Anthony and Michael sit on the six member board of directors : GMAT Problem Solving (PS)
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# Anthony and Michael sit on the six member board of directors

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Anthony and Michael sit on the six member board of directors [#permalink]

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23 Jan 2008, 06:12
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Question Stats:

30% (02:39) correct 70% (01:42) wrong based on 603 sessions

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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%

OPEN DISCUSSION OF THIS QUESTION IS HERE: anthony-and-michael-sit-on-the-six-member-board-of-directors-102027.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 31 Jul 2012, 02:10, edited 2 times in total.
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23 Jan 2008, 06:36
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A

$$ratio=100%*\frac{C^4_1}{C^6_3}=100%*\frac{4*3*2}{6*5*4}=20%$$
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23 Jan 2008, 07:19
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The answer is C. Let's look at a committee where Mike belongs (he's person # 1) on that committee. When we choose person #2, there is a 1/5 probability that it's going to be Anthony (then it doesn't matter who the third person is), and 4/5 probability that it's going to be someone else (then person #3 is going to be Anthony with probability 1/4). Total probability = 1/5+4/5*1/4 = 2/5
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23 Jan 2008, 07:49
I think you missed something...

$$p=\frac26*(\frac15*\frac11+\frac45*\frac14)+ \frac46*(\frac25*\frac14)=\frac{16}{120}+\frac{8}{120}=\frac15$$
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23 Jan 2008, 08:07
I don't think I missed anything. The answer has to be over 1/5.

Let's take any committee with Mike. If we take at random another person one that committee (one out of 2 remaining members), the probability that he will be Albert is 1/5 (there are 5 people besides Mike) which means that the probability is going to be at least 1/5. Obviously, the second person can also be Mike if the first person isn't.
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23 Jan 2008, 08:15
maratikus wrote:
I don't think I missed anything. The answer has to be over 1/5.

Let's take any committee with Mike. If we take at random another person one that committee (one out of 2 remaining members), the probability that he will be Albert is 1/5 (there are 5 people besides Mike) which means that the probability is going to be at least 1/5. Obviously, the second person can also be Mike if the first person isn't.

The probability to take any committee with Mike is 1/2.....
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23 Jan 2008, 08:25
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The question was: what percent of all the possible subcommittees that include Michael also include Anthony?

It's a conditional probability question: what is the probability of Anthony being a committee if Mike belongs there. If you'd like we can look at a situation with 4 people, 2 committees.
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23 Jan 2008, 08:36
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Ahh....You seem to be right. I still have troubles with understanding of problems...
Thanks
+1
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23 Jan 2008, 09:14
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kazakhb wrote:
maratikus wrote:
The answer is C. Let's look at a committee where Mike belongs (he's person # 1) on that committee. When we choose person #2, there is a 1/5 probability that it's going to be Anthony (then it doesn't matter who the third person is), and 4/5 probability that it's going to be someone else (then person #3 is going to be Anthony with probability 1/4). Total probability = 1/5+4/5*1/4 = 2/5

Maratikus, why did you multiply 4/5 by 1/4? what does it mean? can you explain....

Let's say we have M (Mike), A (Anthony), B, C, D, E.

we put together Mike's group. M is the first person. The second person we put in the group can either be A or someone else (B, C, D, E). Probability that it's going to be A is 1/5 and if that happens, we succeed because both M and A are in the same group. If it's not A (say it's B) - that happens with probability with 4/5, then we have M and B in the group and we have to pick M out of M, C, D, E (which happens with probability 1/4) to succeed. That's why total probability is 1/5 (which corresponds with A picked right away) + 4/5*1/4 (corresponds picking someone else first and then A) = 2/5

Does that help?
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23 Jan 2008, 09:42
maratikus wrote:
kazakhb wrote:
maratikus wrote:
The answer is C. Let's look at a committee where Mike belongs (he's person # 1) on that committee. When we choose person #2, there is a 1/5 probability that it's going to be Anthony (then it doesn't matter who the third person is), and 4/5 probability that it's going to be someone else (then person #3 is going to be Anthony with probability 1/4). Total probability = 1/5+4/5*1/4 = 2/5

Maratikus, why did you multiply 4/5 by 1/4? what does it mean? can you explain....

