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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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Question Stats:

49% (01:00) correct
51% (01:28) wrong based on 241 sessions

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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Expert's post

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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

1

This post was BOOKMARKED

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!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

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%

Lampard and Terry sit on the six member board [#permalink]
10 Aug 2015, 07:33

Lampard and Terry 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 Terry also include Lampard?

a. 20 b. 30 c. 40 d. 50 e. 60 _________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos ¯\_(ツ)_/¯

Re: Lampard and Terry sit on the six member board [#permalink]
10 Aug 2015, 07:37

VenoMfTw wrote:

Lampard and Terry 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 Terry also include Lampard?

a. 20 b. 30 c. 40 d. 50 e. 60

All Possible Subcommittees that has terry in it = T _ _ --> 5c2 = 10 Possible subcommittees that include Terry also include Lampard --> T L _ --> 4c1 = 4

Percentage = ( 4 /10 ) * 100 = 40

Option C _________________

I'm happy, if I make math for you slightly clearer And yes, I like kudos ¯\_(ツ)_/¯

Re: Anthony and Michael sit on the six-member board of directors [#permalink]
10 Aug 2015, 07:47

Expert's post

VenoMfTw wrote:

Lampard and Terry 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 Terry also include Lampard?

a. 20 b. 30 c. 40 d. 50 e. 60

PLEASE SEARCH BEFORE POSTING A QUESTION. _________________

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