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

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Manager
Joined: 15 Aug 2013
Posts: 247
Re: There are 9 people in the room. There are two pairs of  [#permalink]

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21 Apr 2014, 17:53
Bunuel wrote:
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 = 1/17.

Hi Bunuel,

I am reading what you're writing but I just cant grasp it. How is this different than the other problems where we try to keep the numerator and denominator in the same units, i.e: people and not pairs. Is P = $$\frac{(9C4)(1C1)}{(9C2)}$$ not saying that i want to pick ANY member out of the 4 siblings and then I can only choose 1 member after choosing that ANY member?
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Joined: 30 Oct 2013
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Re: There are 9 people in the room. There are two pairs of  [#permalink]

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09 Jun 2014, 03:03
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?

I need to choose two people from a group of 9 . I will choose 1 the probablitiy is 100% (9/9), then I have to choose the second person in such a manner that he is not the sibling of the 1 I chose, therefor amongst the 8 I am still left with, the probability becomes 7/8 (excluding the one sibling of the first). Please correct my approach and preferably donot provide me with an alternative solution. It would be of great help.
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Joined: 02 Sep 2009
Posts: 51218
Re: There are 9 people in the room. There are two pairs of  [#permalink]

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09 Jun 2014, 08:10
ankitarihaan 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?

I need to choose two people from a group of 9 . I will choose 1 the probablitiy is 100% (9/9), then I have to choose the second person in such a manner that he is not the sibling of the 1 I chose, therefor amongst the 8 I am still left with, the probability becomes 7/8 (excluding the one sibling of the first). Please correct my approach and preferably donot provide me with an alternative solution. It would be of great help.

The point is that if you choose a person which has no sibling then the probability that the second person won't be his/her sibling would be 1. That's why this approach does not work.

Correct approach is here: there-are-9-people-in-the-room-there-are-two-pairs-of-58609.html#p1357845

Hope it helps.
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Re: There are 9 people in the room. There are two pairs of  [#permalink]

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09 Sep 2018, 23:49
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Re: There are 9 people in the room. There are two pairs of &nbs [#permalink] 09 Sep 2018, 23:49

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