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# MGMAT Test1:#29 In a room filled with 7 people, 4 people

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Senior Manager
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MGMAT Test1:#29 In a room filled with 7 people, 4 people [#permalink]  29 Mar 2008, 11:55
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MGMAT Test1:#29

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?
A. 5/21
B. 3/7
C. 4/7
D. 5/7
E. 16/21

I know OA and OE but didn't get it ( it is confusing for me). I appreciate different approaches and views
Explanations plz, thx.
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Re: PS_Probability_friendship [#permalink]  29 Mar 2008, 15:45
I will solve it using Combination approach.
We have 4 people with 1 friend, so total ways for that = 4C1 = 4
We have 3 people with 2 friend, so total ways for that = 3C2 = 3

Since friendship is mutually exclusive so people above two group will not be friends of each others.

Total ways of selecting 2 people from 7 people = 7C2 = 21

So Probability = 4 * 3 / 21 = 4/7
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Re: PS_Probability_friendship [#permalink]  29 Mar 2008, 17:33
I agree with abhijit_sen's answer. Here's another way to look at things:

It may be easier to understand this by drawing arranging 7 letters A to G in a circle. You'll see that there is no way to achieve the "4 have exactly 1 friend and 3 have exactly 2 friends" requirement without drawing a triangle among 3 people and using 2 lines to pair up the remaining 4 people. For the sake of this argument, let ABC be the triangle, DE be a pair, and FG the last pair.

Now find the number of ways of there not being a pair of friends selected.
Let's say A is chosen first, there are only 4 people that can be chosen that won't a friend. This also applies for B and C, so altogether, there are 12 ways for this to happen.

(Notice that we don't do need to do the reverse where we choose D,E,F or G first because the would only end up choosing A, B, or C. We've already handled the case and the order doesn't matter)

Total ways = 7C2

Hence, 12/21 is the probability which simplifies to 4/7. Answer C
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Re: PS_Probability_friendship [#permalink]  29 Mar 2008, 23:36
Expert's post
$$P=1-(\frac47*\frac16+\frac37*\frac26)=\frac{16}{21}$$
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Re: PS_Probability_friendship [#permalink]  30 Mar 2008, 00:47
agree with Walker solution, it just says friend in a room, not necessary to break out into 2 grps, i think that is the confusing part. C is only correct when we make above assumption
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Re: PS_Probability_friendship [#permalink]  30 Mar 2008, 09:01
Hi, this is a long routed solution but it may help to see the logic.
Lets label the members as A,B,C,D,E,F,G
We begin by counting the relationships

B BC,BD,BE,BF,BG =5 total (we do not count BA because the relationship is mutual and we would be counting it twice)
C CD,CE,CF,CG = 4 total
D DE,DF,DG =3
E EF,EG =2
F FG =1
Therefore the total number of relationships is 6+5+4+3+2+1 =21
Criteria 1
If 4 people have exactly one friend then the number of relationships is 2
eg AB ,CD (A is a friend of B and B is a friend of A (1 relationship between 2 people), C is a friend of D and D is a friend of C (2nd relationship between 2 other people)Therefore we have 2 relationships between 4 people.=2

Criteria 2

3 people have exactly 2 friends in the room

CD,CE,DE (C is a friend of D and E (1 relationship), D is a friend of E and C, (2nd relationship) E is a friend of C and D(3rd relationship)) So we have 3 relationships in total that satisfy the above criteria =3

The probability that the two random individuals are friends =(3+2)/21 =5/21
The probability that the two random individuals chosen are not friends = 1- (5/21)
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Re: PS_Probability_friendship [#permalink]  30 Mar 2008, 11:42
I am grateful to u all for excellent ideas. Thank u, folks.
I don't have to provide OA and OE, as it is exactly the same with that of Ventivish.
Re: PS_Probability_friendship   [#permalink] 30 Mar 2008, 11:42
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