Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 25 Aug 2016, 03:39

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Anthony and Michael sit on the six-member board of directors

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern
Joined: 19 Sep 2010
Posts: 26
Followers: 0

Kudos [?]: 76 [3] , given: 0

Anthony and Michael sit on the six-member board of directors [#permalink]

### Show Tags

01 Oct 2010, 08:39
3
This post received
KUDOS
23
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

48% (02:08) correct 52% (01:32) wrong based on 387 sessions

### HideShow timer Statistics

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%
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 34432
Followers: 6254

Kudos [?]: 79439 [16] , given: 10016

Re: Probability - MGMAT Test [#permalink]

### Show Tags

01 Oct 2010, 08:42
16
This post received
KUDOS
Expert's post
20
This post was
BOOKMARKED
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.
_________________
Intern
Joined: 19 Sep 2010
Posts: 26
Followers: 0

Kudos [?]: 76 [0], given: 0

Re: Probability - MGMAT Test [#permalink]

### Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 34432
Followers: 6254

Kudos [?]: 79439 [1] , given: 10016

Re: Probability - MGMAT Test [#permalink]

### Show Tags

01 Oct 2010, 09:26
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.
_________________
Intern
Joined: 19 Sep 2010
Posts: 26
Followers: 0

Kudos [?]: 76 [0], given: 0

Re: Probability - MGMAT Test [#permalink]

### Show Tags

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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 34432
Followers: 6254

Kudos [?]: 79439 [0], given: 10016

Re: Probability - MGMAT Test [#permalink]

### Show Tags

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}.
_________________
Intern
Joined: 06 Nov 2010
Posts: 22
Followers: 0

Kudos [?]: 31 [0], given: 16

Re: Probability - MGMAT Test [#permalink]

### Show Tags

21 Jan 2011, 16:54
Bunuel, can you advise why do we have to divide 6C3 * 3C3 by 2!, in the fourth approach ?
Math Expert
Joined: 02 Sep 2009
Posts: 34432
Followers: 6254

Kudos [?]: 79439 [0], given: 10016

Re: Probability - MGMAT Test [#permalink]

### Show Tags

21 Jan 2011, 17:14
Expert's post
5
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.

Dividing a group into subgroups:
combinations-problems-95344.html
split-the-group-101813.html
9-people-and-combinatorics-101722.html
ways-to-divide-99053.html
combination-and-selection-into-team-106277.html
ways-to-split-a-group-of-6-boys-into-two-groups-of-3-boys-ea-105381.html
_________________
Senior Manager
Joined: 13 Aug 2012
Posts: 464
Concentration: Marketing, Finance
GMAT 1: Q V0
GPA: 3.23
Followers: 25

Kudos [?]: 384 [0], given: 11

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

### Show Tags

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: $$\frac{6!}{3!3!}* \frac{3!}{3!}= 20$$
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, $$=4*2 = 8$$

Final calculation: $$8/20 = 4/10 = 40%$$

Answer: C
_________________

Impossible is nothing to God.

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11046
Followers: 510

Kudos [?]: 134 [0], given: 0

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

### Show Tags

20 Jan 2014, 10:45
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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11046
Followers: 510

Kudos [?]: 134 [0], given: 0

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

### Show Tags

26 Mar 2015, 09:10
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.
_________________
Intern
Joined: 25 May 2014
Posts: 23
GPA: 3.55
Followers: 0

Kudos [?]: 2 [0], given: 13

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

### Show Tags

08 Jun 2015, 09:24
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! ?
Director
Joined: 18 Oct 2014
Posts: 906
Location: United States
GMAT 1: 660 Q49 V31
GPA: 3.98
Followers: 74

Kudos [?]: 150 [0], given: 64

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

### Show Tags

08 Jun 2015, 19:43
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
_________________

I welcome critical analysis of my post!! That will help me reach 700+

Manager
Joined: 14 Mar 2014
Posts: 144
Followers: 0

Kudos [?]: 119 [0], given: 108

Lampard and Terry sit on the six member board [#permalink]

### Show Tags

10 Aug 2015, 08:33
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)_/¯

Manager
Joined: 14 Mar 2014
Posts: 144
Followers: 0

Kudos [?]: 119 [0], given: 108

Re: Lampard and Terry sit on the six member board [#permalink]

