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Friends-Probability

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Friends-Probability [#permalink]

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New post 07 Nov 2008, 02:57
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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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Re: Friends-Probability [#permalink]

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New post 07 Nov 2008, 03:09
Select the first person from the group

If the person is in (4 people have exactly 1 friend) group, the probability that the other is NOT a friend is maybe from 5 out of 6 people left

P1 = 4/7 * 5/6

If the person is in (3 people have exactly 2 friend) group, the probability that the other is NOT a friend is maybe from 4 out of 6 people left

P2 = 3/7 * 4/6

So P = P1 + P2 = 32/42 = 16/21

Answer E

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Re: Friends-Probability [#permalink]

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New post 07 Nov 2008, 03:24
What if the 4 freind group and the 3-friend group overlap?
they may or may not have common friends.
how do we put that into equation?

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Re: Friends-Probability [#permalink]

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New post 07 Nov 2008, 04:38
prasun84 wrote:
What if the 4 freind group and the 3-friend group overlap?
they may or may not have common friends.
how do we put that into equation?



That is not possible. The question specifically mentions "4 persons having exactly 1 friend and 3 persons having exactly 2 friends".

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Re: Friends-Probability [#permalink]

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New post 07 Nov 2008, 04:54
sorry but im still trying to figure out the arrangement.
let A, B, C, D, E, F, and G be the 7 guys.

4 people exactly having 1 friend wud mean AB, AC, AD,AE.
that makes B,C,D,E have just A as friend.(is this assumption correct?)

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Re: Friends-Probability [#permalink]

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New post 07 Nov 2008, 10:49
prasun84 wrote:
sorry but im still trying to figure out the arrangement.
let A, B, C, D, E, F, and G be the 7 guys.

4 people exactly having 1 friend wud mean AB, AC, AD,AE.
that makes B,C,D,E have just A as friend.(is this assumption correct?)


But, to make it simple, the question gives the following assumption
(Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend).

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Re: Friends-Probability [#permalink]

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New post 07 Nov 2008, 11:11
scthakur wrote:
prasun84 wrote:
sorry but im still trying to figure out the arrangement.
let A, B, C, D, E, F, and G be the 7 guys.

4 people exactly having 1 friend wud mean AB, AC, AD,AE.
that makes B,C,D,E have just A as friend.(is this assumption correct?)


But, to make it simple, the question gives the following assumption
(Assuming that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend).


So does it mean that in the group of 4 people having only 1 friend, can i assume that we just need 4 people to make up this group (instead of 5) because if A is a friend of B, and B is also a friend of A.
I am very confused with this type of question.

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Re: Friends-Probability [#permalink]

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New post 07 Nov 2008, 13:43
Attachment:
diag A.JPG
diag A.JPG [ 6.93 KiB | Viewed 1112 times ]

see inline diagram above, it'd help visualize this a little better. I'm more or less repeating what lylya4 said, but with a graphic - that's all

The group of 4 has exactly 1 friend. Mutual friends are counted twice, coz A is a Friend(B) and B is a Friend(A). The question stem says so..

In the group of 3, each one has exactly 2 friends. as you can see in the attached graphic.

Now for solving the probability of picking 2 out of this group, who'll not be friends.

From the graphic it is evident that for the 2 picks to be not friends, they have to be one each from their respective groups.

Combination #1 - 1st pick from Group(4) and 2nd pick from either Group(3) or the other 2 in Group(4)..so Probability = 4/7 * 5/6 = 20 / 42

Combination #2 = 1st pick from Group(3) and 2nd pick from either Group(4) ..
so Probability = 3/7 * 4/6 = 12 / 42

Adding the 2 probabilities yields... (20+12)/42 = 16/21. So answer is E

Hope I made some friends unlike the 2 picks from the above groups who won't talk to each other :-D
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Re: Friends-Probability [#permalink]

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New post 08 Nov 2008, 05:04
masuhari wrote:
Attachment:
diag A.JPG

see inline diagram above, it'd help visualize this a little better. I'm more or less repeating what lylya4 said, but with a graphic - that's all

The group of 4 has exactly 1 friend. Mutual friends are counted twice, coz A is a Friend(B) and B is a Friend(A). The question stem says so..

In the group of 3, each one has exactly 2 friends. as you can see in the attached graphic.

