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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]
03 Nov 2012, 16:36
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]
13 Dec 2012, 14:08
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. _________________
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Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]
15 May 2013, 00:26
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]
15 May 2013, 00:33
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]
14 Jan 2014, 20:53
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
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.
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]
29 Jan 2014, 10:33
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".
Re: In a room filled with 7 people, 4 people have exactly 1 [#permalink]
10 Jul 2015, 19:46
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)
In a room filled with 7 people, 4 people have exactly 1 [#permalink]
11 Dec 2015, 02:50
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.
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