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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 ## A box contains 10 balls numbered from 1 to 10 inclusive. If ##### This topic has 3 expert replies and 2 member replies ### Top Member ## A box contains 10 balls numbered from 1 to 10 inclusive. If ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and replaces it, and then Jane removes a ball at random, what is the probability that both women removed the same ball? A. 1/100 B. 1/90 C. 1/45 D. 1/10 E. 41/45 OA D Source: Magoosh ### Top Member Legendary Member Joined 29 Oct 2017 Posted: 936 messages Followed by: 4 members For Ann or For Jane TOTAL cases = 10 For Ann favorable cases = 10 For Ann probability of picking the ball = 10/10 = 1 Now that she has replaced the ball TOTAL cases remains the same. For Jane favorable case = 1 (If Ann had picked the ball number 2, then Jane has to pick only that ball, so only one choice for jane, hence only one fav case) Probability for Jane picking the same ball = 1/10. D is the right answer. ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2653 messages Followed by: 18 members Upvotes: 43 BTGmoderatorDC wrote: A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and replaces it, and then Jane removes a ball at random, what is the probability that both women removed the same ball? A. 1/100 B. 1/90 C. 1/45 D. 1/10 E. 41/45 Since Ann can pick any ball and Jane has only a 1/10 chance to match it, the probability that they select the same ball is: 1 x 1/10 = 1/10 Alternate Solution: Many students think that the answer is 1/10 x 1/10 = 1/100, thinking that each woman has a 1/10 probability of picking a particular ball. Let’s look at the “long” solution to this problem to clarify the correct answer of 1/10. Consider the ball marked “1.” The probability that Ann picks this ball is 1/10, and so the probability that Jane also picks this ball is 1/10. The probability that both women will pick the ball marked “1” is, therefore, 1/10 x 1/10 = 1/100. Now consider the ball marked “2.” Again, the probability that Ann picks this ball is 1/10, and the probability that Jane also picks this ball is 1/10. Thus, the probability that both women will pick the ball marked “2” is, 1/10 x 1/10 = 1/100. We continue this for the balls marked 3, 4, 5, 6, 7, 8, 9, and 10. For each ball, the probability will be 1/100. Since there are 10 balls, the probability that Ann and Jane will pick the same ball is, therefore, 10 x 1/100 = 10/100 = 1/10. Answer: D ### GMAT/MBA Expert GMAT Instructor Joined 09 Oct 2010 Posted: 1449 messages Followed by: 32 members Upvotes: 59 BTGmoderatorDC wrote: A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and replaces it, and then Jane removes a ball at random, what is the probability that both women removed the same ball? A. 1/100 B. 1/90 C. 1/45 D. 1/10 E. 41/45 Source: Magoosh $$?\,\, = \,\,P\left( {{\rm{same}}\,\,{\rm{ball}}\,\,{\rm{twice}}\,{\rm{,}}\,{\rm{with}}\,\,{\rm{replacement}}} \right)$$ Exactly 1 ball extracted in the second extraction (out of 10 equiprobable possibilities) will be the same as (any given) ball extracted in the first extraction. Therefore the answer is 1/10, immediately. Another approach: $${\rm{Total}} = 10 \cdot 10\,\,{\rm{pairs}}\,\,\,\left( {{1^{{\rm{st}}}}\,,\,\,{2^{{\rm{nd}}}}} \right)\,{\rm{ extracted}}\,,\,\,{\rm{all}}\,\,{\rm{equiprobable}}$$ $${\rm{Favorable}}\,\,{\rm{ = }}\,\,{\rm{10}} \cdot {\rm{1}}\,\,{\rm{pairs}}\,{\rm{among}}\,{\rm{the}}\,{\rm{above}}\,\,\,\,\,\left[ {\,\left( {{1^{{\rm{st}}}} = {\rm{any}}\,\,{\rm{ball}}\,,\,\,{2^{{\rm{nd}}}} = {\rm{same}}\,\,{1^{{\rm{st}}}}\,\,{\rm{ball}}} \right)\,} \right]\,$$ $$? = {{10} \over {100}} = {1 \over {10}}$$ These solutions follows the notations and rationale taught in the GMATH method. Regards, Fabio. _________________ Fabio Skilnik :: GMATH method creator ( Math for the GMAT) English-speakers :: https://www.gmath.net Portuguese-speakers :: https://www.gmath.com.br ### Top Member Master | Next Rank: 500 Posts Joined 15 Oct 2009 Posted: 329 messages Upvotes: 27 BTGmoderatorDC wrote: A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and replaces it, and then Jane removes a ball at random, what is the probability that both women removed the same ball? A. 1/100 B. 1/90 C. 1/45 D. 1/10 E. 41/45 OA D Source: Magoosh How many ways to take a distinct pair of numbers that don't match out of 10 numbers ? 10!/2!8! = 45. This is like without replacement in order to avoid a matching pair. How many ways to take a distinct pair out of 10 numbers with replacement ? 10^2/2 = 50. This includes the above plus matching pairs. To find the number of matching pairs, subtract the first answer from the second: 50-45 =5 Probability of selecting a matching pair is then 5/50 = D,1/10 ### GMAT/MBA Expert GMAT Instructor Joined 08 Dec 2008 Posted: 12924 messages Followed by: 1248 members Upvotes: 5254 GMAT Score: 770 BTGmoderatorDC wrote: A box contains 10 balls numbered from 1 to 10 inclusive. If Ann removes a ball at random and replaces it, and then Jane removes a ball at random, what is the probability that both women removed the same ball? A. 1/100 B. 1/90 C. 1/45 D. 1/10 E. 41/45 P(Ann and Jane remove same ball) = P(Ann removes ANY ball AND Jane's ball matches Ann's ball) = P(Ann removes ANY ball) x P(Jane's ball matches Ann's ball) = 1 x 1/10 = 1/10 Answer: D Aside: Once Jane removes her ball (and then replaces it), we have 10 balls, and 1 of them is the one that Jane picked. So, P(Jane's ball matches Ann's ball) = 1/10 Cheers, Brent _________________ Brent Hanneson – Creator of GMATPrepNow.com Use my video course along with Sign up for free Question of the Day emails And check out all of these free resources GMAT Prep Now's comprehensive video course can be used in conjunction with Beat The GMAT’s FREE 60-Day Study Guide and reach your target score in 2 months! • Free Veritas GMAT Class Experience Lesson 1 Live Free Available with Beat the GMAT members only code • Free Trial & Practice Exam BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • Magoosh Study with Magoosh GMAT prep Available with Beat the GMAT members only code • Get 300+ Practice Questions 25 Video lessons and 6 Webinars for FREE Available with Beat the GMAT members only code • 1 Hour Free BEAT THE GMAT EXCLUSIVE Available with Beat the GMAT members only code • FREE GMAT Exam Know how you'd score today for$0

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