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Anthony and Michael sit on the sixmember board of directors for compa
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Updated on: 08 Aug 2019, 09:13
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Anthony and Michael sit on the six member board of directors for company X. If the board is to be split up into 2 threeperson subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony? (A) 20% (B) 30% (C) 40% (D) 50% (E) 60%
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Originally posted by kamilaak on 23 Jan 2008, 07:12.
Last edited by Bunuel on 08 Aug 2019, 09:13, edited 3 times in total.
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Re: Anthony and Michael sit on the sixmember board of directors for compa
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01 Oct 2010, 08:42



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Re: Anthony and Michael sit on the sixmember board of directors for compa
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05 Nov 2009, 07:20
Anthony and Michael sit on the sixmember board of directors for company X. If the board is to be split up into 2 threeperson subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony? A. 20% B. 30% C. 40% D. 50% E. 60% First approach: Let's take the group with Michael: there is a place for two other members and one of them should be taken by Anthony, as there are total of 5 people left, hence there is probability of \(\frac{2}{5}=40\%\). Second approach: Again in Michael's group 2 places are left, # of selections of 2 out of 5 is \(C^2_5=10\) = total # of outcomes. Select Anthony  \(C^1_1=1\), select any third member out of 4  \(C^1_4=4\), total # \(=C^1_1*C^1_4=4\)  total # of winning outcomes. \(P=\frac{# \ of \ winning \ outcomes}{total \ # \ of \ outcomes}=\frac{4}{10}=40\%\) Third approach: Michael's group: Select Anthony as a second member out of 5  1/5 and any other as a third one out of 4 left 4/4, total \(=\frac{1}{5}*\frac{4}{4}=\frac{1}{5}\); Select any member but Anthony as second member out of 5  4/5 and Anthony as a third out of 4 left 1/4, \(total=\frac{4}{5}*\frac{1}{4}=\frac{1}{5}\); \(Sum=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}=40\%\) Fourth approach: Total # of splitting group of 6 into two groups of 3: \(\frac{C^3_6*C^_3}{2!}=10\); # of groups with Michael and Anthony: \(C^1_1*C^1_1*C^1_4=4\). \(P=\frac{4}{10}=40\%\) Answer: C.Hope it helps.
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Re: Anthony and Michael sit on the sixmember board of directors for compa
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23 Jan 2008, 10:14
kazakhb wrote: maratikus wrote: The answer is C. Let's look at a committee where Mike belongs (he's person # 1) on that committee. When we choose person #2, there is a 1/5 probability that it's going to be Anthony (then it doesn't matter who the third person is), and 4/5 probability that it's going to be someone else (then person #3 is going to be Anthony with probability 1/4). Total probability = 1/5+4/5*1/4 = 2/5 Maratikus, why did you multiply 4/5 by 1/4? what does it mean? can you explain.... thanks in advance Let's say we have M (Mike), A (Anthony), B, C, D, E. we put together Mike's group. M is the first person. The second person we put in the group can either be A or someone else (B, C, D, E). Probability that it's going to be A is 1/5 and if that happens, we succeed because both M and A are in the same group. If it's not A (say it's B)  that happens with probability with 4/5, then we have M and B in the group and we have to pick M out of M, C, D, E (which happens with probability 1/4) to succeed. That's why total probability is 1/5 (which corresponds with A picked right away) + 4/5*1/4 (corresponds picking someone else first and then A) = 2/5 Does that help?




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Re: Anthony and Michael sit on the sixmember board of directors for compa
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23 Jan 2008, 08:19
The answer is C. Let's look at a committee where Mike belongs (he's person # 1) on that committee. When we choose person #2, there is a 1/5 probability that it's going to be Anthony (then it doesn't matter who the third person is), and 4/5 probability that it's going to be someone else (then person #3 is going to be Anthony with probability 1/4). Total probability = 1/5+4/5*1/4 = 2/5



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Re: Anthony and Michael sit on the sixmember board of directors for compa
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23 Jan 2008, 08:49
I think you missed something... \(p=\frac26*(\frac15*\frac11+\frac45*\frac14)+ \frac46*(\frac25*\frac14)=\frac{16}{120}+\frac{8}{120}=\frac15\)
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Re: Anthony and Michael sit on the sixmember board of directors for compa
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23 Jan 2008, 09:07
I don't think I missed anything. The answer has to be over 1/5.
Let's take any committee with Mike. If we take at random another person one that committee (one out of 2 remaining members), the probability that he will be Albert is 1/5 (there are 5 people besides Mike) which means that the probability is going to be at least 1/5. Obviously, the second person can also be Mike if the first person isn't.



