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 500,000 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:

Re: In a room filled with 7 people, 4 people [#permalink]

Show Tags

03 Nov 2012, 17:36

Bunuel wrote:

Archit143 wrote:

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

Here is how i approached the problem and got the wrong answer can you pls correct me.

Total outcome = 7C2 = 21

Prob of getting exactly 2 sibbling = [4C1 * 3C1] + 3C2 i.e. [ 1 from 4 * 1 from 3] + [ 2 out of 3, since the group of 3 has 2 sibbling] = [4*3] + 3 =15

Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

Show Tags

13 Dec 2012, 15:08

well, this is how i approached it...

Total 7 people - 1,2,3,4,5,6,7 4 people have 1 sibling - [1,2];[3,4] - 2 single-sibling groups 3 people have 2 siblings - [5,6,7] - 1 two-siblings group

Total ways to select 2 people, 7C2 = 21. Ways to select only siblings : 2C1( ways to select 1 group from 2 single-sibling groups) + 3C2( ways to select 2 people from 2 sibling-group) = 2+3= 5

Probability that NO siblings selected = 1- 5/21 = 16/21.

Hence, D.

Please let me know if my approach is flawed.
_________________

What happens with me is less significant than what happens within me! Consider giving Kudos, appreciation is divine.

Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

Show Tags

15 May 2013, 01:26

WarriorGmat wrote:

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

A) 5/21 B) 3/7 C) 4/7 D) 5/7 E) 16/21

If A is a sibling of B, then B is also a sibling of A. If A is a sibling of B, and B is a sibling of C, that means A is also a sibling of C.

For first four people to have exactly one sibling each means two pairs of siblings. For last 3 people to have exactly two siblings each means one triplet of siblings.

Now, to select two individuals from a group of 7, total no. of ways it can be done = 7C2 = 21

Cases if siblings are selected : 1. Pair 1 2. Pair 2 3,4,5 = 2 individuals from triplet of siblings = 3C2 = 3 ways.

So, there are 5 cases in which selected individuals are siblings.

Therefore, probability of two individuals selected are NOT sibligs is (21-5)/21 = 16/21

Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

Show Tags

15 May 2013, 01:33

mkdureja wrote:

WarriorGmat wrote:

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

A) 5/21 B) 3/7 C) 4/7 D) 5/7 E) 16/21

If A is a sibling of B, then B is also a sibling of A. If A is a sibling of B, and B is a sibling of C, that means A is also a sibling of C.

For first four people to have exactly one sibling each means two pairs of siblings. For last 3 people to have exactly two siblings each means one triplet of siblings.

Now, to select two individuals from a group of 7, total no. of ways it can be done = 7C2 = 21

Cases if siblings are selected : 1. Pair 1 2. Pair 2 3,4,5 = 2 individuals from triplet of siblings = 3C2 = 3 ways.

So, there are 5 cases in which selected individuals are siblings.

Therefore, probability of two individuals selected are NOT sibligs is (21-5)/21 = 16/21

Hi mkdureja, Can you please elaborate on this please? Thanks.

Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

Show Tags

15 May 2013, 02:01

1

This post received KUDOS

Hi,

Que: In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

Case 1: 4 people have exactly 1 sibling : Let A, B, C, D be those 4 people If A is sibling of B, then B is also a sibling of A. Basically, they have to exist in pairs Therefore, these four people compose of two pairs of siblings.

Case 2: 3 people have exactly 2 siblings : This must obviously be a triplet of siblings. Let them be E, F, G. If E is sibling of F, and also is sibling of G., it means F and G are also siblings of each other. E is sibling of exactly 2 : F and G F is sibling of exactly 2 : E and G G is sibling of exactly 2 : E and F This group has three pairs of siblings.

Now, selecting two individuals out of the group of 7 people has: 1) Two Pairs, as in Case 1. 2) Three Pairs, as in Case 2. So, 5 cases out of a total of 7C2 = 21 cases.

Hope it clarifies. If you have any further doubt, please point to exactly where you are having a problem understanding it.

In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room. If two individuals are selected from the room at random, what is the probability that those two individuals are NOT siblings?

A. 5/21 B. 3/7 C. 4/7 D. 5/7 E. 16/21

It is first important to understand the problem. So let us first assume a specific case.

