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A committee that includes 6 members is about to be divided
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A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member? A. 10% B. 20% C. 25% D. 40% E. 50%
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Originally posted by sudai on 21 Apr 2010, 01:10.
Last edited by reto on 02 Sep 2015, 04:51, edited 1 time in total.
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Re: A committee that includes 6 members is about to be divided
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21 Apr 2010, 10:25
Committee #1: 50% chance that David will be a member, leaving 2 possible spots the other 5 people. Michael's chances of being one of those 2 is 2/5 or 40%. Chances of David and Michael being on this committee is 50%x40%=20%.
Committee #2: Same.
20% + 20% = 40%.




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Re: A committee that includes 6 members is about to be divided
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21 Apr 2010, 23:38
sudai wrote: A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member? Here if we first calculate total number of subcommittes possible for 1st subcommittee  selection of any 3 out of 6 = 6C3 = 20 fOR 2nd sub committee  3 out of remaining 3 = 1 total subcommittees possible are 20+1 = 21 Now if we take michael and david as person (so both will be in one committee) total possible committee possible are 5C3 = 10 10 is definately not 40% of 21 Where am i going wrong? sometimes i really get confused !!!!
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Re: A committee that includes 6 members is about to be divided
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22 Apr 2010, 11:46
A / B <==> _ _ _ / _ _ _ where A and B are two committees. For A, 2 places are filled by David and Michel, third can be select in 4c1 ways. For B , 3 out of remaining three can be selected in 3c3 total required ways = 4c1*3c3*2 ( we have multiplied with 2 , as they can be part of either 2) To calculate total ways : For A , 3 places can be filled by 6c3 and remaining by 3c3 , here we wont multiply by 2, as there is no specification. for B, 3 out of remaining three can be selected in 3c3 Now probability = \(\frac{(total required ways)}{(total ways)}\) = \(\frac{4c1*3c3*2}{6c3*3c3}\) = \(\frac{4*2}{6c3}\) = \(\frac{4*2*3!*3!}{6!}\) = \(\frac{4*2*3!}{6*5*4}\) = \(\frac{2}{5}\) 2/5 = 2*100/5 % = 40%
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Re: A committee that includes 6 members is about to be divided
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23 Apr 2010, 00:32
i got it now, even if we assume both david and michael as one person we are left with 4 and now we have to choose 1 out of 3 and for 2nd committee remaining 3 out of 3. Because david and michael can be in any of the committe so multiply t by 2 As i said sometimes i get confused in simple solution. i should do quant with fresh mind not when i am tired
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Re: A committee that includes 6 members is about to be divided
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10 Sep 2012, 00:51
sudai wrote: A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member? I found another method more helpful and understandable but missing from this topic hence contributing . Hope it helps total number of 3 person , 2 sub committees \(\frac{C^6_3 * C^3_3}{2!}\) = 10 ( we divide by 2! as the two subcommittees order is not important we do not have \(1^{st}\) subcommittee and \(2^{nd}\) sub committee. Bunuel can enlighten further for beginners in this area ). now suppose M and D are members of a sub committee then the remaining person can be chosen in \(c^4_1\) way , so \(C^1_1 * C^1_1 *C^4_1\)= 4 total ways a sub committee can form with M and D as members hence percentage \(\frac{4}{10} *100\) = 40%
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Re: A committee that includes 6 members is about to be divided
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10 Sep 2012, 05:54
sudai wrote: A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member? Count how many subcommittees Michael is a member of: M * *  means Michael plus two other persons; for the two stars we can choose from 5 members, therefore 5C2=5*4/2=10 different subcommittees. Count how many subcommittees both Michael and David are members of: M D *  means both Michael and David are in the same subcommittee, plus one extra person: for the star we can choose from the 4 remaining members, therefore 4 different subcommittees. This gives 4/10 = 40%.
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Re: A committee that includes 6 members is about to be divided
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10 Sep 2012, 06:09
This is equivalent to: We have 6 chairs divided in 2 Groups: A and B each with 3 chairs. If Michael sits on the first chair in the group A, what is the posibility that David sits on any of another 2 available chairs in Group A?
So: A: M _ _ / B: _ _ _ There are 2 chairs in Group A for favorable outcomes and 5 available chairs as total possible outcomes. The answer is 2/5



