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:

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

Show Tags

01 Nov 2009, 10:23

7

This post received KUDOS

37

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

48% (02:48) correct
52% (01:58) wrong based on 1005 sessions

HideShow timer Statistics

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?

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

As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4). As there are 3 people with exactly 2 siblings each: we have one triple of siblings (5-6-7).

Solution #1: # of selections of 2 out of 7 - \(C^2_7=21\); # of selections of 2 people which are not siblings - \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings)+\(C^1_2*C^1_3\) (one from first pair of siblings*one from triple)+ \(C1^_2*C^1_3\)(one from second pair of siblings*one from triple) \(=4+6+6=16\).

\(P=\frac{16}{21}\)

Solution #2: # of selections of 2 out of 7 - \(C^2_7=21\); # of selections of 2 siblings - \(C^2_3+C^2_2+C^2_2=3+1+1=5\);

4 have exactly 1 sibling which can mean: 1-2 are siblings 3-4 are siblings

3 have exactly 2 siblings which can mean: 5-6-7 are siblings

Let's start with the group of 4: The probability of picking 1 is (1/7) The probability of not getting a sibling pair is (5/6) because the only other sibling is 2 Therefore, if 1 is selected first the probability of not getting a sibling pair is 5/42 Multiply that by 4 because the probability is the same whether you start with 1-2-3-4 so you get 20/42 for the first 4 people

Now let's go to the group of 3: The probability of picking 5 is (1/7) The probability of not getting a sibling pair is (4/6) which the non-sibling pair is 1-2-3-4 Therefore the probability is 4/42. Multiply that probability by 3, which represent 5-6-7 so the probability is 12/42

Now you have two probabilities: 12/42 and 20/42 add both and you get 32/42 or 16/21

1) we have 4/7 probability of getting 1-sibling person and 5/6 probability of not getting his or her sibling. 2) we have 3/7 probability of getting 2-siblings person and 4/6 probability of not getting his or her sibling.

I also get the same result 16/21, but my approach is a bit different. There are 7C2 =21 ways to select 2 people in a group of 7. There are 1+1+3 = 5 ways to select 2 people who are siblings. So the probability of selecting 2 people who are not siblings is : 1-5/1 =16/21

@ Walker: I also tried to solve the problem in the way that you suggested, but i can't get myself clear why the two events mentioned are mutually exclusive. I thought the following events are mutually exclusive for this problem, and i arrived with the following solution 1) 1 person from 2-sibling group and 1 person from 1-sibling group, which is the same with your second event: we have 3/7 probability of getting 2-siblings person and 4/6 probability of not getting his or her sibling. 2) No one from the 2-sibling group. we have 4/7 probability of getting 1-sibling person and 2/6 probability of not getting his or her sibling and not within one of 2-sibling group. p = 3/7*4/6 + 4/7*2/6 = 20/42 (this is not within the answers, so i guess it is wrong, but could you please help me where I got wrong with this). Many thanks.

@ Walker: I also tried to solve the problem in the way that you suggested, but i can't get myself clear why the two events mentioned are mutually exclusive. I thought the following events are mutually exclusive for this problem, and i arrived with the following solution 1) 1 person from 2-sibling group and 1 person from 1-sibling group, which is the same with your second event: we have 3/7 probability of getting 2-siblings person and 4/6 probability of not getting his or her sibling. 2) No one from the 2-sibling group. we have 4/7 probability of getting 1-sibling person and 2/6 probability of not getting his or her sibling and not within one of 2-sibling group. p = 3/7*4/6 + 4/7*2/6 = 20/42 (this is not within the answers, so i guess it is wrong, but could you please help me where I got wrong with this). Many thanks.

In your approach you need to add one more event:

3) 1 person from 1-sibling group and next person from 2-sibling group: 4/7*3/6

p = 20/42 + 12/42 = 16/21

Actually, you have three combinations: 1) 21 2) 11 3) 12. _________________

In a room filled with 7 people, 4 have 1 sibling and 3 have two siblings. If two people at selected at random what is the prob that they are not siblings.

Answer is 6/21

Can someone explain.

Out of 7 people 4 have 1 sibling So these 4 form 2 pairs of siblings 3 have 2 siblings So all these 3 are siblings to each other.

So the 7 people are like A1A2, B1B2, C1C2C3 where A1 and A2 are siblings to each other , B1 and B2 are siblings to each other and C1,C2 and C3 are siblings to each other.

Probability of selecting 2 people who are not siblings = 1 - Probability of selecting 2 people who are siblings = 1 - \((C^2_1+C^3_2)/C^7_2\) (Select either A1A2/B1B2 or any 2 out of C1C2C3) = 1- 5/21 = 16/21 _________________

___________________________________ Please give me kudos if you like my post

total possibilities 2C7=21 possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5 probability 1-5/21

vittar..can you please explain 2C3.....

As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4). As there are 3 people with exactly 2 siblings each: we have one triple of siblings (5-6-7).

Let's calculate the probability of opposite event and subtract it from 1. Opposite event would be that chosen 2 individuals are siblings.

