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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

With my little knowledge, I will try to explain.

If I have understood, you have below understanding:

If we have chosen first guy on a particular team, for the next guy there is only one option. That is nothing but the same team. And for first guy there are two ways to go in any team. So According to you there can be following teams

M A _ || _ _ _ M _ _ || A _ _ A _ _ || M _ _ _ _ _ || M A _

PS: I am only considering combinations here.

With this understanding, you figured out that there can be 2 possible ways for M and A to be in single team and total ways could be 4. = 50%

But, One thing I would like you to give a thought on..

Lets say other four people are K, L, M, N

Now, for the first combination,

M A K || LMN

and

M A L || KMN

are different teams altogether even though M and A are in same team.

So basically for this first Choice there are 4 other combinations possible.

Now for the second choice there are M _ _ 4C2 combinations possible.

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

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12 Jun 2017, 01:43

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 \(\frac{2}{5}=40\%\).

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.

Hi Bunuel, I did this one with your 4th approach. I cannot understand your 1st approach. Can you please elaborate that one? Thanks
_________________

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.

Hi Bunuel, I did this one with your 4th approach. I cannot understand your 1st approach. Can you please elaborate that one? Thanks

Forget about splitting the group. Consider this, Michael is choosing 2 from a group of 5 and the question asks about the probability that one of the chosen ones will be Anthony. The probability in this case would simply be 2/5.
_________________

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%

Let’s first determine the number of ways 2 three-person subcommittees can be formed from 6 people. The number of ways 3 people can be selected from 6 people for the first committee is 6C3 = (6 x 5 x 4)/(3 x 2) = 20. The number of ways 3 people can be selected from remaining 3 people for the second committee is 3C3 = 1. Thus, the number of ways 2 three-person subcommittees can be formed from 6 people is 20 x 1 = 20, if the order of selecting the committees matters. However, since the order of selecting the committees doesn’t matter, we have to divide by 2! = 2. Thus, the number of ways 2 three-person subcommittees can be formed from 6 people is 20/2 = 10.

Since only a total 10 committees can be formed, we can list all of these committees and see how many of them have Anthony and Michael on the same committee. We can let A be Anthony, M be Michae,l and B, C, D, and E be the other 4 people.

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

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15 Jun 2017, 09:58

A 20% As when you select Michael the other two places can be filled by 5x4=20 ways. And when anthony is also selected so last position can be filled by 4 ways. So 4/20 is 20%

I understand why 40% is the right answer but I have a question on why this approach doesn't work

Considering there are 2 subcommities A and B Both Anthony and Michael can be in subcommity A or B

Therefore, we have: Anthony in A and Michael in A Anthony in B and Michael in A Anthony in A and Michael in B Anthony in B and Michael in B

They are in the same group in 2 of the 4 situations. Therefore 50%.

What part of this logic here is incorrect? Thanks!

Note what the question is: what percent of all the possible subcommittees that include Michael also include Anthony

So out of all subcommittees that include Michael, how many include Anthony too?

All subcommittees that include Michael = 1*5C2 = 10 Out of these 10, how many will have Anthony too? Michael, Anthony and the third person can be picked in 4 ways (out of 4 remaining people) = 4

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

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21 Jun 2017, 02:32

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%

If you are able to visualise the problem, you can solve it very easily. Lets take all the subcommittees which Michael is a part of. It obviously has Michael already and the remaining 2 ppl can be any 2 out of the 5 remaining people which includes Anthony. So, the probability of Anthony being included in the same sub committee will be 2/5 and hence the required answer will be 2/5 or 40%.

The other way of looking at the problem is to first find the total number of different sub committees which michael can be a part of. It will be selecting michael which is 1 way and selecting 2 other people in 5C2 ways. So, the total number of subcommittees michael is a part of will be 10. If we have to include Anthony as well, then there will be 4 different sub committees with each of the four other guys. So, the required percentage will be 4/10 or 40%.

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

Show Tags

03 Aug 2017, 03:48

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 \(\frac{2}{5}=40\%\).

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.

Hi Bunuel,

While forming 3 groups of 3 people each from 9 people, we divide 9c3*6c3*3c3 by 3! because order does not matter, but in this case, why don't we divide 1*1*4c1/3! ?