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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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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.



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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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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
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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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
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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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.
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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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.
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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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?
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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19 May 2014, 00:49
\(2*\frac{1}{7}*\frac{5}{6}+2*\frac{1}{7}*\frac{5}{6}+3*\frac{1}{7}*\frac{4}{6}\) \(\frac{32}{42} = \frac{16}{21}\)
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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08 Oct 2014, 04:16
Can anyone please help me on where I went wrong?
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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08 Oct 2014, 04:43
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.
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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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
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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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.
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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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



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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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11 Dec 2015, 03:50
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 This is how I solved. Case 1: The group of 4 have exactly 1 sibling. So maximum 2 people can be picked from that group who are not siblings. Case 2: The group of 3 has 2 siblings each. So maximum only 1 person from the group can be chosen. Total number of ways in which we can pick people is 7C2 = 21. Case 1  \(2C1 * 2C1\) ways Case 2  \(3C1 * 4C1\) ways Probability  \(\frac{2C1 * 2C1 + 4C1 * 3C1}{21} = \frac{16}{21}\) Experts, please confirm if my approach is correct +Kudos if this helped!
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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12 Aug 2017, 11:49
lagomez wrote: You have people 1234567
4 have exactly 1 sibling which can mean: 12 are siblings 34 are siblings
3 have exactly 2 siblings which can mean: 567 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 1234 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 nonsibling pair is 1234 Therefore the probability is 4/42. Multiply that probability by 3, which represent 567 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 It's easier to use the complement. Not sure why people are trying so hard here to use the choose formula when it makes the question 1000X harder 4/7 chance to pick a person with 1 sibling * 1/6 chance you get a sibling or 4/42 3/7 change to pick a 2 person sibling * 2/6 or 6/42 Total of 10/42 of getting a sibling. We get to use the simple or rule not the generalized since the things are mutually exclusive. And we aren't drawing twice just once. So either a 4 person is picked first or 3 person. So that simplifies to 5/21 then 1 minus this is 16/21 for the odds that they are not siblings. 5/21 they are.



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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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11 Aug 2018, 05:28
I have used alphabet analogy to solve this Q.
Consider three groups, each containing same alphabets.
Group 1: AA Group 2: BB Group 3: CCC
We need to select 2 distinct alphabets.
TOTAL OUTCOMES: ways of selecting 2 alphabets out of 7 without any restriction is 7C2= 21
FAVORABLE OUTCOMES: based on restriction of selecting distinct alphabets, we have following possibilities:
1 each from G1&G2: 2C1 x 2C1= 4 1 each from G1&G3: 2C1 x 3C1= 6 1 each from G2&G3: 2C1 x 3C2= 6 Total favorable outcomes: 16
Probability: 16/21
Ans E
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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03 Oct 2018, 17:53
PiyushK wrote: \(2*\frac{1}{7}*\frac{5}{6}+2*\frac{1}{7}*\frac{5}{6}+3*\frac{1}{7}*\frac{4}{6}\) \(\frac{32}{42} = \frac{16}{21}\) this one made the most sense to me. We have two groups of ppl: A B C D E F G starting from A, chance of picking the first person: 1/7 after A, the chance of not picking a sibling: 5/6 Repeat this for B, C, and D 2*(1/7)(5/6) If we pick E F G first then, we know the first chance is 1/7, but second chance is 4/6 hence (1/7)(4/6) repeat this for E F G hence 3x so total should 2*(1/7)(5/6)+2*(1/7)(5/6)+3(1/7)(4/6)



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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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10 Feb 2019, 19:31
I think this question can be a lot easier if we realize that within the 4 people (A,B,C,D) who has exactly 1 sibling, there are two 'sets' of siblings  any selection two people within the group of 2 could be siblings with each other. The three people(E,F,G) with exactly 2 siblings, are siblings with each other.
So instead of breaking the 7 people into three groups, we can solve it by dividing the people into two groups, making up 3 cases.
Case 1: ABCD, Pick 2 4C2 = 6. But of the 6 potential sets of siblings, we already know that there are siblings  so subtract 2, we have 4.
Case 2: EFG, pick 2 3C2=3. Again, we already know that the three people are siblings with each other, so all three potential sets will be siblings  so subtract 2, we get 0 from this group.
Case 3: Pick one from ABCD, pick another from EFG. No restrictions here, people from 2 sibling group are not the same people from 1 sibling group. 4C1+3C1=12
Nonsibling selections: 4+0+12=16 Total selections: 7C2=21 Part/All = 16/21 Answer E



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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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13 May 2019, 11:36
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 siblingsCase 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 peopleWe 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) = 5/21This means P( not siblings) = 1  5/21= 16/21 Answer: E Cheers, Brent
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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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03 Jun 2019, 13:56
I calculated the probability they would be siblings:
(4/7 *1/6) + (3/7 * 2/6) = 10/42 = 5/21 and then i subtracted this from 1 to get 16/21.



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Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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30 Jun 2019, 07:27
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 (12; 34). As there are 3 people with exactly 2 siblings each: we have one triple of siblings (567).
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  \(C^2_3+C^2_2+C^2_2=3+1+1=5\);
\(P=1\frac{5}{21}=\frac{16}{21}\).
Solution #3: \(P=2*\frac{3}{7}*\frac{4}{6}+2*\frac{2}{7}*\frac{2}{6}=\frac{4}{7}+\frac{4}{21}=\frac{16}{21}\).
Answer: E. Bunuelin solution 1, why are we taking it to 3 sections, after combination from both the groups i.e \(C^1_2*C^1_2\) (one from first pair of siblings*one from second pair of siblings), why cant we select one from triple and then any one from the rest for as they wont be siblings to each other. So i took it like 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) + \(C^1_3*C^1_4\) (one from triple*one from rest) What is wrong in here?? I'm really confused.




Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
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