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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 ## An urn contains 10 balls, numbered from 1 to 10. If 2 balls tagged by: BTGmoderatorLU ##### This topic has 2 expert replies and 1 member reply ### Top Member ## An urn contains 10 balls, numbered from 1 to 10. If 2 balls ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult Source: Veritas Prep An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even? A. 25% B. 37.5% C. 50% D. 62.5% E. 75% The OA is C ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 986 messages Followed by: 27 members Upvotes: 59 BTGmoderatorLU wrote: Source: Veritas Prep An urn contains 10 balls, numbered from 1 to 10. If 2 balls are selected at random with replacement from the urn, what is the probability that the sum of the 2 numbers on the balls will be even? A. 25% B. 37.5% C. 50% D. 62.5% E. 75% There are exactly 4 equiprobable events: E1 = first ball odd , second ball odd E2 = first ball odd, second ball even E3 = first ball even, second ball odd E4 = first ball even, second ball even Exactly two of the 4 events mentioned above are favorable (E1 and E4), hence the correct answer is 50% (C). This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or https://GMATH.com.br (Portuguese version) Course release PROMO : finish our test drive till 30/Dec with (at least) 50 correct answers out of 92 (12-questions Mock included) to gain a 50% discount! ### GMAT/MBA Expert Elite Legendary Member Joined 23 Jun 2013 Posted: 9944 messages Followed by: 493 members Upvotes: 2867 GMAT Score: 800 Hi All, We're told that an urn contains 10 balls, numbered from 1 to 10 and we're told that 2 balls are selected at random WITH replacement. We're asked for the probability that the SUM of the 2 numbers on the balls will be EVEN. This question can be solved in a couple of different ways, including as a straight-probability question. To start, there are two ways to get a sum that is EVEN: (Odd on the first) and (Odd on the second) (Even on the first) and (Even on the second) Since there are an equal number of Odd and Even numbers AND we are replacing the first ball before we pull the second, the probability of pulling one Odd or one Even is the same each time: 1/2 (Odd) and (Odd) = (1/2)(1/2) = 1/4 (Even) and (Even) = (1/2)(1/2) = 1/4 Total probability of ending with a sum that is Even on two balls is 1/4 + 1/4 = 2/4 = 1/2 Final Answer: C GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at Rich.C@empowergmat.com ### Top Member Newbie | Next Rank: 10 Posts Joined 06 Dec 2018 Posted: 7 messages There are 10 balls; 5 of which are even and 5 of which are odd. Probability of drawing 1 even ball is 50% Probability of drawing 1 odd ball is 50% Here are the possible combos: 1. Draw even; Draw Even ==> Adds to Even 2. Draw odd; draw odd ==> adds to even For (1): 50% * 50% For (2): 50% * 50% Then add them up, 25% + 25% = 50% • 5 Day FREE Trial Study Smarter, Not Harder Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0

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