Anthony and Michael sit on the six-member board of directors : GMAT Problem Solving (PS) - Page 2
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 23 Jan 2017, 16:46

### 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

Author Message
TAGS:

### Hide Tags

Current Student
Joined: 02 Apr 2012
Posts: 76
Location: United States (VA)
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34
WE: Consulting (Consulting)
Followers: 1

Kudos [?]: 54 [1] , given: 155

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

### Show Tags

15 Jul 2013, 13:07
1
KUDOS
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?

Probability approach:

(AMX XXX) or (XXX AMX)

AM could be in the first committee or in the second committee.
When AM is in the first committee, we select A first and then the probability of selecting M between the remaining 5 persons is 1/5.

The same for the second committee.
Then:
$$P = \frac{6}{6} * \frac{1}{5} + \frac{6}{6}*\frac{1}{5} = \frac{12}{30} = 40%$$

Reversal probability approach:
AMX XXX
q= probability of A and M to be in different committees.
P = 1 - q = probability A and M together in the same committee.

$$q = \frac{2}{2}*\frac{4}{5}*\frac{3}{4} = \frac{12}{20} = 60%$$

For any of the committes:
$$\frac{2}{2}$$ we select A or M from A or M.
$$\frac{4}{5}$$ probability of picking any of the Xs from the remaining XXXXM
$$\frac{3}{4}$$ probability of picking another X from the remaining XXXM

then P = 1 - q = 1 - 60% = 40%

Combinatorial approach:
AMX XXX

We select the group (AM) from the 2 groups (AM) (XXXX). Then we combine the group AM with each X of the group XXXX:

$$P = \frac{{C^2_1 * C^4_1}}{{C^6_3}} = \frac{{2*4}}{20} = 40%$$

Reversal combinatorial approach:

q= probability of A and M to be in different committees.
P = 1 - q = probability that A and M being together in the same committee.

AM XXXX

We choose A from (AM) $$(C^2_1)$$ and then 2 more X (XX) from the remaining (XXXX) $$(C^4_2)$$:

$$q = \frac{{C^2_1 * C^4_2}}{{C^6_3}}= \frac{{12}}{20} = \frac{12}{20} = 60%$$

P = 1 - q = 1 - 60% = 40%
_________________

Encourage cooperation! If this post was very useful, kudos are welcome
"It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

Math Expert
Joined: 02 Sep 2009
Posts: 36618
Followers: 7100

Kudos [?]: 93574 [0], given: 10578

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

### Show Tags

15 Jul 2013, 13:14
Maxirosario2012 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?

Probability approach:

(AMX XXX) or (XXX AMX)

AM could be in the first committee or in the second committee.
When AM is in the first committee, we select A first and then the probability of selecting M between the remaining 5 persons is 1/5.

The same for the second committee.
Then:
$$P = \frac{6}{6} * \frac{1}{5} + \frac{6}{6}*\frac{1}{5} = \frac{12}{30} = 40%$$

Reversal probability approach:
AMX XXX
q= probability of A and M to be in different committees.
P = 1 - q = probability A and M together in the same committee.

$$q = \frac{2}{2}*\frac{4}{5}*\frac{3}{4} = \frac{12}{20} = 60%$$

For any of the committes:
$$\frac{2}{2}$$ we select A or M from A or M.
$$\frac{4}{5}$$ probability of picking any of the Xs from the remaining XXXXM
$$\frac{3}{4}$$ probability of picking another X from the remaining XXXM

then P = 1 - q = 1 - 60% = 40%

Combinatorial approach:
AMX XXX

We select the group (AM) from the 2 groups (AM) (XXXX). Then we combine the group AM with each X of the group XXXX:

$$P = \frac{{C^2_1 * C^4_1}}{{C^6_3}} = \frac{{2*4}}{20} = 40%$$

Reversal combinatorial approach:

q= probability of A and M to be in different committees.
P = 1 - q = probability that A and M being together in the same committee.

AM XXXX

We choose A from (AM) $$(C^2_1)$$ and then 2 more X (XX) from the remaining (XXXX) $$(C^4_2)$$:

$$q = \frac{{C^2_1 * C^4_2}}{{C^6_3}}= \frac{{12}}{20} = \frac{12}{20} = 60%$$

P = 1 - q = 1 - 60% = 40%

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%

OPEN DISCUSSION OF THIS QUESTION IS HERE: anthony-and-michael-sit-on-the-six-member-board-of-directors-102027.html
_________________
Re: Anthony and Michael sit on the six-member board of directors   [#permalink] 15 Jul 2013, 13:14

Go to page   Previous    1   2   [ 22 posts ]

Similar topics Replies Last post
Similar
Topics:
5 Michael, Steve and Tyler shared a box of cookies. Michael ate 1/8 of 5 07 Jul 2016, 05:06
1 Anthony and Michael sit on the six member board od directors 11 27 Sep 2009, 01:51
75 Anthony and Michael sit on the six-member board of directors 23 01 Oct 2010, 07:39
30 Anthony and Michael sit on the six-member board of directors 11 04 Nov 2009, 16:28
59 Anthony and Michael sit on the six member board of directors 43 23 Jan 2008, 06:12
Display posts from previous: Sort by