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Friends Probability Question

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by newgmattest » Tue May 31, 2011 4:21 am
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

OA : E
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by Frankenstein » Tue May 31, 2011 4:50 am
Hi,
Let the friends be 1,2,3,4,5,6,7. According to the problem, 1,2 are friends, 3,4 are friends and 5,6,7 are friends.
Number of ways of picking 2 so that they a re friends is picking(1,2)+picking(3,4)+picking 2out of (5,6,7) = 1+1+3C2 = 5
Total number of ways of picking 2 guys from the group of 7 = 7C2 = 21.
So, probability that the friends picked are friends is 5/21.
Thus, the probability that those two individuals are NOT friends is 1-5/21 =16/21

Hence, answer E

Cheers!
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by Anurag@Gurome » Tue May 31, 2011 5:05 am
newgmattest 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

OA : E
Probability of choosing 2 people who are not friends out of the 4 with one friend = 4/7 * 5/6
Probability of choosing 2 people who are not friends out of the 3 with two friends = 3/7 * 4/6
Therefore, required probability = 4/7 * 5/6 + 3/7 * 4/6 = 32/42 = 16/21

The correct answer is E.
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by newgmattest » Tue May 31, 2011 5:13 am
Hi Anurag. Thanks for reply. Can you please explain how we derived second multiple in each pair?
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by Anurag@Gurome » Tue May 31, 2011 6:47 am
newgmattest wrote:Hi Anurag. Thanks for reply. Can you please explain how we derived second multiple in each pair?
Let us assume that A, B, C, D, E, F, G are 7 people.
Now possible combinations so that 4 have exactly 1 friend and 3 have exactly 2 friends.

A-B
B-A
C-D
D-C
E-F-G
F-E-G
G-E-F

If we choose one of the 4 people with one other friend, then we have 5/6 chance of not picking their 2nd friend.
Similarly, if we choose one of the 3 people with 2 friends, we have 4/6 chance of not picking one of their 2 friends.

Hope that helps.
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