Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 27 May 2017, 02:11

### 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

# Events & Promotions

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

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

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

### Hide Tags

Intern
Joined: 15 Apr 2007
Posts: 37
Followers: 0

Kudos [?]: 25 [5] , given: 0

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

### Show Tags

23 Jan 2008, 07:12
5
KUDOS
18
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

30% (02:40) correct 70% (01:43) wrong based on 631 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%

OPEN DISCUSSION OF THIS QUESTION IS HERE: anthony-and-michael-sit-on-the-six-member-board-of-directors-102027.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 31 Jul 2012, 03:10, edited 2 times in total.
CEO
Joined: 17 Nov 2007
Posts: 3586
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 575

Kudos [?]: 3982 [1] , given: 360

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

23 Jan 2008, 07:36
1
KUDOS
Expert's post
A

$$ratio=100%*\frac{C^4_1}{C^6_3}=100%*\frac{4*3*2}{6*5*4}=20%$$
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Director
Joined: 01 Jan 2008
Posts: 624
Followers: 5

Kudos [?]: 184 [6] , given: 1

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

23 Jan 2008, 08:19
6
KUDOS
1
This post was
BOOKMARKED
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: 3586
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 575

Kudos [?]: 3982 [0], given: 360

Re: Power Prep: Subcommittee [#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+ | PrepGame

Director
Joined: 01 Jan 2008
Posts: 624
Followers: 5

Kudos [?]: 184 [0], given: 1

Re: Power Prep: Subcommittee [#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: 3586
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 575

Kudos [?]: 3982 [0], given: 360

Re: Power Prep: Subcommittee [#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+ | PrepGame

Director
Joined: 01 Jan 2008
Posts: 624
Followers: 5

Kudos [?]: 184 [4] , given: 1

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

23 Jan 2008, 09:25
4
KUDOS
1
This post was
BOOKMARKED
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.
CEO
Joined: 17 Nov 2007
Posts: 3586
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 575

Kudos [?]: 3982 [1] , given: 360

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

23 Jan 2008, 09:36
1
KUDOS
Expert's post
Ahh....You seem to be right. I still have troubles with understanding of problems...
Thanks
+1
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Director
Joined: 01 Jan 2008
Posts: 624
Followers: 5

Kudos [?]: 184 [6] , given: 1

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

23 Jan 2008, 10:14
6
KUDOS
2
This post was
BOOKMARKED
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?
Manager
Joined: 01 Jan 2008
Posts: 223
Schools: Booth, Stern, Haas
Followers: 2

Kudos [?]: 58 [0], given: 2

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

23 Jan 2008, 10:42
maratikus wrote:
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?

yes, thanks, I have some problems with prob, perm and comb that is why I asked about it, it might be seen like easy stuff but for me it is another new discovery, so great thanks +1
CEO
Joined: 17 Nov 2007
Posts: 3586
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 575

Kudos [?]: 3982 [2] , given: 360

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

24 Jan 2008, 08:52
2
KUDOS
Expert's post
kazakhb wrote:
Can we somehow solve it using combination formulas?

$$ratio=100%*\frac{C^4_1}{C^5_2}=100%*\frac{4*2}{5*4}=40%$$

kazakhb wrote:
BTW, when do we use combination/permutations with probability vs. when it's just probability?

Majority of probability problems have a few ways. I think that both ways are good.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

CEO
Joined: 29 Mar 2007
Posts: 2562
Followers: 21

Kudos [?]: 453 [1] , given: 0

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

24 Jan 2008, 09:30
1
KUDOS
kamilaak wrote:
Anthony and Michael sit on the six member board od directors for compnay X. If the board is to be split up into 2 three-person sybcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

A. 20%
B. 30%
C. 40%
D. 50%
E. 60%

Let me give a crack at it. Lemme see if this makes sense. (I didn't get this right when I first did it, I guessed B, but maybe this way will work).

We have 6!/3!3! ways to divide the groups so 20 ways. then we have four ways of picking MAX, MXA or AMX AXM

But we have two groups so its 8 ways total.

8/20 = 2/5 or 40%.

Let me know if this is an appropriate way to solve this. Thx
CEO
Joined: 17 Nov 2007
Posts: 3586
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 575

Kudos [?]: 3982 [0], given: 360

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

24 Jan 2008, 09:54
GMATBLACKBELT, maybe I'm wrong but your way seems to be incorrect.

GMATBLACKBELT wrote:

We have 6!/3!3! ways to divide the groups so 20 ways. then we have four ways of picking MAX, MXA or AMX AXM

But we have two groups so its 8 ways total.

8/20 = 2/5 or 40%.

Let me know if this is an appropriate way to solve this. Thx

1. MAX, MXA or AMX AXM is the same committee. A committee is not sensitive to permutation.
You have really 4 ways: MAX,MAY,MAZ,MAW.

2. the question is "what percent of all the possible subcommittees" but not "ways to divide".

