Author 
Message 
TAGS:

Hide Tags

VP
Status: Final Lap Up!!!
Affiliations: NYK Line
Joined: 21 Sep 2012
Posts: 1004
Location: India
GMAT 1: 410 Q35 V11 GMAT 2: 530 Q44 V20 GMAT 3: 630 Q45 V31
GPA: 3.84
WE: Engineering (Transportation)

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?
5/21 3/7 4/7 5/7 16/21 Merging similar topics. Please refer to the solutions on page 1 (foe example: inaroomfilledwith7people4peoplehaveexactly87550.html#p645861) Hope it helps. BUNEL 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



Senior Manager
Joined: 13 Aug 2012
Posts: 436
Concentration: Marketing, Finance
GPA: 3.23

Re: In a room filled with 7 people, 4 people have exactly 1
[#permalink]
Show Tags
12 Nov 2012, 05:38
Ways to select the those 4 with their sibling? 4/7 x 1/6 = 4/42 Ways to select those 3 with one of their 2 siblings? 3/7 x 2/6 = 6/42 P = 1  (4/42 + 6/42) = 1  10/42 = (42 10)/42 = 32/42 = 16/21 P = 16/21 Answer: E
_________________
Impossible is nothing to God.



Intern
Joined: 11 Dec 2012
Posts: 3

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 singlesibling groups 3 people have 2 siblings  [5,6,7]  1 twosiblings group Total ways to select 2 people, 7C2 = 21. Ways to select only siblings : 2C1( ways to select 1 group from 2 singlesibling groups) + 3C2( ways to select 2 people from 2 siblinggroup) = 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.



Intern
Joined: 04 Jan 2013
Posts: 13
Location: India
Concentration: Finance
GMAT Date: 08262013
GPA: 2.83
WE: Other (Other)

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 (215)/21 = 16/21



Intern
Joined: 09 Feb 2013
Posts: 7

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 (215)/21 = 16/21 Hi mkdureja, Can you please elaborate on this please? Thanks.



Intern
Joined: 04 Jan 2013
Posts: 13
Location: India
Concentration: Finance
GMAT Date: 08262013
GPA: 2.83
WE: Other (Other)

Re: In a room filled with 7 people, 4 people have exactly 1
[#permalink]
Show Tags
15 May 2013, 02:01
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.



Math Expert
Joined: 02 Sep 2009
Posts: 49320

In a room filled with 7 people, 4 people have exactly 1
[#permalink]
Show Tags
05 Jul 2013, 02:26



Director
Joined: 17 Dec 2012
Posts: 636
Location: India

Re: In a room filled with 7 people, 4 people have exactly 1
[#permalink]
Show Tags
05 Jul 2013, 22:07
reply2spg 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 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=218. 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 15/21= 16/21
_________________
Srinivasan Vaidyaraman Sravna Holistic Solutions http://www.sravnatestprep.com
Holistic and Systematic Approach



Intern
Joined: 10 Dec 2013
Posts: 20
Location: India
Concentration: Technology, Strategy
GPA: 3.9
WE: Consulting (Consulting)

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}.
P=2*\frac{3}{7}*\frac{4}{6}+2*\frac{2}{7}*\frac{2}{6}=\frac{4}{7}+\frac{4}{21}=\frac{16}{21}.
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: inaroomfilledwith7people4peoplehaveexactly87550.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.



Manager
Joined: 19 Mar 2012
Posts: 114

Re: Probability
[#permalink]
Show Tags
26 Jan 2014, 11:37
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 5728 times ]
_________________
Feel Free to Press Kudos if you like the way I think .



Math Expert
Joined: 02 Sep 2009
Posts: 49320

Re: Probability
[#permalink]
Show Tags
27 Jan 2014, 01:28
282552 wrote: 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.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 19 Mar 2012
Posts: 114

Re: Probability
[#permalink]
Show Tags
27 Jan 2014, 06:33
Bunuel wrote: 282552 wrote: 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 .



Math Expert
Joined: 02 Sep 2009
Posts: 49320

Re: Probability
[#permalink]
Show Tags
27 Jan 2014, 06:36
282552 wrote: Bunuel wrote: 282552 wrote: 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?
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Intern
Joined: 10 Dec 2013
Posts: 20
Location: India
Concentration: Technology, Strategy
GPA: 3.9
WE: Consulting (Consulting)

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?



Director
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 841
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)

Re: In a room filled with 7 people, 4 people have exactly 1
[#permalink]
Show Tags
19 May 2014, 00:49
\(2*\frac{1}{7}*\frac{5}{6}+2*\frac{1}{7}*\frac{5}{6}+3*\frac{1}{7}*\frac{4}{6}\) \(\frac{32}{42} = \frac{16}{21}\)
_________________
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".



Intern
Joined: 14 Feb 2014
Posts: 7
GMAT 1: 550 Q45 V21 GMAT 2: 690 Q49 V35

Re: In a room filled with 7 people, 4 people have exactly 1
[#permalink]
Show Tags
08 Oct 2014, 04:16
Can anyone please help me on where I went wrong?
Attachments
File comment: Can anyone please help me on where I went wrong?
probability.JPG [ 40.76 KiB  Viewed 4934 times ]



Math Expert
Joined: 02 Sep 2009
Posts: 49320

In a room filled with 7 people, 4 people have exactly 1
[#permalink]
Show Tags
08 Oct 2014, 04:43



Manager
Status: A mind once opened never loses..!
Joined: 05 Mar 2015
Posts: 214
Location: India
MISSION : 800
WE: Design (Manufacturing)

Re: In a room filled with 7 people, 4 people have exactly 1
[#permalink]
Show Tags
30 Jun 2015, 00:43
Hi everyone I was stuck with this question b/w the no of pairs of siblings and the answer I thought i should post this. May be it helps someone out dr. Happy learning
Attachments
sib.jpg [ 149.93 KiB  Viewed 4164 times ]
_________________
Thank you
+KUDOS
> I CAN, I WILL <



Senior Manager
Status: Math is psychological
Joined: 07 Apr 2014
Posts: 421
Location: Netherlands
GMAT Date: 02112015
WE: Psychology and Counseling (Other)

In a room filled with 7 people, 4 people have exactly 1
[#permalink]
Show Tags
08 Jul 2015, 10:00
I am posting a visual solution. To begin with, there are 7!/2!*5! = 21 total possibile pairings. After finding the nosibling pairings (see image below) we end up with 16/21.
Attachments
siblings.png [ 26.7 KiB  Viewed 3185 times ]



Manager
Joined: 10 Mar 2013
Posts: 223
GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39

Re: In a room filled with 7 people, 4 people have exactly 1
[#permalink]
Show Tags
10 Jul 2015, 20:46
x = P(1sibling person) = 4*p(1sibling person) y = P(2sibling person) = 3*p(2sibling person)
p(1sibling person)=1/7*1/6 (1/6 because we have 1 sibling in the 6 remaining people) p(2sibling person)=1/7*2/6 (2/6 because we have 2 siblings in the 6 remaining people)
Answer = 1xy = 16/21
E




Re: In a room filled with 7 people, 4 people have exactly 1 &nbs
[#permalink]
10 Jul 2015, 20:46



Go to page
Previous
1 2 3
Next
[ 46 posts ]