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Re: A committee that includes 6 members is about to be divided
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10 Sep 2012, 06:52
sstasic wrote: This is equivalent to: We have 6 chairs divided in 2 Groups: A and B each with 3 chairs. If Michael sits on the first chair in the group A, what is the posibility that David sits on any of another 2 available chairs in Group A?
So: A: M _ _ / B: _ _ _ There are 2 chairs in Group A for favorable outcomes and 5 available chairs as total possible outcomes. The answer is 2/5 That's an excellent reinterpretation of the question! How did you come up with it?
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Re: A committee that includes 6 members is about to be divided
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10 Sep 2012, 07:37
sudai wrote: A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member? There are 6 members, whom we need to divide into 2 sub committees. So, in each committee we'll have 3 members. If we form committees with the help of combination, we'll get \(^6C 3\)\(= 20\) number of committees. Now, let's see how we'll get 10 numbers of unique committees assume we have following members. a, b, c, x, y, and z. The one of committees will be formed by choosing a, b and c. Naturally x,y and z are left out. Similarly, when we choose x,y and z to form a committee a,b and c are left out. This means unique number of committees \(= 20/2\)\(=10\) Now, let's find the committees in which M and D are members If M and D are already selected we need to select one more member out of 4 members. This means we can select another member in 4 ways. So, % of committees in which both M and D are present \(= 4/10\) \(= 40 %\)
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Re: A committee that includes 6 members is about to be divided
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11 Sep 2012, 03:13
sudai wrote: A committee that includes 6 members is about to be divided into 2 subcommittees with 3 members each. On what percent of the possible subcommittees that Michael is a member of is David also a member? Similar question to practice: anthonyandmichaelsitonthesixmemberboardofdirectors102027.htmlHope it helps.
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Re: A committee that includes 6 members is about to be divided
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25 Sep 2012, 06:04
Total number of Members 6 Size of the new sub committees. 3 x2 ( two sub committees)  Let these be Committee A and B Chances of Micheal being in Comm. A = 1/2 Total no of spots left in comm A = 2 , Members that can possibly take these spots = 5 ( Micheal ie. the sixth member has all ready taken his spot in comm. A ) therefore David has 2/5 chance of being in committee A ... Multiplying the two (to figure out the chances of both of them being in the same committee at the same time) we get ( 2/5 ) x 1/2 = 1/5 or 20 % ... Therefore , Chances that both Micheal and David will be together in the FIRST committee are 20% . They have equal chance ie. 20% of being together in Sub Comm. B (instead of A) therefore the Answer is 20x2 = 40 % ....
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Re: A committee that includes 6 members is about to be divided
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10 Oct 2012, 04:51
In school i was taught to resolve problems in the form of A+B=?. It become much more interesting when i found in real life practical uses for theoretical concepts (especially through computer programming of industrial machinery to obtain a specific real product). The Gmat approach of math is similar to real life one: it takes a life experience and expects you to resolve it. It's like driving a car and somebody asks how much more will we be driving. There is no given equation. To give an answer you first try to find what elements you need: you look at your driving speed and try to get an idea of the distance left. You might have a GPS and get the distance there, you might know the total trip length and setted a trip counter on your board when you left your home. You take the easiest information you have. Maybe you wait a few moments to get in front of a board with distance info and you might have there your destination. Then you use theory concept that Time=Distance / Speed, use the actual numbers and solve the question. Gmat is placing a questions and lets you choose from theoretical concepts you need to answer it. Since every math theorem is demonstrated using simpler previous theorems, it means that you can solve a problem in the most rudimental way (by counting each possible outcome in a probability question) or use derivative concepts for general outcome (combinations for example). There is a multitude of ways to resolve a problem; some of them are faster than others. Sometime counting each possible outcome is faster than using general formulas. The idea is to see what the equation is and if it can be simplified. Maybe 1 (probability for wrong outcomes) is easier to calculate in probability question. When you take the approach of calculating the opposite numbers, you are actually rephrasing the question. The equation would be the same for various other life examples, so try to rephrase it. Take if the current question: ”On what percent of the possible subcommittees that Michael is a member of is David also a member?”. The part with “on what percent of the possible subcommittees” is equivalent to “what is the probability”. The other part just asks David to be in the same Committee with Michael. So: “what is the probability that Michal is in the same committee as David”. The outcome will be the same if you change the layout of the problem. Let’s say that there is a club with 6 members and there should be formed a committee of 3 members: president, treasurer and secretary. Michael is president. What is the probability that David would be in the committee (either as secretary or treasurer)? The information that the rest of the members form or do not form a committee (for a given project for example) is useless. The actual question comes to the probability of occupying one of the 2 given vacant positions in the committee by a specific member from other 5 members of the club! :D (< this is the easier form of the problem I can get to). EvaJager wrote: sstasic wrote: This is equivalent to: ... That's an excellent reinterpretation of the question! How did you come up with it?



