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# In a room filled with 7 people, 4 people have exactly 1

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Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
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In a room filled with 7 people, 4 people have exactly 1 [#permalink]  04 Nov 2006, 02:28
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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 (Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

5/21
3/7
4/7
5/7
16/21
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Impossible is nothing

Senior Manager
Joined: 01 Oct 2006
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Kudos [?]: 20 [0], given: 0

Here we have 2 groups one 4 members having one friend each and the other with 3 members having 2 friends each. probablity of selecting a group is 1/2
for the first we can select 2 people such that they are friends in 4*1=4 ways
Then for second group we get 3*2=6

sum =1/2 *(4+6)=5

no of ways to select 2 persons at random is 7c2=21

Probability that couple selected are friends is 5/21
Not friends =1-5/21=16/21
What is OA?

Also does anyone has a simpler method to solve
Using the golden pair rule i was able to eliminate 3 of the five answers but taking the right between those took time
Manager
Joined: 01 Nov 2006
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This is easier if you don't use counting rules. Do it sequentially.

P(friends) = P(first pick someone with 1 friend)*P(second you pick their friend) + P(first pick someone with two friends)*P(second you pick one of their two friends)

= 4/7*1/6 + 3/7*2/6 = 10/42 = 5/21
Director
Joined: 05 Feb 2006
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joeydvivre wrote:
This is easier if you don't use counting rules. Do it sequentially.

P(friends) = P(first pick someone with 1 friend)*P(second you pick their friend) + P(first pick someone with two friends)*P(second you pick one of their two friends)

= 4/7*1/6 + 3/7*2/6 = 10/42 = 5/21

Forgot to do the last step 1-5/21.... GMAT may catch you on this.....
Manager
Joined: 01 Nov 2006
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Yougsheath did all that.

Actually, I thought that this was an interesting problem because the counting rules way is harder than the conditional probabilties, which is much more usual in real probability problems. The GMAT probability problems are almost all combinatorics problems, not probability problems.

And, I'm about 25 years past the GMAT being able to do anything to me.
Senior Manager
Joined: 23 May 2005
Posts: 266
Location: Sing/ HK
Followers: 1

Kudos [?]: 21 [0], given: 0

I actually had to list it all down all the "friendships" one by one... you can imagine how long it took me to answer this question...
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Impossible is nothing

Senior Manager
Joined: 05 Oct 2006
Posts: 267
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Kudos [?]: 6 [0], given: 0

very good qus.i got the total ways = 10 but forgot to see that the 2 chosen groups can only be selected one at a time.
that's why i was getting 10/21 and hence 11/21 as my answer.

tx guys.good discussion and that's what we r here for.
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