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Anthony and Michael sit on the six-member board of directors for compa 17 Dec 2015, 04:32
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Bunuel
Barkatis
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%

Answer: C.

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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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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Anthony and Michael sit on the six-member board of directors for compa 15 Jun 2016, 11:34
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Barkatis
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%
Show more

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%

Answer:
Show Spoiler
C
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Anthony and Michael sit on the six-member board of directors for compa 14 Jun 2017, 16:25
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Barkatis
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%
Show more

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.

1) A-B-C, D-E-M
2) A-B-D, C-E-M
3) A-B-E, C-D-M
4) A-B-M, C-D-E
5) A-C-D, B-E-M
6) A-C-E, B-D-M
7) A-C-M, B-D-E
8) A-D-E, B-C-M
9) A-D-M, B-C-E
10) A-E-M, B-C-D

We can see that from the 10 committees that can be formed, 4 of them (in bold) include both Anthony and Michael. Thus, the probability is 4/10 = 40%

Answer: C
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Anthony and Michael sit on the six-member board of directors for compa 15 Jun 2017, 10:45
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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!
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Anthony and Michael sit on the six-member board of directors for compa 15 Jun 2017, 11:00
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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!
Show more

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

Hence 4/10
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Anthony and Michael sit on the six-member board of directors for compa 25 May 2022, 01:27
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Backward logic approach:
Probability of selecting Antony = 1-Probability of not selecting Antony
So, 1-(4/5*3/4)=2/5
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Anthony and Michael sit on the six-member board of directors for compa 13 Feb 2023, 22:11
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Total ways of selecting 3 people from 6 people = 6C3 = 20
M and A constitute a set as they will always be together.

Total committees to be formed = 2

Selecting any one committee from 2 committees = 2C1 = 2
Select 1 person from the remaining 4 people = 4C1 = 4

Required ways = 4C1 * 2C1 = 8
Total way = 6C3 = 20

8/20 = 2/5
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Anthony and Michael sit on the six-member board of directors for compa 12 Apr 2023, 15:04
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Bunuel
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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Can you explain the line "# of groups with Michael and Anthony: \(C^1_1*C^1_1*C^1_4=4\)."

how did you arrive at C^1 and C^4
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Anthony and Michael sit on the six-member board of directors for compa 13 Apr 2023, 00:10
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adgarg
Bunuel
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.


Can you explain the line "# of groups with Michael and Anthony: \(C^1_1*C^1_1*C^1_4=4\)."

how did you arrive at C^1 and C^4
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1 way to choose Michael, 1 way to choose Anthony, and 4 ways to choose the third member of the subcommittee from the remaining 4 members.
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Anthony and Michael sit on the six-member board of directors for compa 14 Apr 2023, 12:36
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There are 20 possibilities in creating two sub-committees. When you fix Michael and Anthony in the first committee, there are 4 ways to pick the 3rd person, hence 4 ways the first committee can be formed. Same goes for the 2nd committee: if Michael and Anthony are both in the 2nd committee, there are 4 ways to form it. So 4 ways to form the first committee, 4 ways to form the second committee makes it 8 ways to form two committees in which both Michael and Anthony are in one. 8/20 = 40%.
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Anthony and Michael sit on the six-member board of directors for compa 21 Apr 2023, 19:46
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Another alternative approach:

Assume that both the persons are in the same group.
We need to select any 1 group from the 2 available groups = 2C1 = 2

For the remaining person, we need to select just 1 person from 4 persons = 4C1 = 4

Ways to select 3 persons from 6 persons = 6C3 = 20

Answer = (4*2)/20 = 8/20 = 40%
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Anthony and Michael sit on the six-member board of directors for compa 08 Apr 2024, 03:44
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kamilaak

Given: Anthony and Michael sit on the six member board of directors for company X.
Asked: 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?

Total possible sub-committees = 6C3 = 20
All the possible subcommittees that include Michael also include Anthony = 4C1 *2= 4*2

The percent of all the possible subcommittees that include Michael also include Anthony = 8/20 = 2/5 = 40%

IMO C
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Anthony and Michael sit on the six-member board of directors for compa 07 Oct 2025, 12:56
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There are 3 people each in 2 sub committees.
Total no of ways to form SC = Choose 3 people from 6 for SC1 . Rest 3 are automatically selected for SC2. Ways = \(6C3\) =20.
No of ways in which M and A are together :
They are together in SC1 - only 1 person from remaining 4 has to be chosen - \(4C1\) = 4
They are together in SC2 - 3 people have to selected from remaining 4 for SC1 - \(4C3\) =4. (Doesn't matter who is the 3rd person in SC2 as they are automatically selected after 3 are chosen to SC1)

Probability = \frac{(4+4)}{20} =\frac{2}{5} =40%
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Anthony and Michael sit on the six-member board of directors for compa 05 Jan 2026, 21:04
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Should we not include possible cases of which of the two committee will it be i.e. 4/10 * 2c1 ?
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Anthony and Michael sit on the six-member board of directors for compa 05 Jan 2026, 22:53
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Should we not include possible cases of which of the two committee will it be i.e. 4/10 * 2c1 ?
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No. The question counts 3-person subcommittees containing Michael, not labeled “Committee 1 vs Committee 2.” Multiplying by 2 would double count.
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