CAT Probability

[Helios]

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

Any other thought except the MGMAT explanation?

[jwinawer]

Hi Helios,

Please post the Manhattan GMAT explanation or a summary of it, so we can be sure that when we answer here in the forum we will not give the same explanation

-Jon

[Blue_Lotus]

There are 7 people A,B,C,D,E,F,G
selecting 2 people can be done in 7C2 ways = 21 -----------(1)


Now we know that 3 people , let us call A,B,C each have EXCATLY 2 friends each
they form three FRIENDSHIP GROUP AB,BC and CA ---(2)


We know that 4 people (D,E,F,G)have 1 friend each
they form 4 combination DE , ED, FG,GF

BUT DE and ED are the same group of people,
This means these 4 people make 2 friendship groups only. ---(3)

From 2 and 3 we know that there are 5 friendship groups



P=(No. of favourable out comes )/ ( Total Number of outcomes)

No. of favourable out comes = group of people who are NOT friends
In 21 groups 5 are friendship group , so (21- 5 = 16) are non friendship group

p = 16/21

Answer is E


{
NOTE: you may ask in equation (2) , why I have assumed A B C to be mutual friends and not
AE,AF , BC, BG, CB, CD. [BC and CB are the same , so it is 5 pairs]

The Point is that you have to connect 7 people such that 3 have EXACTLy 3 friends
and 4 have 1 friend. you may do it any way you like.
but you will observe that there are only 5 friendship groups.
}

[jwinawer]

ok, here is a slightly different way to think about the problem. instead of pairs, think of one person at a time.

first, you pick one person from the room. there are 7 people. 4 have exaclty one friend, and 3 have exactly 2 friends.

so this means that there is a 4/7 chance that our first pick has one friend, and a 3/7 chance that he has 2 friends. so let's break that down:

situation A, one friend: 4/7 probability
situation B, two friends: 4/7 probability

now, for each situation, what is the probability that the second person is a friend?

situation A: out of the remaining 6 people, only one can be the friend. so for situation A, there is 4/7 * 1/6 chance of getting friends.

sitatuation B: out of the remaining 6 people, only two can be friends. so for situation B, there is 3/7 * 2/6 chance of getting friends.

now we add: 4/7 * 1/6 + 3/7*2/6 = 5/21

there is 5/21 chance of getting friends, and hence 16/21 chance that we don't get friends.