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Manhattan Gmat Probability question

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Senior Manager
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Manhattan Gmat Probability question [#permalink]

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New post 01 Aug 2006, 00:34
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Hi guys try this out :)


In a room filled with 7 people, 4 people have exactly 1 friend in the room and 3 people have exactly 2 friends in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?
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New post 01 Aug 2006, 00:40
Total ways of selecting ppl. 7C2 = 21
Let
ABC(2friends)
CD(1friends)
DE(1friends)
ways of selecting those who are friends.
3C2+2C2+2C2 = 3+1+1 = 5

1-5/21 = 16/21
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New post 01 Aug 2006, 00:44
Man you guys are fast! :-D yup that's the correct answer
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New post 01 Aug 2006, 00:48
Let the friends be A B C D E F G
Probability of choosing 2 people = 7C2 = 21
Group of friends could be
AB CD EF FG EG

Hence there are 5 groups of friends.

We can choose friends from any of this group.
Hence Probability of choosing friends = 5/21

Probability of not choosing friends = 1 - 5/21 = 16/21
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New post 05 Aug 2006, 20:59
That's a great probability question!
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New post 05 Aug 2006, 21:18
jaynayak wrote:
Let the friends be A B C D E F G
Probability of choosing 2 people = 7C2 = 21
Group of friends could be
AB CD EF FG EG

Hence there are 5 groups of friends.

We can choose friends from any of this group.
Hence Probability of choosing friends = 5/21

Probability of not choosing friends = 1 - 5/21 = 16/21



good approach

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New post 06 Aug 2006, 01:13
Only possibility of group of friends is
12 - Both friends
34 - Both friends
456 - Each friend of the other

Total cases = 7C2 = 21

Cases when either 1 or 2 is selected as first person = 2 * 5 = 10
Cases when either 3 or 4 is selected as first person = 2 * 3 = 6

So Prob = 16/21
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  [#permalink] 06 Aug 2006, 01:13
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Manhattan Gmat Probability question

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