3. we should use 5!/2!3! instead of 6!/3!3! in calculation of probability because we should only take committees with Michael.

Therefore 4/10
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Director
Joined: 01 Jan 2008
Posts: 624
Followers: 5

Kudos [?]: 184 [0], given: 1

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

24 Jan 2008, 09:56
kamilaak wrote:
OA is 40%. (c)
Can we somehow solve it using combination formulas?

We need to pick 2 people out of 5: C(5,2) = 10
We need to pick Anthony out of a group of 1 person: C(1,1) = 1, and one person out of 4: C(4,1)=4

the answer C(1,1)*C(4,1)/C(5,2) = 2/5
Senior Manager
Joined: 06 Jul 2006
Posts: 293
Location: SFO Bay Area
Schools: Berkeley Haas
Followers: 2

Kudos [?]: 67 [3] , given: 0

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

24 Jan 2008, 10:14
3
KUDOS
It should be A (20%).

Lets assume M and A to be one entity. Now we need to select one more member to make a committee of 3 which has to have both M and A. We can select another (any) member from remaining 4 members in 4c1 = 4 ways.

We have 6c3 ways of selecting 3 members from a group of 6, and that should be it. After you have selected 3 members, the other 3 members left will be part of the other committee, there is no choice left for them. Hence 20 ways.

4 / 20 = 1 / 5 = 20%.
_________________

-------------------------------------------------------------
When you come to the end of your rope, tie a knot and hang on.

Director
Joined: 01 Jan 2008
Posts: 624
Followers: 5

Kudos [?]: 184 [0], given: 1

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

24 Jan 2008, 11:44
suntaurian wrote:
It should be A (20%).
We have 6c3 ways of selecting 3 members from a group of 6, and that should be it. ... Hence 20 ways.
4 / 20 = 1 / 5 = 20%.

We should select 2 people out of 5 because Michael is already a member of a committee under consideration. Therefore, there 10 ways and probability is 2/5.
Senior Manager
Joined: 06 Jul 2006
Posts: 293
Location: SFO Bay Area
Schools: Berkeley Haas
Followers: 2

Kudos [?]: 67 [0], given: 0

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

24 Jan 2008, 11:59
It does not say that Micheal is already part of the committee, its just says...find all ways of making "subcommittees that include Michael also include Anthony? "....Its just another way of putting where Michjael and Anthony are part of the same subcommittee.

Still stand by A..20%
_________________

-------------------------------------------------------------
When you come to the end of your rope, tie a knot and hang on.

Director
Joined: 01 Jan 2008
Posts: 624
Followers: 5

Kudos [?]: 184 [1] , given: 1

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

24 Jan 2008, 16:07
1
KUDOS
suntaurian wrote:
It does not say that Micheal is already part of the committee, its just says...find all ways of making "subcommittees that include Michael also include Anthony? "....Its just another way of putting where Michjael and Anthony are part of the same subcommittee.

Still stand by A..20%

Our universe consists of all subcommittees that include Michael (not all subcommittees). The question is how many of those subcommittees also include Anthony.
VP
Joined: 22 Oct 2006
Posts: 1438
Schools: Chicago Booth '11
Followers: 9

Kudos [?]: 192 [0], given: 12

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

25 Jan 2008, 12:11
ok C

Total Arrangements 6!=720

We figure out the number of possibilties when Mike sits in seat 1:

If he sits in seat 1, there are 5!=120 seats left. However Anthony can sit in only 2 seats out of the 5 so we multiply 2/5 * 120 = 48.

Since Michael has 6 different positions we get 48*6 = 288

288/720 = 40%
Manager
Joined: 15 Jul 2008
Posts: 206
Followers: 3

Kudos [?]: 61 [0], given: 0

Re: Power Prep: Subcommittee [#permalink]

### Show Tags

22 Aug 2008, 11:56
walker wrote:
A

$$ratio=100%*\frac{C^4_1}{C^6_3}=100%*\frac{4*3*2}{6*5*4}=20%$$

Walker.. in the denominator you should be taking 6C3 / 2! ... I learnt this from one of your recent posts.
6C3 alone takes the order of the two committees into account. divide by 2! to eliminate that effect
Re: Power Prep: Subcommittee   [#permalink] 22 Aug 2008, 11:56

Go to page    1   2   3    Next  [ 44 posts ]

Similar topics Replies Last post
Similar
Topics:
2 Anthony and Michael sit on the six member board od directors 11 10 Jul 2012, 22:03
1 Anthony and Michael sit on the six-member board of directors for compa 13 14 Oct 2010, 14:34
85 Anthony and Michael sit on the six-member board of directors 25 07 May 2017, 03:47
32 Anthony and Michael sit on the six-member board of directors 11 26 Oct 2013, 01:43
17 Anthony and Michael sit on the six-member board of directors 21 15 Jul 2013, 14:14
Display posts from previous: Sort by

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

 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®.