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In a room with 7 people, 4 people have exactly 1 friend in [#permalink]
08 Dec 2007, 12:00

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Question Stats:

54% (03:42) correct
46% (02:16) wrong based on 137 sessions

In a room 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 Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

Re: PS: Combinatoric [#permalink]
18 Jan 2008, 07:00

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marcodonzelli wrote:

right....let's solve it with combs:

How many different subsets of two people can be formed from the seven people? 7C2 = 21

How many of these subsets contain two friends? please give me a hand in doing this...

you are right that there are 21 possible ways of choosing a pair. I suggest visualizing a graph of friendships. 1 and 2 are friends; 3 and 4 are friends; 5, 6 and 7 are friends (3 pairs here: 5-6, 6-7, 5-7). Therefore, there are 5 total pairs of friends. Then probability of choosing a pair who are not friends is 1-5/21 = 16/21. I hope that helps.

(4/7)(5/6) + (3/7)(4/6) if you choose one of the 4 with one other friend, then you have a 5/6 chance of not picking their friend 2nd. If you choose one of the 3 with 2 friends, you have a 4/6 chance of not picking one of their friends second. Add them up.

Re: PS: Combinatoric [#permalink]
14 Aug 2008, 11:13

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Expert's post

\(4*C_1^6\) - for each person out of 4 we have only one friend out of remained people (7-1=6) \(3*C_2^6\) - for each person out of 3 we have two friends out of remained people (7-1=6)

In a room 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 Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends? 1)5/21 2)3/7 3)4/7 4)5/7 5)16/21

Soln Assuming the people to be A,B,C,D,E,F,G

We can arrange in this way such that 3 people have two friends while 4 have 1 friend A B B A C D E D C E E C D F G G F

First column represents the 7 people, second and third columns represent their friends. Thus A,B,F,G each have one friend while C,D,E have two friends

Probability when two people are chosen where they are not friends is = (5 + 5 + 2 + 2 + 2)/7C2 = 16/21

(4/7)(5/6) + (3/7)(4/6) if you choose one of the 4 with one other friend, then you have a 5/6 chance of not picking their friend 2nd. If you choose one of the 3 with 2 friends, you have a 4/6 chance of not picking one of their friends second. Add them up.

20/42 + 12/42 32/42 = 16/21

E. 16/21

excuse me, can you explain me this passage in detail? why do we have to sum probabilities up? don't we need to multiply them? we need that both conditions apply. thanks a lot

Re: PS: Combinatoric [#permalink]
17 Jan 2008, 04:51

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We don't need to multiply them together because we're only choosing one pair of people.

= probability of choosing 2 people who aren't friends out of the 4 with one friend = probability of choosing 2 people who aren't friends out of the 3 with two friends

Each one represents a separate group of people so adding them together gives you the total probability of selecting two people who aren't friends.

7-t58525 -- here's another post on this topic, but it looks like they handled it exactly the same way

Re: PS: Combinatoric [#permalink]
18 Jan 2008, 04:17

eschn3am wrote:

We don't need to multiply them together because we're only choosing one pair of people.

= probability of choosing 2 people who aren't friends out of the 4 with one friend = probability of choosing 2 people who aren't friends out of the 3 with two friends

Each one represents a separate group of people so adding them together gives you the total probability of selecting two people who aren't friends.

7-t58525 -- here's another post on this topic, but it looks like they handled it exactly the same way

right....let's solve it with combs:

How many different subsets of two people can be formed from the seven people? 7C2 = 21

How many of these subsets contain two friends? please give me a hand in doing this...

Can you explain this a little more please? I understand the denominator - pick two people from 7. I think the numerator is the different combinations of people who ARE friends. But how did you get this?

Re: In a room with 7 people, 4 people have exactly 1 friend in [#permalink]
23 Apr 2013, 05:47

alexperi wrote:

In a room 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 Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends? 1)5/21 2)3/7 3)4/7 4)5/7 5)16/21

1. We will find the reverse case, i.e., 2 people who are selected, are friends. 2. There are two groups here. People with 1 friend and people with 2 friends. 3. Friends can be selected as follows: (i) Among the 4 people each with just 1 friend, there will be 2 pairs of friends. (ii) Among the 3 people each with 2 friends, there will be 3 pairs of friends. 4. So 5 cases of friends can be selected. 5. The total number of ways of selecting 2 people is\(7C2 = 21\) 6. The number of cases when the 2 selected people are not friends \(= 21-5 = 16\) 7. Probability that the two slected people are not friends\(= 16/21\) The answer is therefore E. _________________

Re: In a room with 7 people, 4 people have exactly 1 friend in [#permalink]
23 Apr 2013, 05:51

Expert's post

alexperi wrote:

In a room 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 Jane is Paul's friend, Paul is Jane's friend). If two individuals are selected from the room at random, what is the probability that those two individuals are NOT friends?

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...