X*(1 - Y) = X - XY P(Lose A, Win B) = (1 - X)*Y = Y - XY The probability that either of these events happens is the sum of their individual probabilities, so P(Winning exactly once) = X - XY + Y - XY = X + Y - 2XY Answer: D _________________ Scott Woodbury-Stewart Founder and CEO scott@targettestprep.com See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. 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