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# There are 9 people in the room. There are two pairs of

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There are 9 people in the room. There are two pairs of [#permalink]  17 Jan 2008, 07:10
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There are 9 people in the room. There are two pairs of siblings within that group. If we choose two people, what is the probability that they will not be siblings?
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Re: prob - pairs [#permalink]  17 Jan 2008, 07:38
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$formdata=1-(\frac{4}{9}*\frac{1}{8})=+1-\frac{4}{72}=\frac{68}{72}+=+\frac{17}{18}$
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Re: prob - pairs [#permalink]  17 Jan 2008, 08:41
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another ways:

1. p=\frac{C^9_2-C^2_1}{C^9_2}=1-\frac{2}{36}=\frac{17}{18}

or

2. p=\frac{P^9_2-P^2_1*P^2_2}{P^9_2}=1-\frac{4}{72}=\frac{17}{18}

or

3. p=\frac49*\frac78+\frac59*\frac88=\frac{68}{72}=\frac{17}{18}

probability questions always have a few ways

bmwhype2, thanks for your question
+1
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PROMO: Are you an exiting GMAT ToolKit (iOS) user? Get GMAT ToolKit 2 (iOS) for only $0.99 (read more) Math: GMAT Math Book ||| General: GMATTimer ||| Chicago Booth: Slide Presentation The People Who Are Crazy Enough to Think They Can Change the World, Are the Ones Who Do. CEO Joined: 21 Jan 2007 Posts: 2797 Location: New York City Followers: 5 Kudos [?]: 132 [0], given: 4 Re: prob - pairs [#permalink] 17 Jan 2008, 10:06 thanks, just wanted to be very sure of my answer. _________________ You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson CEO Joined: 29 Mar 2007 Posts: 2618 Followers: 13 Kudos [?]: 142 [0], given: 0 Re: prob - pairs [#permalink] 17 Jan 2008, 14:40 bmwhype2 wrote: There are 9 people in the room. There are two pairs of siblings within that group. If we choose two people, what is the probability that they will not be siblings? we have XXXXXBBCC We have to take into the account when we have XX, XB, XC So 5/9*4/8 5/9*2/8 and 5/9*2/8 note we need to multiply these by 2 (except the first) b/c we can have BX and CX. 20+20+20 --> 60/72 --> 30/36 --> 15/18 then next is BC 2/9*2/8 --> 4/72 2/36 1/18 --> multiply this by 2 b/c of CB so 2/18+15/18 --> 17/18 SVP Joined: 04 May 2006 Posts: 1946 Schools: CBS, Kellogg Followers: 10 Kudos [?]: 168 [0], given: 1 Re: prob - pairs [#permalink] 17 Jan 2008, 22:37 eschn3am wrote: $formdata=1-(\frac{4}{9}*\frac{1}{8})=+1-\frac{4}{72}=\frac{68}{72}+=+\frac{17}{18}$ Can you give a brief explanation? Appreciated! _________________ Director Joined: 12 Jul 2007 Posts: 875 Followers: 8 Kudos [?]: 135 [0], given: 0 Re: prob - pairs [#permalink] 18 Jan 2008, 05:47 9 people are standing in a room. Included are 2 pairs of siblings (so 4 people, each with on sibling) What is the probability of picking two people that are NOT siblings. Start with 1, because that's 100% probability. From here we'll subtract the chances of getting a pair of siblings. What we have left is the probability of choosing 2 people who are NOT siblings. Now we start our choosing. We have 9 people to choose from for our 1st choice. Out of these 9 people we must choose one of the 4 with siblings. If we choose one of the 5 people without siblings then our second choice won't matter at all because they have no relatives we can pick up. $formdata=\frac{4}{9}$ = probability of picking 1 of 4 people with siblings from a group of 9 Now we have to find how many ways we can choose that persons sibling once we have chosen one of the 4. We have 8 people left to choose from and since there are pairs of siblings, each of the 4 people only has ONE sibling. Thus, there is only 1 person we can choose out of the remaining 8. $formdata=\frac{1}{8}$ probability of picking the sibling of the first person we selected Put it all together: $formdata=1-(\frac{4}{9}*\frac{1}{8})=+1-\frac{4}{72}=\frac{68}{72}+=+\frac{17}{18}$ Director Joined: 01 Jan 2008 Posts: 635 Followers: 3 Kudos [?]: 114 [2] , given: 1 Re: prob - pairs [#permalink] 18 Jan 2008, 07:32 2 This post received KUDOS here is another solution. The number of ways to pick 2 people from a group of 9 is equal to 9*(9-1)/2 = 36 there are only 2 pairs of siblings in the room. Therefore, probability of choosing a pair of siblings is 2/36 = 1/18. Probability of choosing a pair who are not siblings is 1-1/18 = 17/18 SVP Joined: 07 Nov 2007 Posts: 1842 Location: New York Followers: 20 Kudos [?]: 291 [0], given: 5 Re: prob - pairs [#permalink] 24 Aug 2008, 15:41 bmwhype2 wrote: There are 9 people in the room. There are two pairs of siblings within that group. If we choose two people, what is the probability that they will not be siblings? = 1- (2C2*2/9C2) = 1-1/18 = 17/18 _________________ Your attitude determines your altitude Smiling wins more friends than frowning Senior Manager Joined: 06 Jul 2007 Posts: 288 Followers: 3 Kudos [?]: 24 [3] , given: 0 Re: prob - pairs [#permalink] 23 Mar 2009, 11:57 3 This post received KUDOS x2suresh wrote: bmwhype2 wrote: There are 9 people in the room. There are two pairs of siblings within that group. If we choose two people, what is the probability that they will not be siblings? = 1- (2C2*2/9C2) = 1-1/18 = 17/18 Method 1 : total number of ways of choosing 2 people out of 9 = 9C2 = 36 probability of choosing two people so that they are siblings = 2C1 = 2 { since there are two pairs of siblings } required probability = (36 -2) /36 = 17/18. Method 2 : total number of ways = 36 let say the nine people that we have are A, B, C, D, E, S11, S12, S21, S22 where (S11, S12) and (S21, S22) are sibling pairs. number of ways one can choose two people is : choose 2 from A, B, C, D, E . Number of ways = 5C2 = 10. choose 1 from A, B, C, D, E AND 1 from (S11, S12, S21, S22) in 5*4 = 20 ways choose 1 from (S11, S12) and 1 from (S21, S22) in 4 ways . total number of ways = 34 probability = 34/36 = 17/18 Manager Joined: 27 Oct 2008 Posts: 188 Followers: 1 Kudos [?]: 42 [0], given: 3 Re: prob - pairs [#permalink] 27 Sep 2009, 06:36 There are 9 people in the room. There are two pairs of siblings within that group. If we choose two people, what is the probability that they will not be siblings? Soln: probability that they will not be siblings = 1 - Probability that they will be siblings = 1 - (4/9 * 1/8) = 1 - (1/18) = 17/18 Intern Joined: 29 Nov 2009 Posts: 7 Followers: 0 Kudos [?]: 0 [0], given: 54 Re: prob - pairs [#permalink] 29 Nov 2009, 14:33 Can someone correct my logic No siblings can be achieved by 1. selecting 1 of siblings 4/9 * selecting 1 from the other pair of siblings 2/4 + 2. selecting non-sibling 5/9 * selecting anyone from the rest 8/8 so 4/9 * 2/4 + 5/9*8/8 = 2/9 + 5/9 = 7/9 Edit: nevermind. I saw the previous post 2/4 should be 7/8, makes sense. Intern Joined: 20 Apr 2010 Posts: 38 Followers: 0 Kudos [?]: 1 [0], given: 7 Re: prob - pairs [#permalink] 30 Sep 2010, 18:38 Struggling with this one here. I applied the methodology used to arrive at the answer for the married couples. For the denominator, you have 9*8 ways of choosing 2 people. For the numerator, you have 9 ways of choosing the first person, and then 7 ways of choosing the second person (excluding the other sibling pair). So, (9*7)/(9*8) = 7/8. Why does this not work in this situation? Thanks. In the married couple example, we are taking 4 people out of 12 to determine the probability that none of them are married to each other. In this case we are looking for the probability that 2 people are not siblings, so the concept is the same. Senior Manager Joined: 25 Feb 2010 Posts: 457 Followers: 3 Kudos [?]: 36 [0], given: 5 Re: prob - pairs [#permalink] 30 Sep 2010, 20:31 Waooo !!! There are lots of ways to attack a simple question .. Choose which ever you like _________________ GGG (Gym / GMAT / Girl) -- Be Serious Its your duty to post OA afterwards; some one must be waiting for that... Intern Joined: 28 Mar 2011 Posts: 34 Followers: 0 Kudos [?]: 1 [0], given: 0 Re: prob - pairs [#permalink] 12 Jun 2011, 05:28 Hi Walker, With reference to your solution for the married couples problem where in we have to select 4 people out of 6 couples such that none is married to each other,can this problem be tackled in the same way? How will we arrive at the solution? Thanks! regards, CEO Joined: 17 Nov 2007 Posts: 3596 Concentration: Entrepreneurship, Other Schools: Chicago (Booth) - Class of 2011 GMAT 1: 750 Q50 V40 Followers: 231 Kudos [?]: 1300 [0], given: 346 Re: prob - pairs [#permalink] 12 Jun 2011, 06:34 You can use the same approach but it's important to note that not all people have siblings. C^5_2 - the number of options to choose 2 people out of 5 people without siblings C^5_1*C^2_1*C^2_1 - the number of options to choose 1 person out of 5 without siblings, choose 1 pair of siblings out of 2, choose 1 person out of the pair. C^2_1*C^2_1 - the number of options to choose 1 person out of 2 for each pair of siblings. p = \frac{C^5_2+C^5_1*C^2_1*C^2_1+C^2_1*C^2_1}{C^9_2} = \frac{10+20+4}{36} = \frac{34}{36} = \frac{17}{18} _________________ iOS/Android: GMAT ToolKit - The bestselling GMAT prep app | GMAT Club (free) | PrepGame | GRE ToolKit | LSAT ToolKit PROMO: Are you an exiting GMAT ToolKit (iOS) user? Get GMAT ToolKit 2 (iOS) for only$0.99 (read more)
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Re: prob - pairs [#permalink]  12 Jun 2011, 08:52
1 - (2c1+2c1)/9c2

=1- (2/36) = 34/36 = 17/18
Re: prob - pairs   [#permalink] 12 Jun 2011, 08:52
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# There are 9 people in the room. There are two pairs of

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