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There are 9 people in the room. There are two pairs of
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17 Jan 2008, 06:10
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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
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17 Jan 2008, 07:41
another ways: 1. \(p=\frac{C^9_2C^2_1}{C^9_2}=1\frac{2}{36}=\frac{17}{18}\) or 2. \(p=\frac{P^9_2P^2_1*P^2_2}{P^9_2}=1\frac{4}{72}=\frac{17}{18}\) 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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Re: prob  pairs
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18 Jan 2008, 04:47
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. = 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. probability of picking the sibling of the first person we selected Put it all together:




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Re: prob  pairs
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17 Jan 2008, 06:38



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Re: prob  pairs
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17 Jan 2008, 13:40
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



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Re: prob  pairs
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18 Jan 2008, 06:32
here is another solution. The number of ways to pick 2 people from a group of 9 is equal to 9*(91)/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 11/18 = 17/18



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Re: prob  pairs
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24 Aug 2008, 14:41
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) = 11/18 = 17/18



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Re: prob  pairs
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23 Mar 2009, 10:57
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) = 11/18 = 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



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Re: prob  pairs
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27 Sep 2009, 05:36
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



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Re: prob  pairs
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29 Nov 2009, 13:33
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 nonsibling 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.



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Re: prob  pairs
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30 Sep 2010, 17:38
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.



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Re: prob  pairs
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12 Jun 2011, 04:28
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,



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Re: prob  pairs
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12 Jun 2011, 05:34
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. \(p = \frac{C^5_2+C^5_1*C^2_1*C^2_1+C^2_1*C^2_1}{C^9_2} = \frac{10+20+4}{36} = \frac{34}{36} = \frac{17}{18}\)
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Re: prob  pairs
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12 Jun 2011, 07:52
1  \((2c1+2c1)/9c2\)
=1 (2/36) = 34/36 = 17/18



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Re: There are 9 people in the room. There are two pairs of
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06 Jul 2013, 08:11
My approach : 1 p(1 minus the probability that the people we pick will be siblings) We have 9 people: A1  A2  B1  B2  C  D  E  F  G Probability of picking siblings: If we first pick A1, Probability of picking A1 \(p=\frac{1}{9}\frac\) (we pick 1 between 9) Then, probability of picking the sibling of A1, A2, after having picked A1: \(p=\frac{1}{8}\frac\) (we pick 1 between the remainder 8 people) * \(\frac{1}{9}\frac\) (conditional probability of picking A2 after having picked A1) = \(\frac{1}{72}\frac\) But, in addition, we could have picked A2 and then A1. Then: \(p=\frac{1}{72}\frac\) *2 = \(\frac{2}{72}\frac\) = \(\frac{1}{36}\frac\) The same if we picked first B1 and secondly B2 or ("or" means "+") if we first picked B2 and secondly B1. Then, the total probability of picking siblings is: \(p=\frac{1}{36}\frac\) * 2 = \(\frac{2}{36}\frac\) = \(\frac{1}{18}\frac\) Then, 1  p = \(\frac{18}{18}\frac\)  \(\frac{1}{18}\frac\) = \(\frac{17}{18}\frac\) Is right this logic? I am understanding the problem correctly



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Re: There are 9 people in the room. There are two pairs of
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06 Jul 2013, 08:28
Maxirosario2012 wrote: My approach : 1 p(1 minus the probability that the people we pick will be siblings) We have 9 people: A1  A2  B1  B2  C  D  E  F  G Probability of picking siblings: If we first pick A1, Probability of picking A1 \(p=\frac{1}{9}\frac\) (we pick 1 between 9) Then, probability of picking the sibling of A1, A2, after having picked A1: \(p=\frac{1}{8}\frac\) (we pick 1 between the remainder 8 people) * \(\frac{1}{9}\frac\) (conditional probability of picking A2 after having picked A1) = \(\frac{1}{72}\frac\) But, in addition, we could have picked A2 and then A1. Then: \(p=\frac{1}{72}\frac\) *2 = \(\frac{2}{72}\frac\) = \(\frac{1}{36}\frac\) The same if we picked first B1 and secondly B2 or ("or" means "+") if we first picked B2 and secondly B1. Then, the total probability of picking siblings is: \(p=\frac{1}{36}\frac\) * 2 = \(\frac{2}{36}\frac\) = \(\frac{1}{18}\frac\) Then, 1  p = \(\frac{18}{18}\frac\)  \(\frac{1}{18}\frac\) = \(\frac{17}{18}\frac\) Is right this logic? I am understanding the problem correctly Absolutely, though not the most elegant way of solving.
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Re: There are 9 people in the room. There are two pairs of
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06 Jul 2013, 10:34
Thank you Bunuel! Yes, it is not the fastest approach



