Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 12 Oct 2008
Posts: 337

In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
Updated on: 30 Jun 2019, 10:07
Question Stats:
45% (02:33) correct 55% (02:27) wrong based on 1733 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? A. 5/21 B. 3/7 C. 4/7 D. 5/7 E. 16/21
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by reply2spg on 01 Nov 2009, 10:23.
Last edited by Bunuel on 30 Jun 2019, 10:07, edited 2 times in total.
Renamed the topic.




Math Expert
Joined: 02 Sep 2009
Posts: 60610

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
01 Nov 2009, 10:51
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.
_________________




VP
Joined: 05 Mar 2008
Posts: 1333

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
01 Nov 2009, 19:28
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




CEO
Joined: 17 Nov 2007
Posts: 2966
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
02 Dec 2009, 21:27
For me it is pretty straightforward: two mutually exclusive events: 1) we have 4/7 probability of getting 1sibling person and 5/6 probability of not getting his or her sibling. 2) we have 3/7 probability of getting 2siblings person and 4/6 probability of not getting his or her sibling. p=4/7*5/6+3/7*4/6 = 16/21 maybe this will be helpful:  Probability
_________________
HOT! GMAT Club Forum 2020  GMAT ToolKit 2 (iOS)  The OFFICIAL GMAT CLUB PREP APPs, musthave apps especially if you aim at 700+



Kaplan GMAT Instructor
Joined: 21 Jun 2010
Posts: 65
Location: Toronto

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
05 Jul 2010, 14:52
Four people have exactly one sibling and three people have exactly two siblings.
Think about how this can be done...the only way to organize is this is to have two pairs of siblings and one trio of siblings. In other words:
AB (one sibling pair) CD (another sibling pair) EFG (a trio of siblings)
This way each of A and B are siblings, each of C and D are siblings, so we have four people each with exactly one sibling. And each of E, F, and G are siblings so we have three people each of whom has exactly two siblings.
Because this is a "NOT" probability question, we should see how many ways we CAN select two individuals who are siblings.
Well, we can select the AB pair or the CD pair. So far, that's 2 ways. We can also select any two people from the EFG trio, so that's another 3C2 or 3 ways.
So, there are a total of 2+3 = 5 ways of pulling out 2 individuals who are siblings.
Probability = (#desired)/(#total)
The denominator of the formula is just all the ways we can select any two people from the seven. So we have:
Probability of selecting 2 people who are siblings = 5/7C2 = 5/21
Therefore, the probability of selecting 2 people who are NOT siblings is:
1  5/21 = 16/21
Choose E.



Math Expert
Joined: 02 Sep 2009
Posts: 60610

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
13 Sep 2010, 19:57
harithakishore wrote: total possibilities 2C7=21 possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5 probability 15/21
vittar..can you please explain 2C3..... 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). 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 567, \({C^2_2}\) is the # of ways to choose 2 siblings out of siblings 12, and \(C^2_2\) is the # of ways to choose 2 siblings out of siblings 34; \(P=1\frac{5}{21}=\frac{16}{21}\). You can check other approaches in my first post. Hope it's clear.
_________________



Director
Joined: 17 Dec 2012
Posts: 622
Location: India

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#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 Magical Logicians https://magicallogicianssatgmatgre.business.siteHolistic and Holy Approach



Intern
Affiliations: ACCA
Joined: 17 Apr 2010
Posts: 30
Schools: IMD, Insead, LBS, IE, Cambridge, Oxford

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
11 Aug 2010, 01:59
total possibilities 2C7=21 possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5 probability 15/21



Intern
Joined: 17 Jun 2010
Posts: 16

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
05 Jul 2010, 17:42
That is one way, I think it is easier to calculate the probabilities than the combinations.
there is 4/7 chance of selecting a person with one sibling initially with a 5/6 probability the second person will not be a sibling.
There is a 3/7 Chance of selecting a person with 2 siblings initially with 4/6 probability the second person will not be a sibling.
Simply adding these probabilities together gives you the answer.
4/7 * 5/6 + 3/7 * 4/6 = 20/42 + 12/42 = 32/42 = 16/21



Manager
Joined: 20 Mar 2010
Posts: 64

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
11 Aug 2010, 00:06
Jinglander wrote: 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



Math Expert
Joined: 02 Sep 2009
Posts: 60610

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
14 Jun 2012, 00:28
Rice wrote: Bunuel wrote: alchemist009 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
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}. \(P=2*\frac{3}{7}*\frac{4}{6}+2*\frac{2}{7}*\frac{2}{6}=\frac{4}{7}+\frac{4}{21}=\frac{16}{21}\). 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#p645861Hope it's clear.
_________________



Senior Manager
Joined: 13 Aug 2012
Posts: 394
Concentration: Marketing, Finance
GPA: 3.23

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



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 9095
Location: United States (CA)

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
18 Sep 2019, 09:54
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
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



Intern
Joined: 02 Dec 2009
Posts: 6

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
02 Dec 2009, 22:50
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 : 15/1 =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 2sibling group and 1 person from 1sibling group, which is the same with your second event: we have 3/7 probability of getting 2siblings person and 4/6 probability of not getting his or her sibling. 2) No one from the 2sibling group. we have 4/7 probability of getting 1sibling person and 2/6 probability of not getting his or her sibling and not within one of 2sibling 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.



Senior Manager
Joined: 23 Oct 2010
Posts: 317
Location: Azerbaijan
Concentration: Finance

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
01 Apr 2012, 01:16
u have 2 groups (a b x y) and (c d e) note, that a=b (siblings) c=d=e (siblings) to find no two identical siblings, u need to find 2 no siblings in group 1 , or 1 sibling from group 1 and 1 sibling from group 2 1.to find 2 no siblings in group 1 (a b x y) u have 4 such combinations  (a;x) (a;y) (b;x) (b;y) 2. 1 sibling from group 1 and 1 sibling from group 2  4C1*3C1=4*3=12 (4+12)/7C2=16/21 hope it helps,but if u have any question, please, ask.
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth



Math Expert
Joined: 02 Sep 2009
Posts: 60610

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
01 Jul 2012, 03:15
imhimanshu wrote: Hi Bunuel, 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 1i.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 2Similarly, 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 3Last 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 (12; 34). As there are 3 people with exactly 2 siblings each: we have one triple of siblings (456).
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 doublecount some cases. Hope it's clear.
_________________



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 sibling in the
[#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: 60610

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
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.
_________________



Manager
Joined: 31 Dec 2016
Posts: 65

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
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.



GMAT Club Legend
Joined: 12 Sep 2015
Posts: 4224
Location: Canada

Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
Show Tags
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
_________________
Test confidently with gmatprepnow.com




Re: In a room filled with 7 people, 4 people have exactly 1 sibling in the
[#permalink]
13 May 2019, 11:36



Go to page
1 2 3
Next
[ 44 posts ]



