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Anthony and Michael sit on the six-member board of directors [#permalink]
 01 Oct 2010, 08:39
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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 three-person 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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Re: Probability - MGMAT Test [#permalink]
01 Oct 2010, 08:42
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Barkatis wrote: Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person 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 2/5=40%. Second approach:Again in Michael's group 2 places are left, # of selections of 2 out of 5 5C2=10 - total # of outcomes. Select Anthony - 1C1=1, select any third member out of 4 - 4C1=4, total # =1C1*4C1=4 - total # of winning outcomes. P=# of winning outcomes/# of outcomes=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=1/5*4/4=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=4/5*1/4=1/5; Sum=1/5+1/5=2/5=40% 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% Answer: C. Hope it helps.
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Re: Probability - MGMAT Test [#permalink]
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: Probability - MGMAT Test [#permalink]
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: Probability - MGMAT Test [#permalink]
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: Probability - MGMAT Test [#permalink]
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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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. NEW!!!
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Anthony and Michael sit on the six-member board [#permalink]
30 Nov 2010, 07:35
Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
20%
30%
40%
50%
60%
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Re: Probability - MGMAT Test [#permalink]
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: Probability - MGMAT Test [#permalink]
21 Jan 2011, 17:14
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Re: Anthony and Michael sit on the six-member board of directors [#permalink]
28 Dec 2012, 02:39
Barkatis wrote: Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
A. 20%
B. 30%
C. 40%
D. 50%
E. 60% Here is my approach: I first counted the possible creation of 2 subcommittees without restriction: \frac{6!}{3!3!}* \frac{3!}{3!}= 20Then, I now proceed to counting the number of ways to create committee with Michael and Anthony together. M A _ + _ _ _ = \frac{4!}{1!3!} * \frac{1!}{1!} = 4MA could be in group#1 or group#2. Thus, =4*2 = 8Final calculation: 8/20 = 4/10 = 40%Answer: C
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Re: Anthony and Michael sit on the six-member board of directors
[#permalink]
28 Dec 2012, 02:39
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