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tricky probability

by runzun » Fri Dec 09, 2011 10:22 pm
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%
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by rijul007 » Fri Dec 09, 2011 11:42 pm
Total no of ways of splitting 6 miembers into 2 committees => 6C3 => 20

If Micheal and Anthony are in the same team.. they need just one more member..
Total no of ways of making up a subcommittee => 4

%age = 4/20 * 100 => 20

Option A
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by gunjan1208 » Sat Dec 10, 2011 1:54 am
Thanks Rijul,

I could just understand 6C3 but I did not understand how to proceed further.

So I will take it as slot fill further:

1*1*4=4 (This is because at the third place any of the four person can show up.

Right?
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by Bharath Krishnan » Sat Dec 10, 2011 5:22 am
gunjan1208 wrote:Thanks Rijul,

I could just understand 6C3 but I did not understand how to proceed further.

So I will take it as slot fill further:

1*1*4=4 (This is because at the third place any of the four person can show up.

Right?
@ gunjan - with respect to your understanding of who will be filling up the 3rd slot to be a part of the subcommitte consisting of Michael and Anthony, this is how it should be done.

Number of ways splitting the 6 members into 2 committees => 6C3 = 20 (total)
Since Michael and Anthony are already part of the team, I require to choose only 1 person from the remaining 4, thus 4C1 ways.

Probability = 4C1/6C3 = 4/20 = 1/5

Percent of the possible subcommittees that include Michael and Anthony = (1/5)*100 = 20%
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by GMATGuruNY » Sat Dec 10, 2011 5:56 am
runzun 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%
Since the 6 people are being split into two 3-member committees, we know that Michael will serve on a committee.
The only question is whether Anthony will be among the pair of people chosen to serve with Michael.

Total number of pairs that can be formed from the 5 people aside from Michael = 5C2 = 10.
Total number of people (aside from Michael) that could be combined with Anthony to form a pair = 4.

(Pairs with Anthony)/(total possible pairs) = 4/10 = 40%.

The correct answer is C.

Another approach:
P(1st person chosen to serve with Michael is NOT Anthony) = 4/5.
P(2nd person chosen to serve with Michael is NOT Anthony) = 3/4.
Thus, P(Anthony is NOT chosen to serve with Michael) = 4/5 * 3/4 = 3/5.
Thus, P(Anthony IS chosen to served with Michael) = 1 - 3/5 = 2/5 = 40%.
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by quantskillsgmat » Sat Dec 10, 2011 9:56 pm
I think there are two slots of 6c3 .
for one slot it is 4/20
so total prob is 8/20
so desired percentage is 8/20 x100=40
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by runzun » Sun Dec 11, 2011 7:20 am
GMATGuruNY wrote:
runzun 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%
Since the 6 people are being split into two 3-member committees, we know that Michael will serve on a committee.
The only question is whether Anthony will be among the pair of people chosen to serve with Michael.

Total number of pairs that can be formed from the 5 people aside from Michael = 5C2 = 10.
Total number of people (aside from Michael) that could be combined with Anthony to form a pair = 4.

(Pairs with Anthony)/(total possible pairs) = 4/10 = 40%.

The correct answer is C.

Another approach:
P(1st person chosen to serve with Michael is NOT Anthony) = 4/5.
P(2nd person chosen to serve with Michael is NOT Anthony) = 3/4.
Thus, P(Anthony is NOT chosen to serve with Michael) = 4/5 * 3/4 = 3/5.
Thus, P(Anthony IS chosen to served with Michael) = 1 - 3/5 = 2/5 = 40%.

thanks for the explaination..
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by GMATGuruNY » Sun Dec 11, 2011 8:15 am
th
runzun wrote:
GMATGuruNY wrote:
runzun 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%
thanks for the explaination..
My pleasure.
There is an even quicker way to solve this problem.
Since the board is being split into two 3-member subcommittees, there are 6 positions to be filled by the 6 board members.
Michael will occupy one of these 6 positions.
Of the remaining 5 positions, 2 will be part of Michael's subcommittee.
Thus, Anthony has a 2/5 = 40% chance of being put on Michael's subcommittee.
Last edited by GMATGuruNY on Sun Dec 11, 2011 9:46 pm, edited 1 time in total.
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by GmatMathPro » Sun Dec 11, 2011 8:28 am
For the people who tried to go the 6C3 route, you should be aware that 6C3 overestimates the number of ways 6 people can be broken up into two groups. Suppose the people are ABCDEF. The logic is that once we choose 3 people, the other group will be made up of the remaining three, so we could have such groups as:

ABC DEF
ABD CEF
ABE DCF

and so on..., but eventually we'll get to

DEF ABC
CEF ABD
DCF ABE

All of the ways to break the group of 6 up will be repeated because all of the "remaining three" in the first set will later become the "chosen three" as you work through all of the 6C3=20 groups. Thus, 20 is two times too big. It should be 10.