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Re: Anthony and Michael sit on the sixmember board of directors for compa
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23 Jan 2008, 09:15
maratikus wrote: I don't think I missed anything. The answer has to be over 1/5.
Let's take any committee with Mike. If we take at random another person one that committee (one out of 2 remaining members), the probability that he will be Albert is 1/5 (there are 5 people besides Mike) which means that the probability is going to be at least 1/5. Obviously, the second person can also be Mike if the first person isn't. The probability to take any committee with Mike is 1/2.....
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Re: Anthony and Michael sit on the sixmember board of directors for compa
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23 Jan 2008, 09:25
The question was: what percent of all the possible subcommittees that include Michael also include Anthony?
It's a conditional probability question: what is the probability of Anthony being a committee if Mike belongs there. If you'd like we can look at a situation with 4 people, 2 committees.



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Re: Anthony and Michael sit on the sixmember board of directors for compa
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05 Nov 2009, 07:50
Bunuel  the fourth approach you posted seems to have an error: Fourth approach: Total # of splitting group of 6 into two groups of 3: 6C3*3C3/2!=10 # of groups with Michael and Anthony: 1C1*1C1*4C1=4 P=4/10=40% Total # of splitting group of 6 into two groups of 3: should be 6C3*3C3=20 and not 10. Out of these 20 , there are 10 groups with Michael in it so answer = groups with michael and anthony/ total groups with michael = 4/10 = 40%
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Re: Anthony and Michael sit on the sixmember board of directors for compa
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05 Nov 2009, 09:18
srini123 wrote: Bunuel  the fourth approach you posted seems to have an error:
Fourth approach: Total # of splitting group of 6 into two groups of 3: 6C3*3C3/2!=10 # of groups with Michael and Anthony: 1C1*1C1*4C1=4 P=4/10=40%
Total # of splitting group of 6 into two groups of 3: should be 6C3*3C3=20 and not 10. Out of these 20 , there are 10 groups with Michael in it
so answer = groups with michael and anthony/ total groups with michael = 4/10 = 40% Do we have group #1 and #2? Does it matter in which group Michael and Anthony will be? If no we should divide 6C3*3C3 by 2! and get 10 different split ups of 6 to groups of 3 when order of groups doesn't matter. # of groups with Michael and Anthony together: 1C1*1C1*4C1=4 P=4/10=40%. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\). \(\frac{6!}{(3!)^2*2!}=10\) OR another way: In our case \(\frac{6C3*3C3}{2!}=10.\)
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Re: Anthony and Michael sit on the sixmember board of directors for compa
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05 Nov 2009, 13:35
Bunuel wrote: srini123 wrote: Bunuel  the fourth approach you posted seems to have an error:
Fourth approach: Total # of splitting group of 6 into two groups of 3: 6C3*3C3/2!=10 # of groups with Michael and Anthony: 1C1*1C1*4C1=4 P=4/10=40%
Total # of splitting group of 6 into two groups of 3: should be 6C3*3C3=20 and not 10. Out of these 20 , there are 10 groups with Michael in it
so answer = groups with michael and anthony/ total groups with michael = 4/10 = 40% Do we have group #1 and #2? Does it matter in which group Michael and Anthony will be? If no we should divide 6C3*3C3 by 2! and get 10 different split ups of 6 to groups of 3 when order of groups doesn't matter. # of groups with Michael and Anthony together: 1C1*1C1*4C1=4 P=4/10=40%. The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\). \(\frac{6!}{(3!)^2*2!}=10\) OR another way: In our case \(\frac{6C3*3C3}{2!}=10.\) Sorry I may be missing something but, if we want to split 6 persons into 2 groups of 3 where order doesn't matter. dont we have 20 groups ? for eg., let A,B,C,D,E,F be 6 people and the groups with 3 people in each will be as below ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF  10 BCD, BCE,BCE, BDE,BDF, BEF  6 CDE, CDE,CEF 3 DEF 1 total 20. Let A be michael for this question , then total groups we have with michael is 10 and 4 groups have Michael and Anthony (assume D=anthony) im sorry im still not getting why 20 is not the total groups irrespective of michael in it or not
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Re: Anthony and Michael sit on the sixmember board of directors for compa
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05 Nov 2009, 14:04
srini123 wrote: Sorry I may be missing something but, if we want to split 6 persons into 2 groups of 3 where order doesn't matter. dont we have 20 groups ? for eg., let A,B,C,D,E,F be 6 people and the groups with 3 people in each will be as below
ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF  10 BCD, BCE,BCE, BDE,BDF, BEF  6 CDE, CDE,CEF 3 DEF 1
total 20.
Let A be michael for this question , then total groups we have with michael is 10 and 4 groups have Michael and Anthony (assume D=anthony)
im sorry im still not getting why 20 is not the total groups irrespective of michael in it or not Now I see what you've meant. But I think it's not quite true and here is where you are making the mistake: you listed # of selections of three out of 6  6C3=20. This is different from splitting the group into TWO groups of three members each: 1. ABC  DEF 2. ABD  CEF 3. ABE  CDF 4. ABF  CDE 5. ACD  BEF 6. ACE  BDF 7. ACF  BDE 8. ADE  BCF 9. ADF  BCE 10. AEF  BCD So, here are all possible ways to split 6 people into two groups of three members each. Let's say Anthony is A and Michael is B. You can see that there are only 4 scenarios when A and B are in one group (1,2,3,4). Hence 4/10.
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Re: Anthony and Michael sit on the sixmember board of directors for compa
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01 Oct 2010, 09:05
Thanks. Just a question: I noticed that you responded very quickly to my posts. Which I thank you for. But I am wondering if it's ok that I post questions that have been obviously posted before. If it's a problem then please tell me how can I check whether a question is already on the forum or not. Thanks.



