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Anthony and Michael sit on the six-member board of directors [#permalink]
01 Oct 2010, 07:39

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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?

Re: Probability - MGMAT Test [#permalink]
01 Oct 2010, 07:42

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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%

Re: Probability - MGMAT Test [#permalink]
01 Oct 2010, 08:05

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.

Re: Probability - MGMAT Test [#permalink]
01 Oct 2010, 08:26

1

This post received KUDOS

Expert's post

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. _________________

Re: Probability - MGMAT Test [#permalink]
04 Oct 2010, 11:56

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 ?

Re: Probability - MGMAT Test [#permalink]
04 Oct 2010, 12:12

Expert's post

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}. _________________

Re: Probability - MGMAT Test [#permalink]
21 Jan 2011, 16:14

Expert's post

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.

Re: Anthony and Michael sit on the six-member board of directors [#permalink]
28 Dec 2012, 01:39

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: \(\frac{6!}{3!3!}* \frac{3!}{3!}= 20\) Then, I now proceed to counting the number of ways to create committee with Michael and Anthony together.

M A _ + _ _ _ = \(\frac{4!}{1!3!} * \frac{1!}{1!} = 4\)

MA could be in group#1 or group#2. Thus, \(=4*2 = 8\)

Re: Anthony and Michael sit on the six-member board of directors [#permalink]
20 Jan 2014, 09:45

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Re: Anthony and Michael sit on the six-member board of directors [#permalink]
26 Mar 2015, 08:10

Hello from the GMAT Club BumpBot!

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Re: Anthony and Michael sit on the six-member board of directors [#permalink]
08 Jun 2015, 08:24

Bunuel wrote:

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.

Bunuel, my question is for your fourth approach: Why don't we have to divide "# of groups with Michael and Anthony: 1C1*1C1*4C1=4" by 2! anymore?

To expound.

For total # of groups: 6C3 to choose 3 people for first group. 3C3 to choose 3 people for second group.

(6C3)(3C3)/2! divide by 2! as order is not important.

For # groups with Michael and Anthony:

1C1 to choose 1st person for group 1, 1C1 to choose 2nd person for group 1, 4C1 to choose 3rd person for group 1 = 4 3C3 to choose 3 people for second group.

(1C1)(1C1)(4C1)(3C3) <-- how come we no longer need to divide this by 2! ?

Re: Anthony and Michael sit on the six-member board of directors [#permalink]
08 Jun 2015, 18:43

Hi!

Just a question here.

Number of ways to Divide 3 people out of 6 are- 6C3= 20 Ways Number of ways by which 1st member is Michal, 2nd is Anthony and 3rd is anyone from remaining four are- 1X1X4= 4 ways So probability= number of desired events/number of total events *100= 4/20*100= 20% This is the probability of Michal and Anthony being present in group 1. Group 2 will also have the same probability of 20%. So total probability is 20+20=40%

Please suggest if I am wrong anywhere. Thanks

gmatclubot

Re: Anthony and Michael sit on the six-member board of directors
[#permalink]
08 Jun 2015, 18:43

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