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Peter is John's friend

Expert replies
by logitech » Tue Nov 25, 2008 10:40 pm
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?

OA [spoiler]16/21[/spoiler]
LGTCH
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Source: — Problem Solving |

by mals24 » Wed Nov 26, 2008 2:38 am
Ok say there are 7 guys A B C D E F G

Now 4 have exactly 1 friend and 3 have exactly 2

So the total number of pairs of friends:

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

That makes it a total of 5 pair of friends.

Total number of ways of selecting 2 people out of 7 = 7C2 = 21

Hence probability of selecting 2 people who are NOT friends:

1-probability of selecting 2 friends = 1-5/21 = 16/21
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by vivek.kapoor83 » Wed Nov 26, 2008 2:45 am
Mals !!
How u arrived on pairs of friends...pls elaborate
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by mals24 » Wed Nov 26, 2008 3:51 am
See there are 7 people in the party

A B C D E F G

Now exactly 4 people have 1 friend

Lets say A B C D have exactly 1 friend

A is B's friend
C is D's friend

Now in the question its given that friendship is a mutual relationship, i.e. if John is Peter's friend, Peter is John's friend

This means if A is B's friend, then B is also A's friend and since we assumed that B has only 1 friend A will be B's only friend right.

So we have 1 pair of friendship A-B
Same logic for C and D

So that gives us 2 pairs, A-B; C-D

Now exactly 3 people have 2 friends

Lets say E F G each has exactly 2 friends. But neither one of them can be friends with A, B, C and D (because they are done with their share of friendship)

So E will have 2 friends F and G
F will have 2 friends E and G
G will have 2 friends E and F

So now we have 3 more pairs.

E = F+G
F = E+G
G = F+E

So in total 5 pairs.

Still confused
:?:
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by logitech » Wed Nov 26, 2008 8:27 am
Mals24 in da house!!! Beautiful solution.
LGTCH
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by jimmiejaz » Wed Nov 26, 2008 9:08 am
Logitech,

What should be the approach to solve these type of questions if the numbers are too large??? I mean here there are 7 people. What if the question asks about 20 people?
Is there any approach we shall follow?
What if i have not yet beat the beast, I know i will beat it!!!!!!!!
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