Then when you do 4/10, you get the correct answer of 40%.

Also, note that this strategy for dividing a group into two groups will work without providing repeats if you're not dividing them into equal groups. For example, if we wanted to divide the group of 6 into a group of 4 and a group of 2, then 6C2=15 is the right answer because in this case, the size of the chosen group is different from the size of the remaining group, so if you list them all out, there's no danger of the remaining groups also becoming the chosen group later on.
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by LalaB » Sun Dec 11, 2011 10:19 am
is my logic ok?

since we need A and M to be in the same sub-committy, we assume the following-

let any sub -committy be like this -

1st place =Anthony ;2nd place= Michael; 3d place =any of the remaining 4 persons of a board of directors

so, we have 1*1*4 (4=6-2)

to get 3 persons our of 6 > 6C3=20 (total)

since we have 2 sub-committies, and Anthony and Michael can be in any of these sub-committies, we have the following-

(2*4)/20=2/5=40%
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by urshohini » Sun Dec 11, 2011 11:57 am
Since the 6 people are being split into two 3-member committees, we know that Michael will serve on a committee.
The only question is whether Anthony will be among the pair of people chosen to serve with Michael.

Total number of pairs that can be formed from the 5 people aside from Michael = 5C2 = 10.
Total number of people (aside from Michael) that could be combined with Anthony to form a pair = 4.

(Pairs with Anthony)/(total possible pairs) = 4/10 = 40%.

The correct answer is C.

Another approach:
P(1st person chosen to serve with Michael is NOT Anthony) = 4/5.
P(2nd person chosen to serve with Michael is NOT Anthony) = 3/4.
Thus, P(Anthony is NOT chosen to serve with Michael) = 4/5 * 3/4 = 3/5.
Thus, P(Anthony IS chosen to served with Michael) = 1 - 3/5 = 2/5 = 40%.
Mitch,
Why aren't we considering the no of ways by which the 2 three member committee are chosen out of 6 people? like 6C3 ways?
Also should not we consider the no of ways by which Michael/Anthony are chosen at first? (like 6 ways), and then go on to solve the no of ways by which the 2nd person chosen is Anthony?
Pl explain.
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by pemdas » Mon Dec 12, 2011 3:01 am
We need to find the subcommittees of three members that include Michael, and the subcommittee which include Michael and Anthony as well.
Michael and Anthony can be both in four subcommittee of three members: Michael+Anthony+ variability of other member (MA-C, MA-D, MA-E, MA-F, C,D,E,F are other members)
Or change the set & arrange -> fix Michael and Anthony in their places, M A _
arrange (without order) four remaining members in one empty slot after M A (4C1, one place four members). With M A there will be 4 ways.

To find in how many groups Michael sits, fix him in his right place, M _ _
two empty slots must be taken by the arrangement (without order) of five remaining members for two slots. With M alone there will be 5C2 or 10 ways

Find the probability 4/10=40%

runzun 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%
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by pemdas » Mon Dec 12, 2011 3:21 am
revised question
Anthony and Michael sit on the 8 member board of directors for company X. If the board is to be split up into 2 four-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
we can check when Michael and Anthony are on the same subcommittees M A _ _ two slots and 6 members, 6C2=6!/(4!2!)=15

Check when Michael is on four-member subcommittees M _ _ _ three slots and 7 members to be arranged, 7C3=7!/(4!*3!)=35
The required probability is 15/35=3/7 [spoiler]~43%[/spoiler]
Last edited by pemdas on Mon Dec 12, 2011 10:07 am, edited 1 time in total.
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by LalaB » Mon Dec 12, 2011 9:24 am
pemdas wrote: agree about this, we can do so. In new q. we have 8-2=6
we can check M A _ one slot and 6 members, 6C1=6!/5!=6 (fine :) )
hey pemdas, thnx for ur help. dont u think that there is smth forgotten here. if u change the number of persons to 8, u will have 2 sub-committies with 4(!) ppl, not 3

still, I feel that I was wrong. I have read all solutions above, and I hope I understand them.

I think I misunderstood the q, and wrongly thought, that M cant be without A.
I thought we need to find out M+A/all, but actually the q stem asked M+A/M+rest (rest=all-M)
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by pemdas » Mon Dec 12, 2011 9:29 am
couldn't get the last one, is it something put in bold?
i'm arranging out of 8, six 6 members except for M A
LalaB wrote:
pemdas wrote: agree about this, we can do so. In new q. we have 8-2=6
we can check M A _ one slot and 6 members, 6C1=6!/5!=6 (fine :) )
hey pemdas, thnx for ur help. dont u think that there is smth forgotten here. if u change the number of persons to 8, u will have 2 sub-committies with 4(!) ppl, not 3

still, I feel that I was wrong. I have read all solutions above, and I hope I understand them.

I think I misunderstood the q, and wrongly thought, that M cant be without A.
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