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Re: Anthony and Michael sit on the sixmember board of directors for compa
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01 Oct 2010, 09:26
Barkatis wrote: Thanks. Just a question: I noticed that you responded very quickly to my posts. Which I thank you for. But I am wondering if it's ok that I post questions that have been obviously posted before. If it's a problem then please tell me how can I check whether a question is already on the forum or not. Thanks. Generally it's a good idea to do the search before posting, for example I would search in PS subforum (there is a search field above the topics) for the word "Anthony", don't think that there are many questions with this name. Though it's not a probelm to post a question that was posted before: if moderators find previous discussions they will merge the topics, copy the solution from there or give a link to it.
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Re: Anthony and Michael sit on the sixmember board of directors for compa
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04 Oct 2010, 12:56
Bunuel, A question just came to my mind concerning the second approach of the solution that you proposed:
total # of winning outcomes Select Anthony: 1C1=1, select any third member out of 4: 4C1=4, total # =1C1*4C1=4
If we say 1C1*4C1 don't we assume that the winning can be either (Anthony,Third member) or (Third member, Anthony) but not both ? shouldn't we multiply 1C1*4C1 by 2 to get the total number of possible combinations ?



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Re: Anthony and Michael sit on the sixmember board of directors for compa
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04 Oct 2010, 13:12
Barkatis wrote: Bunuel, A question just came to my mind concerning the second approach of the solution that you proposed:
total # of winning outcomes Select Anthony: 1C1=1, select any third member out of 4: 4C1=4, total # =1C1*4C1=4
If we say 1C1*4C1 don't we assume that the winning can be either (Anthony,Third member) or (Third member, Anthony) but not both ? shouldn't we multiply 1C1*4C1 by 2 to get the total number of possible combinations ? We are counting # of committees with Anthony and Michael: {M,A,1}; {M,A,2}; {M,A,3}; {M,A,4}. Here {M,A,1} is the same committee as {M,1,A}.
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Re: Anthony and Michael sit on the sixmember board of directors for compa
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21 Jan 2011, 16:54
Bunuel, can you advise why do we have to divide 6C3 * 3C3 by 2!, in the fourth approach ?



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Re: Anthony and Michael sit on the sixmember board of directors for compa
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21 Jan 2011, 17:14
praveenvino wrote: Bunuel, can you advise why do we have to divide 6C3 * 3C3 by 2!, in the fourth approach ? We are dividing by 2! (factorial of the # of groups) because the order of the groups is not important (we don't have the group #1 and the group #2) and we need to get rid of the duplications. Dividing a group into subgroups: combinationsproblems95344.htmlsplitthegroup101813.html9peopleandcombinatorics101722.htmlwaystodivide99053.htmlcombinationandselectionintoteam106277.htmlwaystosplitagroupof6boysintotwogroupsof3boysea105381.html
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Re: Anthony and Michael sit on the sixmember board of directors for compa
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24 Feb 2012, 14:58
Bunuel wrote: Fourth approach: Total # of splitting group of 6 into two groups of 3: 6C3*3C3/2!=10
Do you mind elaborating on that calculation, please? Isn't 6C3 the right calculation to select 3 people from a group of 6? You seem to be multiplying that by 3C3/2! Is there another example that illustrates this  6C3*3C3/2! concept, thanks.




Re: Anthony and Michael sit on the sixmember board of directors for compa
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