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

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

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New post Updated on: 08 Aug 2019, 09:13
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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%

Originally posted by kamilaak on 23 Jan 2008, 07:12.
Last edited by Bunuel on 08 Aug 2019, 09:13, edited 3 times in total.
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 01 Oct 2010, 08:42
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 05 Nov 2009, 07:20
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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%

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: \(C^1_1*C^1_1*C^1_4=4\).

\(P=\frac{4}{10}=40\%\)

Answer: C.

Hope it helps.
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 23 Jan 2008, 10:14
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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....
thanks in advance


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

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New post 23 Jan 2008, 08:19
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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
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 23 Jan 2008, 08:49
I think you missed something...

\(p=\frac26*(\frac15*\frac11+\frac45*\frac14)+ \frac46*(\frac25*\frac14)=\frac{16}{120}+\frac{8}{120}=\frac15\)
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 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.
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 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.....
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 23 Jan 2008, 09:25
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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.
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 05 Nov 2009, 07:50
Bunuel - the fourth approach you posted seems to have an error:

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%

Total # of splitting group of 6 into two groups of 3: should be 6C3*3C3=20 and not 10.
Out of these 20 , there are 10 groups with Michael in it

so answer = groups with michael and anthony/ total groups with michael = 4/10 = 40%
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 05 Nov 2009, 09:18
srini123 wrote:
Bunuel - the fourth approach you posted seems to have an error:

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%

Total # of splitting group of 6 into two groups of 3: should be 6C3*3C3=20 and not 10.
Out of these 20 , there are 10 groups with Michael in it

so answer = groups with michael and anthony/ total groups with michael = 4/10 = 40%


Do we have group #1 and #2? Does it matter in which group Michael and Anthony will be? If no we should divide 6C3*3C3 by 2! and get 10 different split ups of 6 to groups of 3 when order of groups doesn't matter.

# of groups with Michael and Anthony together: 1C1*1C1*4C1=4
P=4/10=40%.

The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).
\(\frac{6!}{(3!)^2*2!}=10\)

OR another way:
In our case \(\frac{6C3*3C3}{2!}=10.\)
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 05 Nov 2009, 13:35
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Bunuel wrote:
srini123 wrote:
Bunuel - the fourth approach you posted seems to have an error:

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%

Total # of splitting group of 6 into two groups of 3: should be 6C3*3C3=20 and not 10.
Out of these 20 , there are 10 groups with Michael in it

so answer = groups with michael and anthony/ total groups with michael = 4/10 = 40%


Do we have group #1 and #2? Does it matter in which group Michael and Anthony will be? If no we should divide 6C3*3C3 by 2! and get 10 different split ups of 6 to groups of 3 when order of groups doesn't matter.

# of groups with Michael and Anthony together: 1C1*1C1*4C1=4
P=4/10=40%.

The number of ways in which mn different items can be divided equally into m groups, each containing n objects and the order of the groups is not important is \(\frac{(mn)!}{(n!)^m*m!}\).
\(\frac{6!}{(3!)^2*2!}=10\)

OR another way:
In our case \(\frac{6C3*3C3}{2!}=10.\)



Sorry I may be missing something but, if we want to split 6 persons into 2 groups of 3 where order doesn't matter. dont we have 20 groups ?
for eg., let A,B,C,D,E,F be 6 people and the
groups with 3 people in each will be as below

ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF - 10
BCD, BCE,BCE, BDE,BDF, BEF - 6
CDE, CDE,CEF -3
DEF -1

total 20.

Let A be michael for this question , then total groups we have with michael is 10 and 4 groups have Michael and Anthony (assume D=anthony)

im sorry im still not getting why 20 is not the total groups irrespective of michael in it or not
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 05 Nov 2009, 14:04
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srini123 wrote:
Sorry I may be missing something but, if we want to split 6 persons into 2 groups of 3 where order doesn't matter. dont we have 20 groups ?
for eg., let A,B,C,D,E,F be 6 people and the
groups with 3 people in each will be as below

ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF - 10
BCD, BCE,BCE, BDE,BDF, BEF - 6
CDE, CDE,CEF -3
DEF -1

total 20.

Let A be michael for this question , then total groups we have with michael is 10 and 4 groups have Michael and Anthony (assume D=anthony)

im sorry im still not getting why 20 is not the total groups irrespective of michael in it or not


Now I see what you've meant. But I think it's not quite true and here is where you are making the mistake: you listed # of selections of three out of 6 - 6C3=20. This is different from splitting the group into TWO groups of three members each:

1. ABC - DEF
2. ABD - CEF
3. ABE - CDF
4. ABF - CDE
5. ACD - BEF
6. ACE - BDF
7. ACF - BDE
8. ADE - BCF
9. ADF - BCE
10. AEF - BCD

So, here are all possible ways to split 6 people into two groups of three members each. Let's say Anthony is A and Michael is B. You can see that there are only 4 scenarios when A and B are in one group (1,2,3,4). Hence 4/10.
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 01 Oct 2010, 09:05
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: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 01 Oct 2010, 09:26
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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: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 04 Oct 2010, 12:56
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: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 04 Oct 2010, 13:12
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: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 21 Jan 2011, 16:54
Bunuel, can you advise why do we have to divide 6C3 * 3C3 by 2!, in the fourth approach ?
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Re: Anthony and Michael sit on the six-member board of directors for compa  [#permalink]

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New post 21 Jan 2011, 17:14
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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 for compa  [#permalink]

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New post 24 Feb 2012, 14:58
Bunuel wrote:
Fourth approach:
Total # of splitting group of 6 into two groups of 3: 6C3*3C3/2!=10


Do you mind elaborating on that calculation, please? Isn't 6C3 the right calculation to select 3 people from a group of 6? You seem to be multiplying that by 3C3/2!

Is there another example that illustrates this - 6C3*3C3/2!- concept, thanks.
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Re: Anthony and Michael sit on the six-member board of directors for compa   [#permalink] 24 Feb 2012, 14:58

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