### Show Tags

10 Aug 2015, 08:37
1
This post was
BOOKMARKED
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)_/¯

Math Expert
Joined: 02 Sep 2009
Posts: 34432
Followers: 6254

Kudos [?]: 79439 [0], given: 10016

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

### Show Tags

10 Aug 2015, 08:47
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.
_________________
Intern
Joined: 26 Jul 2015
Posts: 23
Followers: 0

Kudos [?]: 5 [0], given: 12

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

### Show Tags

11 Aug 2015, 22:16
Bunuel, some day you should take this test and DEMOLISH it.
http://www.matrix67.com/iqtest/

Cheers!

Jawad
Manager
Joined: 19 Oct 2012
Posts: 165
Location: India
Concentration: General Management, Operations
GMAT 1: 660 Q47 V35
GPA: 3.81
WE: Information Technology (Computer Software)
Followers: 1

Kudos [?]: 10 [0], given: 28

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

### Show Tags

07 Oct 2015, 08:11
I want to put my approach out there too. Bunuel, please vet this if you can.

So, the total combinations of making a 3 group out of 6 members: 20.

there are 2 subcommittees. So say, 1st group has Micheal and Anthony and 2nd group doesn't; so: M A __(any of 4 remaining members)__ = 4 possibilities
Similarly, say 2nd group has Micheal and Anthony and 1st group doesnt; so: M A ___(Any of remaining 4 members)___ = 4 possibilities.

%tage having MA in same group = (4+4)/20 x 100 = 40 %
_________________

Citius, Altius, Fortius

Intern
Joined: 26 Sep 2015
Posts: 3
Followers: 0

Kudos [?]: 0 [0], given: 456

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

### Show Tags

07 Dec 2015, 17:13
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.

Hi Bunnuel,

On approach 3, are you implying that the order does matter? By calculating the odds of picking Anthony in the second position and anyone else in the third position AND the odds of picking anyone except Anthony in the second position and Anthony in the 3rd position you are basically establishing that the order does matter (whether Anthony is in the second or 3rd position).

Can you clarify? I am sure there is something wronog in my thought process. Thanks.
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 6829
Location: Pune, India
Followers: 1918

Kudos [?]: 11926 [1] , given: 221

Re: Anthony and Michael sit on the six-member board of directors [#permalink]

### Show Tags

17 Dec 2015, 04:32
1
This post received
KUDOS
Expert's post
jegf1987 wrote:
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.

Hi Bunnuel,

On approach 3, are you implying that the order does matter? By calculating the odds of picking Anthony in the second position and anyone else in the third position AND the odds of picking anyone except Anthony in the second position and Anthony in the 3rd position you are basically establishing that the order does matter (whether Anthony is in the second or 3rd position).

Can you clarify? I am sure there is something wronog in my thought process. Thanks.

Responding to a pm:

No, we know very well that order does not matter when forming groups. Here, you have to form a group/subcommittee so the order in which you pick people is irrelevant. That said, we consider order in method 3 because of the limitations of this particular method. We are using probability. I can find the probability that the "next" guy I pick is Anthony. But how do I find the probability that Anthony is one of the next two guys I pick? For that, I have to use two steps:
- The next one is Anthony or
- Next to next one is Anthony
We add these two probabilities and get the probability that either of the next two guys is Anthony.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Re: Anthony and Michael sit on the six-member board of directors   [#permalink] 17 Dec 2015, 04:32

Go to page    1   2    Next  [ 24 posts ]

Similar topics Replies Last post
Similar
Topics:
3 Michael, Steve and Tyler shared a box of cookies. Michael ate 1/8 of 4 07 Jul 2016, 06:06
1 Anthony and Michael sit on the six member board od directors 11 27 Sep 2009, 02:51
28 Anthony and Michael sit on the six-member board of directors 11 04 Nov 2009, 17:28
51 Anthony and Michael sit on the six member board of directors 43 23 Jan 2008, 07:12
14 Anthony and Michael sit on the six-member board of directors 21 02 Jan 2008, 14:50
Display posts from previous: Sort by

# Anthony and Michael sit on the six-member board of directors

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.