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sparsh.21: Master | Next Rank: 500 Posts; Posts: 129; Joined: Tue Jul 22, 2008 3:53 am; Thanked: 2 times

PS-1

by sparsh.21 » Sat Jan 03, 2009 10:51 am

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. 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

OA

E

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Source: Beat The GMAT — Problem Solving | Sat Jan 03, 2009 10:51 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pm; Location: Vancouver, BC; Thanked: 5254 times; Followed by:1268 members; GMAT Score:770

by Brent@GMATPrepNow » Sat Jan 03, 2009 11:30 am

Nice question.
I always like to begin by finding the denominator because it sometimes sheds light on how to calculate the numerator (not this time though

There are 7 people, so the number of ways to select 2 people is 7C2 = 21

There are two types of people: 1) Those with one friend and 2) those with 2 friends.

1) If a person has 1 friend, then there are 5 people who are not friends with this individual. So, there are 5 possible ways to select a non-friend.
In total there are 4 people with 1 friend, so there are a total of 20 possible ways to select non-friends.

2) If a person has 2 friends, then there are 4 people who are not friends with this individual. So, there are 4 possible ways to select a non-friend.
In total there are 3 people with 2 friend,s so there are a total of 12 possible ways to select non-friends.

Add those two values and we get a total of 32 possible ways to select non-friends. BUT, we must divide this number by two, because we have counted every pairing twice.

So, our numerator is 16 and our probability is 16/21

Brent Hanneson - Creator of GMATPrepNow.com

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bsandhyav: Master | Next Rank: 500 Posts; Posts: 116; Joined: Fri Aug 29, 2008 10:22 am; Location: Mumbai,India; Thanked: 1 times

Re: PS-1

by bsandhyav » Sat Jan 03, 2009 11:43 am

Let us consider the 7 people as A B C D E F G

4 people have exactly 1 friend :A<->B C<->D
3 people have exactly 2 friends : E friend of F & G
F friend of E & G
G friend of E & F

So no. of individuals who are not friends = Total pairs - No. pairs who are friends
= 7C2-5
= 16

So probability = no. of individuals who are not friends /no. of pairs of people

= 16/21

Hence E

Quote

woo: Senior | Next Rank: 100 Posts; Posts: 39; Joined: Sat Jan 03, 2009 11:34 pm

maybe it is easier to understand

by woo » Sat Jan 03, 2009 11:45 pm

I think another easier way to think of it is the following.

The probability of selected 2 being not friends is

1 - selected 2 being friends

Therefore, now we try to find the probability of selected 2 being friends.

The probability of picking a person with only 1 friend is 4/7

and picking the friend of the previously picked person is 1/6.

Therefore, the probabiliy is 4/7*1/6=2/21

Also, The probability of picking a person with 2 friends is 3/7

and picking a friend of the previously picked person is 2/6.

Therefore, the probability is 3/7*2/6=1/7

Adding the two probability gives 5/21.

Therefore, the probability of picking 2 who are not friends is

1-5/21=16/21. Therefore E