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

 It is currently 13 Nov 2018, 17:22

### 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 November
PrevNext
SuMoTuWeThFrSa
28293031123
45678910
11121314151617
18192021222324
2526272829301
Open Detailed Calendar
• ### Essential GMAT Time-Management Hacks

November 14, 2018

November 14, 2018

07:00 PM PST

08:00 PM PST

Join the webinar and learn time-management tactics that will guarantee you answer all questions, in all sections, on time. Save your spot today! Nov. 14th at 7 PM PST

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

Author Message
Manager
Joined: 02 Apr 2012
Posts: 69
Location: United States (VA)
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34
WE: Consulting (Consulting)
Re: Anthony and Michael sit on the six-member board of directors  [#permalink]

### Show Tags

15 Jul 2013, 13:07
2
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: 50572
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
_________________
Non-Human User
Joined: 09 Sep 2013
Posts: 8759
Re: Anthony and Michael sit on the six-member board of directors  [#permalink]

### Show Tags

10 Sep 2018, 04:09
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.
_________________
Re: Anthony and Michael sit on the six-member board of directors &nbs [#permalink] 10 Sep 2018, 04:09

Go to page   Previous    1   2   [ 23 posts ]

Display posts from previous: Sort by