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# 10 members of a society, including members A and B, attended a me

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Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 6227
GMAT 1: 760 Q51 V42
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10 members of a society, including members A and B, attended a me  [#permalink]

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06 Mar 2018, 03:58
00:00

Difficulty:

15% (low)

Question Stats:

85% (00:56) correct 15% (01:42) wrong based on 78 sessions

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[GMAT math practice question]

10 members of a society, including members A and B, attended a meeting. 2 of these members were selected to form a committee. What is the probability that the 2 members selected were members A and B?

A. 1/10
B. 1/15
C. 1/20
D. 1/30
E. 1/45

_________________

MathRevolution: Finish GMAT Quant Section with 10 minutes to spare
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"Only $99 for 3 month Online Course" "Free Resources-30 day online access & Diagnostic Test" "Unlimited Access to over 120 free video lessons - try it yourself" CEO Joined: 12 Sep 2015 Posts: 2878 Location: Canada Re: 10 members of a society, including members A and B, attended a me [#permalink] ### Show Tags 06 Mar 2018, 08:06 Top Contributor MathRevolution wrote: [GMAT math practice question] 10 members of a society, including members A and B, attended a meeting. 2 of these members were selected to form a committee. What is the probability that the 2 members selected were members A and B? A. 1/10 B. 1/15 C. 1/20 D. 1/30 E. 1/45 Let's apply some probability rules P(A and B both selected) = P(1st selection is one member of the pair AND 2nd selection is the other member of the pair) = P(1st selection is one one member of the pair) x P(2nd selection is the other member of the pair) = 2/10 x 1/9 = 1/45 Answer: E RELATED VIDEO FROM OUR COURSE _________________ Brent Hanneson – GMATPrepNow.com Sign up for our free Question of the Day emails SVP Joined: 08 Jul 2010 Posts: 2334 Location: India GMAT: INSIGHT WE: Education (Education) Re: 10 members of a society, including members A and B, attended a me [#permalink] ### Show Tags 06 Mar 2018, 08:18 MathRevolution wrote: [GMAT math practice question] 10 members of a society, including members A and B, attended a meeting. 2 of these members were selected to form a committee. What is the probability that the 2 members selected were members A and B? A. 1/10 B. 1/15 C. 1/20 D. 1/30 E. 1/45 Favourable ways of choosing committee = 1 (A and B both) Total ways of selecting a 2 member committee out of 10 members = 10C2 = 45 Probability = 1/45 answer: Option E _________________ Prosper!!! GMATinsight Bhoopendra Singh and Dr.Sushma Jha e-mail: info@GMATinsight.com I Call us : +91-9999687183 / 9891333772 Online One-on-One Skype based classes and Classroom Coaching in South and West Delhi http://www.GMATinsight.com/testimonials.html 22 ONLINE FREE (FULL LENGTH) GMAT CAT (PRACTICE TESTS) LINK COLLECTION Math Revolution GMAT Instructor Joined: 16 Aug 2015 Posts: 6227 GMAT 1: 760 Q51 V42 GPA: 3.82 Re: 10 members of a society, including members A and B, attended a me [#permalink] ### Show Tags 08 Mar 2018, 02:25 => The number of committees of 2 people that include members A and B is equal to the number of ways that 2 people can be selected from 2 people, which is 2C2 = 1. The number different possible committees of 2 people, is equal to the number of ways that 2 people can be selected from 10 people, which is 10C2 = $$\frac{(10*9)}{(1*2)} = 45.$$ The probability that the committee contains members A and B is 2C2 / 10C2 = $$\frac{1}{45}$$. Therefore, the answer is E. Answer: E _________________ MathRevolution: Finish GMAT Quant Section with 10 minutes to spare The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. "Only$99 for 3 month Online Course"
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Joined: 28 Jan 2018
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Re: 10 members of a society, including members A and B, attended a me  [#permalink]

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20 Mar 2018, 17:23
MathRevolution wrote:
[GMAT math practice question]

10 members of a society, including members A and B, attended a meeting. 2 of these members were selected to form a committee. What is the probability that the 2 members selected were members A and B?

A. 1/10
B. 1/15
C. 1/20
D. 1/30
E. 1/45

Probability of choosing both A and B = $$\frac{2}{10} * \frac{1}{9} = \frac{1}{45}$$

The first part of the equation is $$\frac{2}{10}$$ because it can be either A or B, while the 2nd is $$\frac{1}{9}$$ because it's the only remaining member
Re: 10 members of a society, including members A and B, attended a me &nbs [#permalink] 20 Mar 2018, 17:23
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# 10 members of a society, including members A and B, attended a me

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