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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 ## Three boys are ages 4, 6 and 7 respectively. Three girls are tagged by: BTGmoderatorLU ##### This topic has 3 expert replies and 0 member replies ### Top Member ## Three boys are ages 4, 6 and 7 respectively. Three girls are ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult Source: Manhattan Prep Three boys are ages 4, 6 and 7 respectively. Three girls are ages 5, 8 and 9, respectively. If two of the boys and two of the girls are randomly selected and the sum of the selected children's ages is z, what is the difference between the probability that z is even and the probability that z is odd? A. 1/9 B. 1/6 C. 2/9 D. 1/4 E. 1/2 The OA is A. ### GMAT/MBA Expert GMAT Instructor Joined 22 Aug 2016 Posted: 1770 messages Followed by: 28 members Upvotes: 470 BTGmoderatorLU wrote: Source: Manhattan Prep Three boys are ages 4, 6 and 7 respectively. Three girls are ages 5, 8 and 9, respectively. If two of the boys and two of the girls are randomly selected and the sum of the selected children's ages is z, what is the difference between the probability that z is even and the probability that z is odd? A. 1/9 B. 1/6 C. 2/9 D. 1/4 E. 1/2 The OA is A. Since the ages are integers, their sum would be either even or odd. We see that there are three boys and three girls, thus, there are 3*3 = 9 possible sums; out of which few are even and few are odd. Let's first find out the sum = Odd. Sum would be odd if an even adds to odd OR and odd adds to even. 1. Say, we pick one boy with even age, thus, we must pick a girl with odd age. There are two boys with even age (4 and 6) and there are two girls with odd age (5 and 9). > Number of ways to select one out of two boys = 2C1 = 2 > Number of ways to select one out of two girls = 2C1 = 2 Total number of ways = 2*2 = 4 2. Say, we pick one boy with odd age, thus, we must pick a girl with even age. There is only one boy with even age (7) and there is only one girl with odd age (8). > Number of ways to select one out of one boy = 1 > Number of ways to select one out of one girl = 1 Total number of ways = 1*1 = 2 Total number of ways to get odd sum = 4 + 1 = 5 => The probability that z is odd = 5/9 => The probability that z is even = 1 - 5/9 = 4/9 Difference = |4/9 - 5/9| = 1/9 The correct answer: A Hope this helps! -Jay _________________ Manhattan Review GRE Prep Locations: GRE Classes Raleigh NC | GRE Prep Course Singapore | GRE Prep Philadelphia | SAT Prep Classes Toronto | and many more... Schedule your free consultation with an experienced GMAT Prep Advisor! Click here. ### GMAT/MBA Expert GMAT Instructor Joined 25 May 2010 Posted: 15105 messages Followed by: 1859 members Upvotes: 13060 GMAT Score: 790 BTGmoderatorLU wrote: Source: Manhattan Prep Three boys are ages 4, 6 and 7 respectively. Three girls are ages 5, 8 and 9, respectively. If two of the boys and two of the girls are randomly selected and the sum of the selected children's ages is z, what is the difference between the probability that z is even and the probability that z is odd? A. 1/9 B. 1/6 C. 2/9 D. 1/4 E. 1/2 Since only a few cases are possible, we can list them out: 4+6+5+8 = 23 4+6+5+9 = 24 4+6+8+9 = 27 4+7+5+8 = 24 4+7+5+9 = 25 4+7+8+9 = 28 6+7+5+8 = 26 6+7+5+9 = 27 6+7+8+9 = 30 Of the 9 cases above, the four in red are ODD, implying that the remaining five cases are EVEN. Since 5/9 of the cases are even, and 4/9 are odd, we get: P(z is even) - P(z is odd) = 5/9 - 4/9 = 1/9. The correct answer is A. _________________ Mitch Hunt Private Tutor for the GMAT and GRE GMATGuruNY@gmail.com If you find one of my posts helpful, please take a moment to click on the "UPVOTE" icon. Available for tutoring in NYC and long-distance. For more information, please email me at GMATGuruNY@gmail.com. Student Review #1 Student Review #2 Student Review #3 Free GMAT Practice Test How can you improve your test score if you don't know your baseline score? Take a free online practice exam. Get started on achieving your dream score today! Sign up now. ### GMAT/MBA Expert GMAT Instructor Joined 25 Apr 2015 Posted: 2012 messages Followed by: 15 members Upvotes: 43 BTGmoderatorLU wrote: Source: Manhattan Prep Three boys are ages 4, 6 and 7 respectively. Three girls are ages 5, 8 and 9, respectively. If two of the boys and two of the girls are randomly selected and the sum of the selected children's ages is z, what is the difference between the probability that z is even and the probability that z is odd? A. 1/9 B. 1/6 C. 2/9 D. 1/4 E. 1/2 The sum of the two selected boys can be either 4 + 6 = 10, 4 + 7 = 11 or 6 + 7 = 13. Thus, there is a 1/3 probability that the sum of the ages of the two boys will be even and 2/3 probability that the sum of the ages of the two boys will be odd. Similarly, the sum of the two selected girls can be either 5 + 8 = 13, 5 + 9 = 14 or 8 + 9 = 17. Thus, there is a 1/3 probability that the sum of the ages of the two girls will be even and 2/3 probability that the sum of the ages of the two girls will be odd. Now, let’s first find the probability that z is even. Since z is the sum of ages of the selected boys and girls, z can be even if the sum of both the selected boys and selected girls ages are even or if sum of both the selected boys and selected girls ages are odd. The probability that both sums are even is 1/3 x 1/3 = 1/9 and the probability that both sums are odd is 2/3 x 2/3 = 4/9. Thus, there is a 1/9 + 4/9 = 5/9 probability that z is even. Similarly, let’s find the probability that z is odd. z can only be odd of one of the sums is even and the other is odd. Probability that the boys sum is even and girls sum is odd is 1/3 x 2/3 = 2/9. Probability that the boys sum is odd and the girls sum is even is 2/3 x 1/3 = 2/9. Thus, there is a 2/9 + 2/9 = 4/9 probability that z is odd. Finally, the difference between the probabilities that z is even and z is odd is 5/9 - 4/9 = 1/9. 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