Now for solving the probability of picking 2 out of this group, who'll not be friends.

From the graphic it is evident that for the 2 picks to be not friends, they have to be one each from their respective groups.

Combination #1 - 1st pick from Group(4) and 2nd pick from either Group(3) or the other 2 in Group(4)..so Probability = 4/7 * 5/6 = 20 / 42

Combination #2 = 1st pick from Group(3) and 2nd pick from either Group(4) ..
so Probability = 3/7 * 4/6 = 12 / 42

Adding the 2 probabilities yields... (20+12)/42 = 16/21. So answer is E

Hope I made some friends unlike the 2 picks from the above groups who won't talk to each other :-D


Combination #1 - 1st pick from Group(4) and 2nd pick from either Group(3) or the other 2 in Group(4)..so Probability = 4/7 * 5/6 = 20 / 42

I don't understand why 5/6 not 3/6 as we select from a group of 3 people.

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Re: Friends-Probability [#permalink]

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New post 08 Nov 2008, 10:14
5/6 instead of 3/6, coz there are 2 others in the same group(4) who'd be fair game too. They are friends with each other and not necessarily friends with our possible pick. Hence you have to consider picking them as favorable outcome(which totals 3 in group(3) and 2 from group(4)). Ofcourse total # of outcomes would be 6(7-1). HTH...
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Re: Friends-Probability [#permalink]

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New post 08 Nov 2008, 10:51
I started the problem trying to find out what is the probability of two people picked are friends

Total combo's = 7 C 2 = 21

{(a,b) (c,d)} (e,f,g) are such groups that fit the description

2 people can be picked from any of the 3 groups in 3 ways.

2 people out of (a,b) can be picked in 1 way. Similarly in 1 way out of (c,d). 3 c 2 ways from (e,f,g)

I got 9/21 ways for them being friends. How ever, the answer seems to be 5/21 as the correct answer for not being friends is 16/21. Where did I go wrong fellas?

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Re: Friends-Probability [#permalink]

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New post 10 Nov 2008, 05:16
masuhari wrote:
5/6 instead of 3/6, coz there are 2 others in the same group(4) who'd be fair game too. They are friends with each other and not necessarily friends with our possible pick. Hence you have to consider picking them as favorable outcome(which totals 3 in group(3) and 2 from group(4)). Ofcourse total # of outcomes would be 6(7-1). HTH...


Thank you very much masuhari :) It clear now.

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Re: Friends-Probability [#permalink]

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New post 10 Nov 2008, 14:20
icandy wrote:
I started the problem trying to find out what is the probability of two people picked are friends

Total combo's = 7 C 2 = 21

{(a,b) (c,d)} (e,f,g) are such groups that fit the description

2 people can be picked from any of the 3 groups in 3 ways.

2 people out of (a,b) can be picked in 1 way. Similarly in 1 way out of (c,d). 3 c 2 ways from (e,f,g)

I got 9/21 ways for them being friends. How ever, the answer seems to be 5/21 as the correct answer for not being friends is 16/21. Where did I go wrong fellas?


hi icandy,
I think with your way of thinking, you have
1 way of picking 2 people from (a,b)
1 way of picking 2 people from (c,d)
3 way of picking 2 people from (e,f,g) (3!/2!)
Total will be 5 ways
You have a total combo is 21
So the Probability of picking 2 people who are friends is 5/21. So the probability of picking 2 people who are not friends is 16/21.
Is that correct?

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Re: Friends-Probability [#permalink]

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New post 10 Nov 2008, 15:05
icandy wrote:
I started the problem trying to find out what is the probability of two people picked are friends

Total combo's = 7 C 2 = 21

{(a,b) (c,d)} (e,f,g) are such groups that fit the description

2 people can be picked from any of the 3 groups in 3 ways.

2 people out of (a,b) can be picked in 1 way. Similarly in 1 way out of (c,d). 3 c 2 ways from (e,f,g)

I got 9/21 ways for them being friends. How ever, the answer seems to be 5/21 as the correct answer for not being friends is 16/21. Where did I go wrong fellas?


YOU MISSED THESE TWO FRIENDS: (a,b) (c,d).

So total friends = 5

so the desired prob = 1 - 5/21 = 16/21
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Re: Friends-Probability   [#permalink] 10 Nov 2008, 15:05
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