February 23, 2019 February 23, 2019 07:00 AM PST 09:00 AM PST Learn reading strategies that can help even nonvoracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT February 24, 2019 February 24, 2019 07:00 AM PST 09:00 AM PST Get personalized insights on how to achieve your Target Quant Score.
Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 12 Oct 2008
Posts: 443

In a room filled with 7 people, 4 people have exactly 1
[#permalink]
Show Tags
Updated on: 14 Feb 2012, 21:48
Question Stats:
45% (02:32) correct 55% (02:28) wrong based on 1291 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, 09:23.
Last edited by Bunuel on 14 Feb 2012, 21:48, edited 1 time in total.
Edited the question and added the OA




Math Expert
Joined: 02 Sep 2009
Posts: 53066

In a room filled with 7 people, 4 people have exactly 1
[#permalink]
Show Tags
01 Nov 2009, 09:51
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.
_________________
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




VP
Joined: 05 Mar 2008
Posts: 1399

Re: Probability
[#permalink]
Show Tags
01 Nov 2009, 18:28
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: 3420
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: MGMAT CAT  Picking Friends
[#permalink]
Show Tags
02 Dec 2009, 20:27
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 TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  Limited GMAT/GRE Math tutoring in Chicago



Intern
Joined: 02 Dec 2009
Posts: 6

Re: MGMAT CAT  Picking Friends
[#permalink]
Show Tags
02 Dec 2009, 21:50
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.



CEO
Joined: 17 Nov 2007
Posts: 3420
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

Re: MGMAT CAT  Picking Friends
[#permalink]
Show Tags
03 Dec 2009, 04:58
Fiven wrote: @ 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. In your approach you need to add one more event: 3) 1 person from 1sibling group and next person from 2sibling group: 4/7*3/6 p = 20/42 + 12/42 = 16/21 Actually, you have three combinations: 1) 21 2) 11 3) 12.
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android)  The OFFICIAL GMAT CLUB PREP APP, a musthave app especially if you aim at 700+  Limited GMAT/GRE Math tutoring in Chicago



Kaplan GMAT Instructor
Joined: 21 Jun 2010
Posts: 70
Location: Toronto

Re: PS probability 700800
[#permalink]
Show Tags
05 Jul 2010, 13:52
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.



Intern
Joined: 17 Jun 2010
Posts: 16

Re: PS probability 700800
[#permalink]
Show Tags
05 Jul 2010, 16:42
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: 77

Re: Sibling Problem
[#permalink]
Show Tags
10 Aug 2010, 23:06
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
_________________
___________________________________ Please give me kudos if you like my post



Intern
Affiliations: ACCA
Joined: 17 Apr 2010
Posts: 32
Schools: IMD, Insead, LBS, IE, Cambridge, Oxford

Re: Sibling Problem
[#permalink]
Show Tags
11 Aug 2010, 00:59
total possibilities 2C7=21 possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5 probability 15/21



Intern
Joined: 04 Aug 2010
Posts: 21

Re: Probability
[#permalink]
Show Tags
13 Sep 2010, 18:36
total possibilities 2C7=21 possibility that selected siblings = 2C2 + 2C2 + 2C3 = 1 + 1 + 3 =5 probability 15/21
vittar..can you please explain 2C3.....



Math Expert
Joined: 02 Sep 2009
Posts: 53066

Re: Probability
[#permalink]
Show Tags
13 Sep 2010, 18:57
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.
_________________
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: 04 Aug 2010
Posts: 21

Re: Probability
[#permalink]
Show Tags
13 Sep 2010, 18:58
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
There are suppose A B C D E F G members in the room 4 people who have exactly one sibling....A B C D....(A is B`s sibling and viceversa) (C is D`s sibling and viceversa)...now remaning EFG are 3 people who have exactly 2 siblings....(E has F and G as his/her sibling and so on..) there are now 3 different set of siblings (A and B) (C and D);(EFG) Now first selecting 2 people out of 7 is 7C2=21 first sibling pair(A and B)selecting 2 people 2C2=1 second sibling pair (C and D)selecting 2 people2C2=1 third sibling pair (E F G)selecting 2 out of 3 3C2=3
total= 1+1+3=5 but,a/c to formula P(success)1p(fail) here,p(failure)is selecting 2 people who are siblings =5/21(21 is 7C2) =15/21 =16/21 ANS E



Manager
Joined: 08 Oct 2010
Posts: 197
Location: Uzbekistan
Schools: Johnson, Fuqua, Simon, Mendoza
WE 3: 10

Re: Probability
[#permalink]
Show Tags
16 Dec 2010, 05:54
first, find probability of getting two pairs and triplet 4/7*1/6=4/42 and 3/7*2/6=6/42 second, add these two results 4/42 + 6/42 = 10/42 = 5/21 third, subtract the result of addition from 1 1 – 5/21 = 16/21



Director
Joined: 01 Feb 2011
Posts: 654

Re: PSProbability
[#permalink]
Show Tags
14 Aug 2011, 16:04
there are 7 people in the room.
there are 4 ppl with exactly one sibling. lets consider the following 1,3  group1 2,4  group 2 there are 3 people with exactly two siblings 5,6,7  group 3
so the probability of picking 2 non siblings is nothing but picking each from different groups.
= (2c1*2c1 + 2c1*3c1 +3c1*2c1)/7c2 = 16/21
Answer is E.



Senior Manager
Joined: 23 Oct 2010
Posts: 348
Location: Azerbaijan
Concentration: Finance

Re: Toughest MGMAT Math Problems (2/20)
[#permalink]
Show Tags
01 Apr 2012, 00:16
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



Intern
Joined: 08 Oct 2007
Posts: 31

Re: sibling
[#permalink]
Show Tags
13 Jun 2012, 22:25
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.



Math Expert
Joined: 02 Sep 2009
Posts: 53066

Re: sibling
[#permalink]
Show Tags
13 Jun 2012, 23:28
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.
_________________
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



Senior Manager
Joined: 07 Sep 2010
Posts: 258

Re: Probability
[#permalink]
Show Tags
30 Jun 2012, 19:51
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.



Math Expert
Joined: 02 Sep 2009
Posts: 53066

Re: Probability
[#permalink]
Show Tags
01 Jul 2012, 02:15
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.
_________________
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







Go to page
1 2 3
Next
[ 48 posts ]



