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

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.
_________________

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:
_________________

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
_________________

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

Hence 4/10
_________________

Backward logic approach:
Probability of selecting Antony = 1-Probability of not selecting Antony
So, 1-(4/5*3/4)=2/5

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

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

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.
_________________

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%.

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%