1. Assume the 7 people are A,B, C, D, E, F and G 2. Assume the 4 people who have 1 sibling are A, B, C and D 3. Let's assume A's sibling is B. Therefore B's sibling is A. Similarly for C and D. 4. So we are left with E, F and G. Each should have exactly 2 siblings. 5. E's siblings will be F and G. So F's siblings will be E and G and G's siblings will be E and F 6. Now look at the general case. 7.Total number of ways of selecting 2 people out of 7 people is 7C2=21 8. Instead of A, B, C and D assume any 4 people. We can see for every such 4 people assumed, there are 2 cases where the selected 2 will be siblings. In the case we assumed they are A and B or C and D. This gives one of the favorable outcomes 9. Or the 2 people selected being siblings may come out of the 3 siblings. The number of favorable outcomes is 3 as we can see in the specific case they are E and F, or F and G or E and G. 10. The total number of favorable outcomes for the selected two being siblings is 2+3=5. 11. The probability that the two selected are siblings is 5/21. 12, Therefore the probability that the two selected are not siblings is 1-5/21= 16/21
_________________

Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

Show Tags

14 Jan 2014, 21:53

Quote:

Sure.

We have the following siblings: {1, 2}, {3, 4} and {5, 6, 7}.

Now, in order to select two individuals who are NOT siblings we must select EITHER one from {5, 6, 7} and ANY from {1, 2} or {3, 4} OR one from {1, 2} and another from {3, 4}.

3/7 - selecting a sibling from {5, 6, 7}, 4/6 - selecting any from {1, 2} or {3, 4}. Multiplying by 2 since this selection can be don in two ways: the first from {5, 6, 7} and the second from {1, 2} or {3, 4} OR the first from {1, 2} or {3, 4} and the second from {5, 6, 7};

2/7 - selecting a sibling from {1, 2}, 2/6 - selecting a sibling from {3, 4}. Multiplying by 2 since this selection can be don in two ways: the first from {1, 2} and the second from {3, 4} OR the first from {3, 4} and the second from {1, 2}.

Other approaches here: in-a-room-filled-with-7-people-4-people-have-exactly-87550.html#p645861

Hope it's clear.

Hi Bunuel, Can you please explain to me that why when we use simple probability we need to consider both the cases of selection, if we pick a sibling from the first group of 2 siblings first and then we pick a sibling from the second group of 2 siblings and if we pick a sibling from the second group of 2 siblings first and then we pick a sibling from the first group of 2 siblings. While if we use combinomatric approach we just count one case i.e. 2C1*2C1 which is for selecting one each from the 2 different 2 sibling groups.

Hi Bunuel Please correct me where I went wrong. I assumed the 7 siblings in this way.PFA the image. So this scenario also satisfies that B.C.F have 2 siblings and A,D,E ,G have 1 sibling each. Now to select two people that are not siblings : 1)I first select A(1/7) and from rest of the people I can select C,D,E,F,G so the probability of selecting other person that is not sibling is (1/5). And this scenario will be vaild for D,E and G as well So the proability=4*(1/7 * 1/5) 2)Now if I select B (1/7) and from rest of the people I can select ,D,E,F,G so the probability of selecting other person that is not sibling is (1/4). And this scenario will be vaild for B,C,F as well So the proability=3*(1/7 * 1/4) So Total= 4*(1/7 * 1/5) + 3*(1/7 * 1/4)

Which does not lead to a correct answer. Please HELP!!!!!!!!!!! And please pardon my bad drawing skills

Attachments

File comment: My Solution

Questions.jpg [ 20.61 KiB | Viewed 4133 times ]

_________________

Feel Free to Press Kudos if you like the way I think .

Hi Bunuel Please correct me where I went wrong. I assumed the 7 siblings in this way.PFA the image. So this scenario also satisfies that B.C.F have 2 siblings and A,D,E ,G have 1 sibling each. Now to select two people that are not siblings : 1)I first select A(1/7) and from rest of the people I can select C,D,E,F,G so the probability of selecting other person that is not sibling is (1/5). And this scenario will be vaild for D,E and G as well So the proability=4*(1/7 * 1/5) 2)Now if I select B (1/7) and from rest of the people I can select ,D,E,F,G so the probability of selecting other person that is not sibling is (1/4). And this scenario will be vaild for B,C,F as well So the proability=3*(1/7 * 1/4) So Total= 4*(1/7 * 1/5) + 3*(1/7 * 1/4)