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Re: There are 9 people in the room. There are two pairs of
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14 Jul 2013, 15:17
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? The reversal probability approach is the most efficient way to solve. But I explore the other methods just for practice. Tell me your opinions, I have a little doubt on the reversal combinatorial approach.
Probability approach:
XXXXXAABB
\(P= \frac{6}{9} * \frac{1}{2}*2 +\frac{5}{9}*\frac{4}{8} = \frac{12}{18} + \frac{5}{18} = \frac{17}{18}\)
\(\frac{6}{9}*\frac{1}{2}*2\): we choose 6 people out of 9 (XXX XXA, 6/9) and then 1 people out of the other 2 (BB 1/2). Then multiply by 2 because we choose again 6 people out of 9 (XXXXXB) and then 1 people out of the other 2 (AA).
\(\frac{5}{9}*\frac{4}{8}\): then we focus only on the other 5 that are not siblings (XXXXX). We choose this 5 people (XXXXX) out of all 9 (then 5/9) and we combine this one with the remaining 4 of the group (XXXX) of the remaining 8.
Reversal Probability approach: P = 1q (q=probability of choosing one sibling)
\(q= \frac{2}{9} * \frac{1}{8} + \frac{2}{9} * \frac{1}{8} = \frac{4}{72} = \frac{1}{18}\) \(P = 1q = \frac{18}{18}  \frac{1}{18} = \frac{17}{18}\)
\(\frac{2}{9}\): probability of picking the person A from AABBXXXXX \(\frac{1}{8}\): probability of picking the other A from the remaining ABBXXXXX
Then we add the probability of picking B from AABBXXXXX and probability of picking again B from AABXXXXX
Combinatorial approach:
\(P= \frac{{C^6_1 * C^2_1 + C^6_1 * C^2_1 + C^5_2}}{C^9_2}= \frac{{6*2 + 6*2 + 10}}{36} = \frac{34}{36} = \frac{17}{18}\)
\(C^6_1 * C^2_1\): we choose one people from AXXXXX and combine with one people of BB \(C^6_1 * C^2_1\): we pick one people of BXXXXX and combine it with one people of AA \(C^5_2\): we choose 2 people from XXXXX \(C^9_2\): total combinations of 2 from 9.
Reversal Combinatorial approach:
\(q= \frac{{C^2_1 + C^2_1}}{C^9_2}= \frac{{2 + 2}}{72} =\frac{4}{72} = \frac{1}{18}\) \(P = 1q = \frac{18}{18}  \frac{1}{18} = \frac{17}{18}\)
\(C^2_1\): we choose 2 persons from the sibling (AA) \(C^2_1\): we choose 2 persons from the sibling (BB)



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Re: There are 9 people in the room. There are two pairs of
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20 Apr 2014, 17:14
Can someone please explain where i'm going wrong?
I'm using the approach: No Sibling Pair = 1  Probability of sibling pair.
P = \(\frac{(9C4)(1C1)}{(9C2)}\)  why is this wrong? Is this equation not implying that at first, I have the probability of choose 4 people out of 9, then I can only pick 1 so my top is the favorable solution. My denominator is the total number of outcomes.



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There are 9 people in the room. There are two pairs of
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20 Apr 2014, 23:19
russ9 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?
Can someone please explain where i'm going wrong?
I'm using the approach: No Sibling Pair = 1  Probability of sibling pair.
P = \(\frac{(9C4)(1C1)}{(9C2)}\)  why is this wrong? Is this equation not implying that at first, I have the probability of choose 4 people out of 9, then I can only pick 1 so my top is the favorable solution. My denominator is the total number of outcomes. The numerator is wrong. We are not choosing 4 people, we are choosing 2. How many ways to choose a sibling pair from 9 people where there are two pairs of siblings (X, X, Y, Y, A, B, C, D, E)? Only two ways XX and YY. Hence, P = 1  2/(9C2) = 1  1/18 = 17/18.
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