Problem Solving

sudhir3127 wrote:Hi jamesk486..i would be nice if u can post the OA. I request Ian/ Stuart to pitch in and clear the confusion. For me its still 40%.

The answer is certainly 40%, and I think Sudhir's first post above offers the most elegant possible solution here. Pepeprepa's solution is a more 'conventional' solution, and is also very quick- that's the solution I probably would have come up with during a test.

One key here is reading the question carefully (misreading the question is one of the easiest traps to fall into on GMAT questions, and it seems it happened here to a few people): what percent of all the possible subcommittees that include Michael also include Anthony?

We only want to count subcommittees that include Michael; we don't want to count all subcommittees. As Pepeprepa points out, only 5C2 subcommittees include Michael. 6C3 is the total number of subcommittees, whether they include Michael or not.

Can someone please explain why 5C2 is the number of sub-commitees on which Michael can be? Any input would be highly appreciated. Thanks in advance.

hi sorry..the OA is 40%

pranavc wrote:Can someone please explain why 5C2 is the number of sub-commitees on which Michael can be? Any input would be highly appreciated. Thanks in advance.

A subcommittee contains 3 people, and if Michael is one of those people, you need to choose 2 additional people from the remaining 5 people, which you can do in 5C2 ways.

I used Ian Stewart's approach, which he mentioned in the question about Olympians.

Probability that M will be in any group - 1
Probability that A will be in the same group - 2/5
2 - quantity of vacant places, 5 - q-ty of people for these 2 places
2/5-40%

This makes my previous answer incorrect, because 4/6C3 should be multiplied on 2 - M can be in 2 groups.

[quote="4meonly]
All possibilities to make group of 3 from 6 is 6C3 = 6!/(3!3!)=20
I cant find mistake in my reasoning. [/quote]

you have to take all possibilities in which Michael is included. 6C3 gives you all possibilities irrespective of Michael. As per quest we have to take possibilities in which Michael is included and that is 5C2 = 10

AlaaQasem wrote:I think the answer is 20% as well. Here is how I solved it. I'm not sure of the answer, though.

What we need to find is:
(The probability of having Anthony in sub-committee 1 AND the probability of having Michael in sub-committee 1)
OR
(The probability of having Anthony in sub-committee 2 AND the probability of having Michael in sub-committee 2)

Mathematically, this will translate into:
The probability of having Anthony in sub-committee 1 = count of individual to select/count of available individuals to select from = 1/6
The probability of having Michael in sub-committee 1 = count of individual to select/count of available individuals to select from exculding Anthony = 1/5
The probability of having Michael in sub-committee 2 = count of individual to select/count of available individuals to select from exculding the three individuals already assigned to sub-committtee 1 = 1/3
The probability of having Michael in sub-committee 2 = count of individual to select/count of available individuals to select from exculding the three individuals already assigned to sub-committtee 1 and Anthony = 1/2

Now substitute the numbers above:
(1/6)*(1/5)+(1/3)*(1/2) = (1/30) + (1/6) = (1+5)/30 = 6/30 = 1/5 = 20%

several things are wrong with this approach.

first of all, the probability that michael is on the first sub-committee is not 1/6. if you picked the three members of that sub-committee in order, 1/6 would be the probability that michael would be picked first, but it's definitely not the probability that he gets on the first sub-committee altogether, as he could also be picked second or third.

the same thing goes for the probability that michael goes onto the second subcommittee. it appears that you're trying to calculate the probability that this happens if michael isn't already selected for the first subcommittee (because you're assuming that he is one of the 3 "remaining" people) - but that probability would be 1, or 100%, because he would have to be assigned to the second subcommittee if he isn't picked for the first one.

in fact, it's easy to see that the probability that michael gets on the 1st sub-committee is 1/2, as is the probability that he gets onto the second sub-committee, because there are 2 sub-committees of equal size and he must be assigned to one of them. by symmetry, there is no way that those probabilities could be anything other than 1/2 each.

--

in any case, this approach, as well as many of the other approaches posted here, is ignoring the ESSENTIAL FACT that you're only picking from committees that include michael in the first place.
if we call michael M, anthony A, and the others 1, 2, 3, and 4, then there are only 10 possible committees (all equiprobable) that include M:
MA1
MA2
MA3
MA4
M12
M13
M14
M23
M24
M34
(notice that this is 5C2 = (5!)/(2!3!) = 10, because michael is fixed and you're effectively just finding the number of ways of selecting the other 2 members of the subcommittee)
of these, 4 include anthony, so the desired percentage is 4/10 = 40%.

--

also, there's no point in conflating this discussion with "probability" at all, because the problem isn't stated in terms of probability. why introduce such a troublesome concept where it's not even needed?

pepeprepa wrote:For me, with 6C3 you count the total number of possibilities
With 5C2, you count all the commitees in which you have Michael.

If you use 5C2, how do you know that Anthony is one of the 2 that is picked? I understand that you fixed Michael in the equation, but you don't know definitively that when you select your 2 people from the 5 that you'll get Anthony.

Ian Stewart wrote:

pranavc wrote:Can someone please explain why 5C2 is the number of sub-commitees on which Michael can be? Any input would be highly appreciated. Thanks in advance.

A subcommittee contains 3 people, and if Michael is one of those people, you need to choose 2 additional people from the remaining 5 people, which you can do in 5C2 ways.

Ian,

If you assume Michael is fixed on one board, and you then have 2 spots left to fill, why would 5C2 work? Doesn't that mean you have 5 people: A, 1,2, 3,4 who could all be picked for those final two spots? What happens if, when you pick 2 from the 5 available, you select 1 and 2. Then the board would consist of Michael, 1, 2; not M, A, 1 or M, A, 2. Please explain.

Ian Stewart wrote:

sudhir3127 wrote:Hi jamesk486..i would be nice if u can post the OA. I request Ian/ Stuart to pitch in and clear the confusion. For me its still 40%.

The answer is certainly 40%, and I think Sudhir's first post above offers the most elegant possible solution here. Pepeprepa's solution is a more 'conventional' solution, and is also very quick- that's the solution I probably would have come up with during a test.

One key here is reading the question carefully (misreading the question is one of the easiest traps to fall into on GMAT questions, and it seems it happened here to a few people): what percent of all the possible subcommittees that include Michael also include Anthony?

We only want to count subcommittees that include Michael; we don't want to count all subcommittees. As Pepeprepa points out, only 5C2 subcommittees include Michael. 6C3 is the total number of subcommittees, whether they include Michael or not.

Also, is there a way to figure out how many ways Anthony and Michael are in the same group without enumerating the combinations? It seems that everyone has discovered its "4" by enumeration.