Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

4

This post received KUDOS

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

85% (hard)

Question Stats:

54% (03:39) correct
46% (02:15) wrong based on 131 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?

(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.

(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

1

This post was BOOKMARKED

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...

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

4

This post received KUDOS

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.

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

1

This post received KUDOS

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

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?

safe.txmblr Can business make a difference in the great problems that we face? My own view is nuanced. I think business potentially has a significant role to play...

Still 7 months to go to be at Lausanne. But, as Lausanne has a vacancy rate of 0.1% for rental properties, I booked my rental apartment yesterday for...