GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 15 Oct 2019, 04:20

### 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

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

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 15 Apr 2007
Posts: 38
Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

Updated on: 08 Aug 2019, 09:13
9
36
00:00

Difficulty:

95% (hard)

Question Stats:

42% (01:45) correct 58% (02:08) wrong based on 606 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%

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.
Math Expert
Joined: 02 Sep 2009
Posts: 58327
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

01 Oct 2010, 08:42
28
39
Math Expert
Joined: 02 Sep 2009
Posts: 58327
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

05 Nov 2009, 07:20
8
13
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 $$\frac{2}{5}=40\%$$.

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: $$C^1_1*C^1_1*C^1_4=4$$.

$$P=\frac{4}{10}=40\%$$

Hope it helps.
_________________
Senior Manager
Joined: 01 Jan 2008
Posts: 473
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

23 Jan 2008, 10:14
7
3
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....

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?
##### General Discussion
Senior Manager
Joined: 01 Jan 2008
Posts: 473
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

23 Jan 2008, 08:19
6
2
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
CEO
Joined: 17 Nov 2007
Posts: 3082
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

23 Jan 2008, 08:49
I think you missed something...

$$p=\frac26*(\frac15*\frac11+\frac45*\frac14)+ \frac46*(\frac25*\frac14)=\frac{16}{120}+\frac{8}{120}=\frac15$$
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | Limited GMAT/GRE Math tutoring in Chicago
Senior Manager
Joined: 01 Jan 2008
Posts: 473
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

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.
CEO
Joined: 17 Nov 2007
Posts: 3082
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

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.....
_________________
HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | Limited GMAT/GRE Math tutoring in Chicago
Senior Manager
Joined: 01 Jan 2008
Posts: 473
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

23 Jan 2008, 09:25
5
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.
Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 213
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

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%
_________________
Thanks, Sri
-------------------------------
keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip
Math Expert
Joined: 02 Sep 2009
Posts: 58327
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

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 $$\frac{6C3*3C3}{2!}=10.$$
_________________
Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 213
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

05 Nov 2009, 13:35
1
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
_________________
Thanks, Sri
-------------------------------
keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip
Math Expert
Joined: 02 Sep 2009
Posts: 58327
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

05 Nov 2009, 14:04
2
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
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.
_________________
Intern
Joined: 19 Sep 2010
Posts: 21
Re: Anthony and Michael sit on the six-member board of directors for compa  [#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: 58327
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

01 Oct 2010, 09:26
2
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: 21
Re: Anthony and Michael sit on the six-member board of directors for compa  [#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: 58327
Re: Anthony and Michael sit on the six-member board of directors for compa  [#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: 19
Re: Anthony and Michael sit on the six-member board of directors for compa  [#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: 58327
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

21 Jan 2011, 17:14
1
6
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
_________________
Intern
Joined: 24 Feb 2012
Posts: 29
Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

### Show Tags

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 six-member board of directors for compa   [#permalink] 24 Feb 2012, 14:58

Go to page    1   2    Next  [ 32 posts ]

Display posts from previous: Sort by