Which does not lead to a correct answer. Please HELP!!!!!!!!!!! And please pardon my bad drawing skills

So, as per your diagram (A, B), (B, C), (C, D) are siblings but (A, C), (A, D), and (B, D) are not? How? This is not what question implies.
_________________

Hi Bunuel Please correct me where I went wrong. I assumed the 7 siblings in this way.PFA the image. So this scenario also satisfies that B.C.F have 2 siblings and A,D,E ,G have 1 sibling each. Now to select two people that are not siblings : 1)I first select A(1/7) and from rest of the people I can select C,D,E,F,G so the probability of selecting other person that is not sibling is (1/5). And this scenario will be vaild for D,E and G as well So the proability=4*(1/7 * 1/5) 2)Now if I select B (1/7) and from rest of the people I can select ,D,E,F,G so the probability of selecting other person that is not sibling is (1/4). And this scenario will be vaild for B,C,F as well So the proability=3*(1/7 * 1/4) So Total= 4*(1/7 * 1/5) + 3*(1/7 * 1/4)

Which does not lead to a correct answer. Please HELP!!!!!!!!!!! And please pardon my bad drawing skills

So, as per your diagram (A, B), (B, C), (C, D) are siblings but (A, C), (A, D), and (B, D) are not? How? This is not what question implies.

I still do not understand .The question simply says "In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room". and my diagram meets the given criterion.Also is it necessary that if A&B and B& C are siblings then A&C also need to be siblings.
_________________

Feel Free to Press Kudos if you like the way I think .

Hi Bunuel Please correct me where I went wrong. I assumed the 7 siblings in this way.PFA the image. So this scenario also satisfies that B.C.F have 2 siblings and A,D,E ,G have 1 sibling each. Now to select two people that are not siblings : 1)I first select A(1/7) and from rest of the people I can select C,D,E,F,G so the probability of selecting other person that is not sibling is (1/5). And this scenario will be vaild for D,E and G as well So the proability=4*(1/7 * 1/5) 2)Now if I select B (1/7) and from rest of the people I can select ,D,E,F,G so the probability of selecting other person that is not sibling is (1/4). And this scenario will be vaild for B,C,F as well So the proability=3*(1/7 * 1/4) So Total= 4*(1/7 * 1/5) + 3*(1/7 * 1/4)

Which does not lead to a correct answer. Please HELP!!!!!!!!!!! And please pardon my bad drawing skills

So, as per your diagram (A, B), (B, C), (C, D) are siblings but (A, C), (A, D), and (B, D) are not? How? This is not what question implies.

I still do not understand .The question simply says "In a room filled with 7 people, 4 people have exactly 1 sibling in the room and 3 people have exactly 2 siblings in the room". and my diagram meets the given criterion.Also is it necessary that if A&B and B& C are siblings then A&C also need to be siblings.

Do you know what a sibling mean? How can (A, B), (B, C), (C, D) be BROTHERS, and (A, C), (A, D), and (B, D) not to be?
_________________

Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

Show Tags

29 Jan 2014, 11:33

Hi,

Can someone please explain to me that why when we use simple probability we need to consider both the cases of selection i.e. if we pick a sibling from the first group of 2 siblings(A,B) first and then we pick a sibling from the second group of 2 siblings(C,D) and if we pick a sibling from the second group of 2 siblings(C,D) first and then we pick a sibling from the first group of 2 siblings(A,B) While if we use combinomatric approach we just count one case i.e. 2C1*2C1 which is for selecting one each from the 2 different 2 sibling groups

How is A,C and C,A different if we just have to check that whether the two selected are siblings or not? And if it makes a difference why we did not you permutation instead of combinations?

Piyush K ----------------------- Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison Don't forget to press--> Kudos My Articles: 1. WOULD: when to use?| 2. All GMATPrep RCs (New) Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

When you select groups 1 and 2, you have 3*2 choices, when groups 1 and 3 you have 3*2 choices and when groups 2 and 3 you have 2*2 choices: total = 6 + 6 + 4.
_________________

Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]

Show Tags

10 Jul 2015, 20:46

x = P(1-sibling person) = 4*p(1-sibling person) y = P(2-sibling person) = 3*p(2-sibling person)

p(1-sibling person)=1/7*1/6 (1/6 because we have 1 sibling in the 6 remaining people) p(2-sibling person)=1/7*2/6 (2/6 because we have 2 siblings in the 6 remaining people)