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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 ## Bag of coins ##### This topic has 3 expert replies and 0 member replies ## Bag of coins A bag contains 3 gold coins and 7 silver coins. The bag contains nothing else. If 3 coins are selected at random and without replacement from the bag, what is the probability that exactly 2 of the coins are silver? A) 7/40 B) 7/24 C) 3/10 D) 21/40 E) 2/3 ### GMAT/MBA Expert GMAT Instructor Joined 11 Apr 2010 Posted: 1179 messages Followed by: 88 members Upvotes: 447 Probability that exactly 2 coins are silver means the third coin will be gold coin. So, there will be 1 gold coin and 2 silver coins. Hence, the required probability = So, the probability that exactly 2 of the coins are silver is 21/40. The correct answer is (D). _________________ Rahul Lakhani Quant Expert Gurome, Inc. https://www.GuroMe.com On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits 1-800-566-4043 (USA) +91-99201 32411 (India) ### GMAT/MBA Expert GMAT Instructor Joined 19 Oct 2009 Posted: 1302 messages Followed by: 162 members Upvotes: 539 GMAT Score: 800 Note that Rahul is using: Probability = #desirable/#total in which the denominator is "all the ways we can 'pull out' three coins from ten", and in which the numerator is "all the ways we can pull out 2 silver coins from 7 AND 1 gold coin from 3". The question could also have been worded "If 3 coins are selected randomly one after another without replacement..." (If you're not replacing them, it doesn't matter whether you pull the three coins out simultaneously or one after another). _____ We could have also solved by considering order. We want 2 silver and 1 gold, so: (3/10)*(7/9)*(6/8) = 7/40 (G).........(S).....(S) Because there are three arrangements in which we can have 2 silver and 1 gold (SSG, GSS, SGS), we have: 3*(7/40) = 21/40 _________________ Kaplan Teacher in Toronto Free GMAT Practice Test under Proctored Conditions! - Find a practice test near you or live and online in Kaplan's Classroom Anywhere environment. Register today! ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2456 messages Followed by: 18 members Upvotes: 43 pkw209 wrote: A bag contains 3 gold coins and 7 silver coins. The bag contains nothing else. If 3 coins are selected at random and without replacement from the bag, what is the probability that exactly 2 of the coins are silver? A) 7/40 B) 7/24 C) 3/10 D) 21/40 E) 2/3 The number of ways to select 2 silver coins is 7C2 = (7 x 6)/2! = 21. The number of ways to select 1 gold coin is 3C1 =3. The number of ways to select 3 coins is 10C3 = (10 x 9 x 8)/3! = (10 x 9 x 8)/(3 x 2 x 1) = 10 x 3 x 4 = 120. So the probability is (21 x 3)/120 = 63/120 = 21/40. Alternate Solution: There are 3 different ways to select exactly 2 silver and 1 gold coin, without replacement, out of the coins in the bag. The outcomes are: (GSS) or (SGS) or (SSG). The probability of GSS is 3/10 x 7/9 x 6/8 = 21/120. The probability of SGS is 7/10 x 3/9 x 6/8 = 21/120. The probability of SSG is 7/10 x 6/9 x 3/8 = 21/120. Thus, the probability of drawing exactly 2 silver and 1 gold coin is 21/120 x 3 = 63/120 = 21/40. 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. Read our reviews • 5-Day Free Trial 5-day free, full-access trial TTP Quant Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Award-winning private GMAT tutoring Register now and save up to$200

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