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Re: A committee that includes 6 members is about to be divided
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25 Oct 2012, 21:52
Let the committee members be A,B,C,D,E and F, where A and B corresponds to Michael and David. The possible combinations in which 1st three committee member can be formed are  ABC ABD ABE ABF ACD ACE ACF ADE ADF AEF
Rest of the combinations would form a pair to one of the above mentioned combination. i.e, they would form the 2nd committee of one of the above.
Total combinations in which A and B are together = 4 Total number of combinations = 10
So percentage in which A and B are together = 4/10 = 2/5.
May be this method takes a bit longer, but I guess its still helpful in understanding the concept and will avoid confusion.



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Re: A committee that includes 6 members is about to be divided
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28 Oct 2012, 05:39
gurpreetsingh wrote: A / B <==> _ _ _ / _ _ _ where A and B are two committees.
For A, 2 places are filled by David and Michel, third can be select in 4c1 ways.
For B , 3 out of remaining three can be selected in 3c3
total required ways = 4c1*3c3*2 ( we have multiplied with 2 , as they can be part of either 2)
To calculate total ways :
For A , 3 places can be filled by 6c3 and remaining by 3c3 , here we wont multiply by 2, as there is no specification.
for B, 3 out of remaining three can be selected in 3c3
Now probability = \(\frac{(total required ways)}{(total ways)}\)
= \(\frac{4c1*3c3*2}{6c3*3c3}\) = \(\frac{4*2}{6c3}\) = \(\frac{4*2*3!*3!}{6!}\) = \(\frac{4*2*3!}{6*5*4}\) = \(\frac{2}{5}\)
2/5 = 2*100/5 % = 40% Hey Gurpreet, Can you please have a look at the below question. The answer should be 3/5 according to your approach. acommitteeof3hastobeformedrandomlyfromagroupof81051.htmlOnly thing I am confused about is, if we actually need to multiply the numerator by 2. In this question you have done that, if that is correct way of doing it, why is that multiplication is not required in the below question. Bunnel, can you please help too? Thanks
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Re: A committee that includes 6 members is about to be divided
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28 Oct 2012, 07:13
hermit84 wrote: Hey Gurpreet, Can you please have a look at the below question. The answer should be 3/5 according to your approach. acommitteeof3hastobeformedrandomlyfromagroupof81051.htmlOnly thing I am confused about is, if we actually need to multiply the numerator by 2. In this question you have done that, if that is correct way of doing it, why is that multiplication is not required in the below question. Bunnel, can you please help too? Thanks For my solution I assumed A and B committee to be specific. So I multiplied by 2. If you do not want to do so..you can leave it to 4c3 but while calculating you have to divide 6c3 by 2! to remove the duplicate values. again the value would be same..as we will divide by 2! only when we do not care about the committee being specific.
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Re: A committee that includes 6 members is about to be divided
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30 Oct 2012, 14:48
Just answered a similar question:
Here's some clarification for those who are still confused.
First group of 6
     
Gets broken into 2 subcommittees of 3 each:
     
We know Michael is already in one of them:
M     
We just need to fill in one of those two slots next to Michael with an "A" for Anthony.
So we already know Mike is in that slot and that there are 5 remaining choices.
Well, how many ways can we pick Anthony such that he ends up in Michael's group?
Keep in mind that order matters so we can multiply a line of nCr formulas:
[ (Out of 1 available Anthony, pick that 1 Anthony for that 2nd spot) * (Out of the remaining 4, choose any 1 for that 3rd spot) ] =  (Out of the initial 5 remaining people, choose 2 to fill up the 2nd and 3rd slots)
= [ (1C1) * (4C1) ] / (5C2)
= 4 / 10
= 40%
Hope that helps.



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Re: A committee that includes 6 members is about to be divided
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07 Sep 2015, 12:33
IMO, this is not a gmat type question, at least from wording perspective... It's not clear whether the M&D among 6. It was suppose to define that unambiguously. I thought that both are among 6, and if one is in subcategory, the chance would be in the same 50%.
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Re: A committee that includes 6 members is about to be divided
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19 Jan 2016, 11:16
If first team has Michael and David, there will be 4 ways to arrange that team. M, D, _ . (Last member can be one of the remaining four). Second team will be arranged only in one way, order does not matter. So 4 * 1 = 4 ways for first team having M and D.
Repeat for second team = 4 ways.
Total number of possibilities: 6C3 = 20.
So, 8 possible ways out of 20 = 40 percent.




Re: A committee that includes 6 members is about to be divided
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