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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]

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01 Oct 2010, 07:39
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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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Anthony and Michael sit on the six-member board of directors [#permalink]

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01 Oct 2010, 07:42
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Barkatis 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?

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.
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Re: Probability - MGMAT Test [#permalink]

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01 Oct 2010, 08:05
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.

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Re: Probability - MGMAT Test [#permalink]

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01 Oct 2010, 08:26
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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.
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Re: Probability - MGMAT Test [#permalink]

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04 Oct 2010, 11:56
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 ?

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Re: Probability - MGMAT Test [#permalink]

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04 Oct 2010, 12:12
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}.
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Re: Probability - MGMAT Test [#permalink]

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21 Jan 2011, 15:54
Bunuel, can you advise why do we have to divide 6C3 * 3C3 by 2!, in the fourth approach ?

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Re: Probability - MGMAT Test [#permalink]

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21 Jan 2011, 16:14
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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
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Re: Anthony and Michael sit on the six-member board of directors [#permalink]

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28 Dec 2012, 01:39
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Barkatis 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?

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

Here is my approach:

I first counted the possible creation of 2 subcommittees without restriction: $$\frac{6!}{3!3!}* \frac{3!}{3!}= 20$$
Then, I now proceed to counting the number of ways to create committee with Michael and Anthony together.

M A _ + _ _ _ = $$\frac{4!}{1!3!} * \frac{1!}{1!} = 4$$

MA could be in group#1 or group#2. Thus, $$=4*2 = 8$$

Final calculation: $$8/20 = 4/10 = 40%$$

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

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08 Jun 2015, 08:24
Bunuel wrote:
Barkatis 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?

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%

Hope it helps.

Bunuel, my question is for your fourth approach: Why don't we have to divide "# of groups with Michael and Anthony: 1C1*1C1*4C1=4" by 2! anymore?

To expound.

For total # of groups:
6C3 to choose 3 people for first group.
3C3 to choose 3 people for second group.

(6C3)(3C3)/2! divide by 2! as order is not important.

For # groups with Michael and Anthony:

1C1 to choose 1st person for group 1, 1C1 to choose 2nd person for group 1, 4C1 to choose 3rd person for group 1 = 4
3C3 to choose 3 people for second group.

(1C1)(1C1)(4C1)(3C3) <-- how come we no longer need to divide this by 2! ?

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

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08 Jun 2015, 18:43
Hi!

Just a question here.

Number of ways to Divide 3 people out of 6 are- 6C3= 20 Ways
Number of ways by which 1st member is Michal, 2nd is Anthony and 3rd is anyone from remaining four are- 1X1X4= 4 ways
So probability= number of desired events/number of total events *100= 4/20*100= 20%
This is the probability of Michal and Anthony being present in group 1. Group 2 will also have the same probability of 20%. So total probability is 20+20=40%

Please suggest if I am wrong anywhere.
Thanks
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Re: Lampard and Terry sit on the six member board [#permalink]

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10 Aug 2015, 07:37
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VenoMfTw wrote:
Lampard and Terry 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 Terry also include Lampard?

a. 20
b. 30
c. 40
d. 50
e. 60

All Possible Subcommittees that has terry in it = T _ _ --> 5c2 = 10
Possible subcommittees that include Terry also include Lampard --> T L _ --> 4c1 = 4

Percentage = ( 4 /10 ) * 100 = 40

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

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11 Aug 2015, 21:16
Bunuel, some day you should take this test and DEMOLISH it.
http://www.matrix67.com/iqtest/

Cheers!

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

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07 Oct 2015, 07:11
I want to put my approach out there too. Bunuel, please vet this if you can.

So, the total combinations of making a 3 group out of 6 members: 20.

there are 2 subcommittees. So say, 1st group has Micheal and Anthony and 2nd group doesn't; so: M A __(any of 4 remaining members)__ = 4 possibilities
Similarly, say 2nd group has Micheal and Anthony and 1st group doesnt; so: M A ___(Any of remaining 4 members)___ = 4 possibilities.

%tage having MA in same group = (4+4)/20 x 100 = 40 %
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Re: Anthony and Michael sit on the six-member board of directors [#permalink]

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07 Dec 2015, 16:13
Bunuel wrote:
Barkatis 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?

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%

Hope it helps.

Hi Bunnuel,

On approach 3, are you implying that the order does matter? By calculating the odds of picking Anthony in the second position and anyone else in the third position AND the odds of picking anyone except Anthony in the second position and Anthony in the 3rd position you are basically establishing that the order does matter (whether Anthony is in the second or 3rd position).

Can you clarify? I am sure there is something wronog in my thought process. Thanks.

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

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17 Dec 2015, 03:32
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jegf1987 wrote:
Bunuel wrote:
Barkatis 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?

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%

Hope it helps.

Hi Bunnuel,

On approach 3, are you implying that the order does matter? By calculating the odds of picking Anthony in the second position and anyone else in the third position AND the odds of picking anyone except Anthony in the second position and Anthony in the 3rd position you are basically establishing that the order does matter (whether Anthony is in the second or 3rd position).

Can you clarify? I am sure there is something wronog in my thought process. Thanks.

Responding to a pm:

No, we know very well that order does not matter when forming groups. Here, you have to form a group/subcommittee so the order in which you pick people is irrelevant. That said, we consider order in method 3 because of the limitations of this particular method. We are using probability. I can find the probability that the "next" guy I pick is Anthony. But how do I find the probability that Anthony is one of the next two guys I pick? For that, I have to use two steps:
- The next one is Anthony or
- Next to next one is Anthony
We add these two probabilities and get the probability that either of the next two guys is Anthony.
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Re: Anthony and Michael sit on the six-member board of directors [#permalink]

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20 Jan 2016, 15:50
See the attached image fir explanation. Thanks.
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Probability Question : Anthony and Michael sit on the six-member boad [#permalink]

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15 Jun 2016, 09:32
Anthony and Michael sit on the six-member boad of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommitteees that include Micahel and also Anthony?

A) 30%
B) 40%
C) 50%
D) 60%
E) 70%

This is from an old MGMT Practice test.

so 3 person chosen from 6 people. 6C3 would be the total: 20
Because Michael and also Anthony have to be together, 6C3 / 2! would be the probability. I didn't get the correct answer. Help!

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

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15 Jun 2016, 10:34
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Barkatis 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?

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

We can think of this as a probability question: What is the probability that Anthony and Michael are on the same subcommittee?

Now, assume that we're creating subcommittees.
We want to place 6 people in the following spaces:
_ _ _ | _ _ _

First, we place Michael in one subcommittee; it makes no difference which one:
M _ _ | _ _ _

When we go to place Anthony, we see that there are 5 spaces remaining.
2 spaces are on the same subcommittee as Michael.
So the probability that they are on the same subcommittee is 2/5 = 40%

[Reveal] Spoiler:
C

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

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07 May 2017, 02:00
Hi Bunuel,

I understood all the approaches provided by you. Please guide why is the below approach is incorrect?

We have to form 2 committee of 3 persons each.

Anthony can be selected in one of the two committees in 2 ways, but ,as per the condition, Michael should be in the same committee, so once Anthony 's committee is selected, there is only one way of selecting Michael's committee.

p=fav case/ total cases
P= (2*1)/2*2

p=1/2

Thanks a lot

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Re: Anthony and Michael sit on the six-member board of directors   [#permalink] 07 May 2017, 02:00

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