GMAT Club Forum https://gmatclub.com:443/forum/ 

In a room filled with 7 people, 4 people have exactly 1 sibling in the https://gmatclub.com/forum/inaroomfilledwith7people4peoplehaveexactly1siblinginthe8755020.html 
Page 2 of 2 
Author:  Bunuel [ 08 Oct 2014, 03:43 ] 
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the 
keyrun wrote: Can anyone please help me on where I went wrong? 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. 
Author:  BrentGMATPrepNow [ 13 May 2019, 10:36 ] 
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the 
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 First we need to recognize that the given information tells us that the 7 people consist of:  a sibling trio  a sibling pair  and another sibling pair Let's use counting techniques to answer this question For this question, it's easier to find the complement. So P(not siblings) = 1  P(they are siblings) P(they are siblings) = [# of ways to select 2 siblings] / [total # of ways to select 2 people] # of ways to select 2 siblings Case a) 2 siblings from the sibling trio: from these 3 siblings, we can select 2 siblings in 3C2 ways (3 ways) Case b) 2 siblings from first sibling pair: we can select 2 siblings in 2C2 ways (1 way) Case c) 2 siblings from second sibling pair: we can select 2 siblings in 2C2 ways (1 way) So, total number of ways to select 2 siblings = 3+1+1 = 5 total # of ways to select 2 people We have 7 people and we want to select 2 of them We can accomplish this in 7C2 ways (21 ways) So, P(they are siblings) = 5/21 This means P(not siblings) = 1  5/21 = 16/21 Answer: E Cheers, Brent 
Author:  ScottTargetTestPrep [ 18 Sep 2019, 08:54 ] 
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the 
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 Let A, B, C, D, E, F, and G be the 7 people in the room. To satisfy the condition that 4 people have exactly 1 sibling and 3 people have exactly 2 siblings, we can let A and B be siblings (but not to other people), C and D be siblings (but not to other people), and E, F and G are siblings (but not to other people). Let’s consider the probability of how each person is chosen: If A is chosen first, then B can’t be chosen. So the probability is: 1/7 x 5/6 = 5/42 This probability will be the same if B, C, or D is chosen first. If E is chosen first, then neither F nor G can be chosen. So the probability is: 1/7 x 4/6 = 4/42 This probability will be the same if F or G is chosen first. Therefore, the overall probability is: 5/42 x 4 + 4/42 x 3 = 20/42 + 12/42 = 32/42 = 16/21 Alternate Solution: Notice that 2 people can be chosen out of 7 people in 7C2 = 7!/(5!*2!) = (7 x 6)/2 = 21 ways. With A, B, C, D, E, F, and G as above, we see that there are 5 ways to choose a sibling pair: AB, CD, EF, EG and FG. Thus, 21  5 = 16 choices of do not include a sibling pair. Therefore, the probability that the chosen two people are not siblings is 16/21. Answer: E 
Author:  BrentGMATPrepNow [ 09 Dec 2020, 06:12 ] 
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the 
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 First we need to recognize that the given information tells us that the 7 people consist of:  a sibling trio, which we'll call "the TRIPLETS"  a sibling pair, which we'll call "the TWINS"  and another sibling pair, which we'll also call "the TWINS" So, P(two are NOT siblings) = P(1st choice is a triplet AND 2nd choice is not a sibling OR 1st choice is a twin AND 2nd choice is not a sibling) = P(1st choice is a triplet) x P(2nd choice is not a sibling) + P(1st choice is a twin) x P(2nd choice is not a sibling) = (3/7 x 4/6) + (4/7 x 5/6) = 12/42 + 20/42 = 32/42 = 16/21 Answer: E 
Author:  Bambi2021 [ 05 Apr 2021, 01:38 ] 
Post subject:  In a room filled with 7 people, 4 people have exactly 1 sibling in the 
Three groups can be combined in pairs of two groups A, B, C in AB + AC + BC = 3 ways where A = 2 people B = 2 people C = 3 people AB = 2C1*2C1 = 4 different ways to chose two people AC = 2C1*3C1 = 6 different ways BC = 2C1*3C1 = 6 different ways Total number of ways to chose any two people out of seven = 7C2 = 21 ways 16/21 
Author:  rvgmat12 [ 04 Sep 2021, 06:17 ] 
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the 
3 people in one group (each has two siblings) 4 people ( 2 groups of siblings ) Total combinations of 2 people selected not being siblings = 1  (3C2+2)/7C2 = 15/21 = 16/21 E 
Author:  M838TE [ 31 Mar 2022, 11:35 ] 
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the 
