Tough probability question

Question

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?
20%
30%
40%
50%
60%

This question has been posted before, but I'm wondering if the following method is a viable way to get the answer:

Since Michael and Anthony need to be on the same committee, fix their position:

MA4

The "4" stands for the remaining people who can fill the seat on a committee with Michael and Anthony. Since the committee that contains M and A can be arranged in any manner, you must multiply by 3!. So there are 3! * 4= 24 ways to organize each committee. There are 2 possible committees = 48.

Each committee can be arranged in 6*5*4 ways =120

So the answer is: 48/120= 40%

Is this a viable way to get the answer? I'm wondering if multiplying by 3! is INCORRECT, since a committee with MXA is the same as XMA, thus, we'd be double counting.

Can an expert weigh in?

pseudononymous · Answer

I think your method calculates M&A/All rather than M&A/M. The combos for Michael should be in the denominator rather than the total combos. M&A/M = 4/5C2 = 40%

Stockmoose16 · Answer

pseudononymous wrote:I think your method calculates M&A/All rather than M&A/M. The combos for Michael should be in the denominator rather than the total combos.

M&A/M = 4/5C2 = 40%

Also, how did you get M&A being on the same committee as "4"?

It can be XMA, XAM, AMX, AXM, MAX, MXA... that's 6*2 committees =12

Stockmoose16 · Answer

pseudononymous wrote:I think your method calculates M&A/All rather than M&A/M. The combos for Michael should be in the denominator rather than the total combos.

M&A/M = 4/5C2 = 40%

Also, how did you get M&A being on the same committee as "4"?

It can be XMA, XAM, AMX, AXM, MAX, MXA... that's 6*2 committees =12

pseudononymous · Answer

Stockmoose16 wrote:pseudononymous wrote:I think your method calculates M&A/All rather than M&A/M. The combos for Michael should be in the denominator rather than the total combos. M&A/M = 4/5C2 = 40% Also, how did you get M&A being on the same committee as "4"? It can be XMA, XAM, AMX, AXM, MAX, MXA... that's 6*2 committees =12 i was talking about unique combos where order doesn't matter

Stockmoose16 · Answer

pseudononymous wrote:Stockmoose16 wrote:pseudononymous wrote:I think your method calculates M&A/All rather than M&A/M. The combos for Michael should be in the denominator rather than the total combos. M&A/M = 4/5C2 = 40% Also, how did you get M&A being on the same committee as "4"? It can be XMA, XAM, AMX, AXM, MAX, MXA... that's 6*2 committees =12 i was talking about unique combos where order doesn't matter But order does matter when you're doing probability. And this is a probability question. If I say, "what are the chances of picking a red marble and a blue marble from a bag that contains 4 red marbles and 6 blue marbles, you need to calculate the chances of getting a red then a blue, and then the other way around (blue then red). Why do you think order doesn't matter?

pseudononymous · Answer

In this problem, MAX is the same committee as XAM which is why order doesn't matter.

Numerator (4C1=4):
MA1
MA2
MA3
MA4

Denominator (5C2=10):
M12
M13
M14
M15
M23
M24
M25
M34
M35
M45

4/10 = 40%

If you care about order, you can multiply both the numerator and denominator by 3! which will give you the same answer:

4C1*3! / 5C2*3! = 24 / 60 = 40%

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