# of selections of 2 out of 7 - \(C^2_7=21\); # of selections of 2 siblings - \(C^2_3+C^2_2+C^2_2=3+1+1=5\), here \(C^2_3\) is the # of ways to choose 2 siblings out of siblings 5-6-7, \({C^2_2}\) is the # of ways to choose 2 siblings out of siblings 1-2, and \(C^2_2\) is the # of ways to choose 2 siblings out of siblings 3-4;

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

There are suppose A B C D E F G members in the room 4 people who have exactly one sibling....A B C D....(A is B`s sibling and viceversa) (C is D`s sibling and viceversa)...now remaning EFG are 3 people who have exactly 2 siblings....(E has F and G as his/her sibling and so on..) there are now 3 different set of siblings (A and B) (C and D);(EFG) Now first selecting 2 people out of 7 is 7C2=21 first sibling pair----(A and B)--selecting 2 people --2C2=1 second sibling pair (C and D)--selecting 2 people--2C2=1 third sibling pair (E F G)--selecting 2 out of 3 --3C2=3

total= 1+1+3=5 but,a/c to formula P(success)-1-p(fail) here,p(failure)is selecting 2 people who are siblings =5/21(21 is 7C2) =1-5/21 =16/21 ANS E

first, find probability of getting two pairs and triplet 4/7*1/6=4/42 and 3/7*2/6=6/42 second, add these two results 4/42 + 6/42 = 10/42 = 5/21 third, subtract the result of addition from 1 1 – 5/21 = 16/21

there are 4 ppl with exactly one sibling. lets consider the following 1,3 - group1 2,4 - group 2 there are 3 people with exactly two siblings 5,6,7 - group 3

so the probability of picking 2 non siblings is nothing but picking each from different groups.

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?

5/21 3/7 4/7 5/7 16/21

how do i figure it out in simplest manner? please help

Merging similar topics. Please ask if anything remains unclear.

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?

5/21 3/7 4/7 5/7 16/21

how do i figure it out in simplest manner? please help

Merging similar topics. Please ask if anything remains unclear.

Bunuel,

can you explain your #3 approach.

Thanks in advance.

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}.

I have a doubt in approach 1. We have created groups and then we made the selections. i.e selecting one individual from group 1 and another individual from group 2 or select one individual from group 1 and another individual from Group 3 and so on..

However, I would like to ask, can't we dissolve the group and then find the probability.

Lets say, we have total (A,B) (C,D) and (EFG) as groups

Case 1

i.e. Select One individual from Group 1->(A,B) This can be done in 2C1 way and now select the other individual from remaining 5 individuals i.e from C,D,E,F,G - this can be done in 5C1 way. Therefore, total number of ways 2C1*5C1

Case 2

Similarly, select one sibling from (C,D) and select the other sibling from (A,B,E,F,G) ie total number of ways 2C1*5C1

Case 3

Last Case, select one sibling from (E,F,G) and remaining from (A,B,C,D) i.e. 3C1* 4C1

Whats wrong with this approach. Please clarify.

Thanks H

Bunuel 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

As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4). As there are 3 people with exactly 2 siblings each: we have one triple of siblings (4-5-6).

Solution #1: # of selections of 2 out of 7 - \(C^2_7=21\); # of selections of 2 people which are not siblings - \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings)+\(C^1_2*C^1_3\) (one from first pair of siblings*one from triple)+ \(C1^_2*C^1_3\)(one from second pair of siblings*one from triple) \(=4+6+6=16\).

I have a doubt in approach 1. We have created groups and then we made the selections. i.e selecting one individual from group 1 and another individual from group 2 or select one individual from group 1 and another individual from Group 3 and so on..

However, I would like to ask, can't we dissolve the group and then find the probability.

Lets say, we have total (A,B) (C,D) and (EFG) as groups

Case 1

i.e. Select One individual from Group 1->(A,B) This can be done in 2C1 way and now select the other individual from remaining 5 individuals i.e from C,D,E,F,G - this can be done in 5C1 way. Therefore, total number of ways 2C1*5C1

Case 2

Similarly, select one sibling from (C,D) and select the other sibling from (A,B,E,F,G) ie total number of ways 2C1*5C1

Case 3

Last Case, select one sibling from (E,F,G) and remaining from (A,B,C,D) i.e. 3C1* 4C1

Whats wrong with this approach. Please clarify.

Thanks H

Bunuel 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

As there are 4 people with exactly 1 sibling each: we have two pairs of siblings (1-2; 3-4). As there are 3 people with exactly 2 siblings each: we have one triple of siblings (4-5-6).

Solution #1: # of selections of 2 out of 7 - \(C^2_7=21\); # of selections of 2 people which are not siblings - \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings)+\(C^1_2*C^1_3\) (one from first pair of siblings*one from triple)+ \(C1^_2*C^1_3\)(one from second pair of siblings*one from triple) \(=4+6+6=16\).

\(P=\frac{16}{21}\)

Answer: E.

I don't understand the red part at all.

As for your solution it's wrong because when for case 1 you select one from {A,B} and then one from {C,D, E, F, G} you can have pair {A,C}. Next, when for case 2 you select one from {C,D} and then one from {A,B,E,F,G} you can have pair {A,C} again. So, you double-count some cases.

As for your solution it's wrong because when for case 1 you select one from {A,B} and then one from {C,D, E, F, G} you can have pair {A,C}. Next, when for case 2 you select one from {C,D} and then one from {A,B,E,F,G} you can have pair {A,C} again. So, you double-count some cases.

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

prob of getting sibbling = 15 /21

prob of not getting sibbling = 1 - 15/21 = 6/21

gmatclubot

Re: In a room filled with 7 people, 4 people
[#permalink]
03 Nov 2012, 17:36

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...

Rishabh from Gyan one services, India had a one to one interview with me where I shared my experience at IMD till now. http://www.gyanone.com/blog/life-at-imd-interview-with-imd-mba/ ...