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

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Anthony and Michael sit on the six-member board of directors [#permalink] New post 18 Dec 2005, 01:49
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A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 1 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?


A. 20%
B. 30%
C. 40%
D. 50%
E. 60%
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Answer [#permalink] New post 18 Dec 2005, 04:23
Michael sits with the propability of 1/2 in one board, because there are three places to choose out of six candidates.

In the board in which Michaekl sits are now two free spaces to occupy out of five candidates, so that the propability that Anthony is in is 2/5.

P(M n A)= 1/2 * 2/5= 1/5, which is 20%

I think it is A
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 [#permalink] New post 18 Dec 2005, 06:57
If we put M in the team we need 2 more people which can be selected in 5C2 or 10 ways.When M and A are together we need one more person who0 can be selected in 4 ways. Seems like 40%
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Re: PS MGMAT 1-29 [#permalink] New post 18 Dec 2005, 08:08
2 out of 5 will be with Miachel. Everyone has the same chance. So each one's chance to be with Miachel is 2/5. Thus Anthony's chance to be with Miachel is 2/5.
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 [#permalink] New post 18 Dec 2005, 08:44
A group of 3 from 6 can be selected in 6c3 ways.
The combinations in which Michael and Anthony be together is 4c1

Therefore the probability is 4c1/6c3 = 20%
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 [#permalink] New post 19 Dec 2005, 07:12
kishore, please note the stem:
what percent of all the possible subcommittees that include Michael also include Anthony? [quote][/quote]
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 [#permalink] New post 19 Dec 2005, 13:33
BG wrote:
kishore, please note the stem:
what percent of all the possible subcommittees that include Michael also include Anthony?


There are 4 combinations that would put Michael and Antony together.

Dividing the group into 2 subcommittees of 3 members each, would be 6C3. 6C3 = 20.

4/20 is 20%.

Answer is A)
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 [#permalink] New post 19 Dec 2005, 22:52
There is a simple and there is a really complicated way to solving this problem.

Simple way:

subdivide the committee into two three-person teams.

_ _ _ and _ _ _

Fix Michael on any slot of either team.

That leaves 2 spots remaining for Anthony to be on the same team and five total possible spots.

2/5=40%

OA is C.

The complicated way and OE:

It is important to first note that our point of reference in this question is all the possible subcommittees that include Michael. We are asked to find what percent of these subcommittees also include Anthony.

Let's first find out how many possible subcommittees there are that must include Michael. If Michael must be on each of the three person committees that we are considering, we are essentially choosing people to fill the two remaining spots of the committee. Therefore, the number of possible committees can be found by considering the number of different two people groups that can be formed from a pool of 5 candidates (not 6 since Michael was already chosen).

Using the anagram method to solve this combinations question, we assign 5 letters to the various board members in the first row. In the second row, two of the board members get assigned a Y to signify that they were chosen and the remaining 3 get an N, to signify that they were not chosen:

A
B
C
D
E

Y
Y
N
N
N

The number of different combinations of two person committees from a group of 5 board members would be the number of possible anagrams that could be formed from the word YYNNN = 5! / (3!2!) = 10. Therefore there are 10 possible committees that include Michael.

Out of these 10 possible committees, of how many will Anthony also be a member? If we assume that Anthony and Michael must be a member of the three person committee, there is only one remaining place to fill. Since there are four other board members, there are four possible three person committees with both Anthony and Michael. Of the 10 committees that include Michael, 4/10 or 40% also include Anthony.

The correct answer is C.
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 [#permalink] New post 20 Dec 2005, 00:20
# of possible to form 3 person team from 6 people = 6C3 = 20. That leaves another 20 combinations for the other team. Total = 40 possibilites (20 team A combination, 20 team B combination).

If Micheal is in team A, then Anthony has to be in team A, and team A will have only 4 ways to form (due to 4 people remaining).

But Micheal and Anthony can be in team A or team B. So % = (4/20)*2*100% = 40%
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 [#permalink] New post 20 Dec 2005, 01:20
I think that's my problem. I usually see terms as "Michael also Anthony" as dependant, namely "Anthony under the condition that Michael". That's simply wrong in this cases.
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 [#permalink] New post 23 Dec 2005, 20:07
Kishore wrote:
A group of 3 from 6 can be selected in 6c3 ways.
The combinations in which Michael and Anthony be together is 4c1

Therefore the probability is 4c1/6c3 = 20%


Oops!!
I missed out considering Michael and Anthony in another team as well, Hence it is 4/20 + 4/20 ==> 40%
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 [#permalink] New post 24 Dec 2005, 16:05
GMATT73,

I got the simple way..

The complicated way & OE explanation took time to read..
Really a long one :-(
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a smart person knows what to say and
a wise person knows whether or not to say it."

  [#permalink] 24 Dec 2005, 16:05
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