IanStewart Hi, ian. Would you mind explaining approach 3 that bunuel accounted for? I am not sure if I understand the multiplication by 2 for the 2sibling group. Does order matter here? Not sure why picking 12, 34 then the 567 are different than picking 567 first, then [12], [34]. Understanding the concept by picking one person at a time was easier for me: 1/7 to pick any*5/6 because one person is picked and the other person in the group is a sibling that is picked up. The numerator is the total ways of picking a nonsibling pair, order does not matter and position does not matter, then you multiply by 4 because there are 4 diff ppl with 1sibling to pick from. Posted from my mobile device 
Author:  IanStewart [ 31 Mar 2022, 11:53 ] 
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the 
M838TE wrote: Bunuel explained that solution a bit further down on page one of this thread. He first divided the seven people into sibling groups, which is a complication there's no need to think about here. We have a 4/7 probability of picking someone with one sibling first, then a 1/6 chance of picking their only sibling second. We have a 3/7 probability of picking someone with two siblings first, then a 2/6 chance of picking one of their two siblings second. So the probability we get a pair of siblings is (4/7)(1/6) + (3/7)(2/6) = 5/21, and the probability we don't is 1  (5/21) = 16/21. You could also use this approach, but solve directly, by finding the probability right away that you do not pick a sibling on the second selection (then you get (4/7)(5/6) + (3/7)(4/6) = 16/21 and do not need to subtract the result from 1 at the end). Bunuel's method #3 does essentially that, though it considers more cases, because it treats each sibling group separately. 
Author:  M838TE [ 31 Mar 2022, 14:41 ] 
Post subject:  In a room filled with 7 people, 4 people have exactly 1 sibling in the 
IanStewart wrote: M838TE wrote: Bunuel explained that solution a bit further down on page one of this thread. He first divided the seven people into sibling groups, which is a complication there's no need to think about here. We have a 4/7 probability of picking someone with one sibling first, then a 1/6 chance of picking their only sibling second. We have a 3/7 probability of picking someone with two siblings first, then a 2/6 chance of picking one of their two siblings second. So the probability we get a pair of siblings is (4/7)(1/6) + (3/7)(2/6) = 5/21, and the probability we don't is 1  (5/21) = 16/21. You could also use this approach, but solve directly, by finding the probability right away that you do not pick a sibling on the second selection (then you get (4/7)(5/6) + (3/7)(4/6) = 16/21 and do not need to subtract the result from 1 at the end). Bunuel's method #3 does essentially that, though it considers more cases, because it treats each sibling group separately. I actually saw his explanation. But I was confused by the part where he multiply by2. specifically this part, " 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};" Posted from my mobile device 
Author:  JJDa [ 19 Jul 2022, 15:11 ] 
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the 
I found this nice explanation on youtube https://www.youtube.com/watch?v=Lefh_Ywk3qw 
Author:  ThatDudeKnows [ 20 Jul 2022, 06:05 ] 
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the 
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 We have three groups: AA, BB, CCC We can "win" in two ways: 1. We select either A or B and then select someone other than their sibling. \(\frac{4}{7}*\frac{5}{6} = \frac{20}{42}\) OR 2. We select C and then someone other than their sibling. \(\frac{3}{7}*\frac{4}{6} = \frac{12}{42}\) \(\frac{20}{42}+\frac{12}{42} = \frac{32}{42} = \frac{16}{21}\) Answer choice E. We could also have found the probability of "losing" and subtracted from 1. 1. We select either A or B and then select their sibling. \(\frac{4}{7}*\frac{1}{6} = \frac{4}{42}\) OR 2. We select C and then their sibling. \(\frac{3}{7}*\frac{2}{6} = \frac{6}{42}\) \(\frac{4}{42}+\frac{6}{42} = \frac{10}{42} = \frac{5}{21}\) \(1\frac{5}{21} = \frac{16}{21}\) Answer choice E. 
Author:  BIGDAMNGOD [ 01 Oct 2022, 08:43 ] 
Post subject:  Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the 
I think I have a conceptual gap here, Experts please help  My approach  Group A  2 siblings (4 people) Group B  3 siblings (3 people) 4C1* 5C1 = 20 3C1* 4C1 = 12 Total = 7C2 = 21 Why does the total selection 7C2 not work this way? 
Page 2 of 2  All times are UTC  8 hours 
Powered by phpBB © phpBB Group http://www.phpbb.com/ 