Let's say we have M (Mike), A (Anthony), B, C, D, E.

we put together Mike's group. M is the first person. The second person we put in the group can either be A or someone else (B, C, D, E). Probability that it's going to be A is 1/5 and if that happens, we succeed because both M and A are in the same group. If it's not A (say it's B) - that happens with probability with 4/5, then we have M and B in the group and we have to pick M out of M, C, D, E (which happens with probability 1/4) to succeed. That's why total probability is 1/5 (which corresponds with A picked right away) + 4/5*1/4 (corresponds picking someone else first and then A) = 2/5

Does that help?

yes, thanks, I have some problems with prob, perm and comb that is why I asked about it, it might be seen like easy stuff but for me it is another new discovery, so great thanks +1
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24 Jan 2008, 07:52
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kazakhb wrote:
Can we somehow solve it using combination formulas?

$$ratio=100%*\frac{C^4_1}{C^5_2}=100%*\frac{4*2}{5*4}=40%$$

kazakhb wrote:
BTW, when do we use combination/permutations with probability vs. when it's just probability?

Majority of probability problems have a few ways. I think that both ways are good.
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24 Jan 2008, 08:30
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kamilaak wrote:
Anthony and Michael sit on the six member board od directors for compnay X. If the board is to be split up into 2 three-person sybcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?

A. 20%
B. 30%
C. 40%
D. 50%
E. 60%

Let me give a crack at it. Lemme see if this makes sense. (I didn't get this right when I first did it, I guessed B, but maybe this way will work).

We have 6!/3!3! ways to divide the groups so 20 ways. then we have four ways of picking MAX, MXA or AMX AXM

But we have two groups so its 8 ways total.

8/20 = 2/5 or 40%.

Let me know if this is an appropriate way to solve this. Thx
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24 Jan 2008, 08:54
GMATBLACKBELT, maybe I'm wrong but your way seems to be incorrect.

GMATBLACKBELT wrote:

We have 6!/3!3! ways to divide the groups so 20 ways. then we have four ways of picking MAX, MXA or AMX AXM

But we have two groups so its 8 ways total.

8/20 = 2/5 or 40%.

Let me know if this is an appropriate way to solve this. Thx

1. MAX, MXA or AMX AXM is the same committee. A committee is not sensitive to permutation.
You have really 4 ways: MAX,MAY,MAZ,MAW.

2. the question is "what percent of all the possible subcommittees" but not "ways to divide".

3. we should use 5!/2!3! instead of 6!/3!3! in calculation of probability because we should only take committees with Michael.

Therefore 4/10
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24 Jan 2008, 08:56
kamilaak wrote:
OA is 40%. (c)
Can we somehow solve it using combination formulas?

We need to pick 2 people out of 5: C(5,2) = 10
We need to pick Anthony out of a group of 1 person: C(1,1) = 1, and one person out of 4: C(4,1)=4

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24 Jan 2008, 09:14
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It should be A (20%).

Lets assume M and A to be one entity. Now we need to select one more member to make a committee of 3 which has to have both M and A. We can select another (any) member from remaining 4 members in 4c1 = 4 ways.

We have 6c3 ways of selecting 3 members from a group of 6, and that should be it. After you have selected 3 members, the other 3 members left will be part of the other committee, there is no choice left for them. Hence 20 ways.

4 / 20 = 1 / 5 = 20%.
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24 Jan 2008, 10:44
suntaurian wrote:
It should be A (20%).
We have 6c3 ways of selecting 3 members from a group of 6, and that should be it. ... Hence 20 ways.
4 / 20 = 1 / 5 = 20%.

We should select 2 people out of 5 because Michael is already a member of a committee under consideration. Therefore, there 10 ways and probability is 2/5.
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24 Jan 2008, 10:59
It does not say that Micheal is already part of the committee, its just says...find all ways of making "subcommittees that include Michael also include Anthony? "....Its just another way of putting where Michjael and Anthony are part of the same subcommittee.

Still stand by A..20%
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24 Jan 2008, 15:07
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suntaurian wrote:
It does not say that Micheal is already part of the committee, its just says...find all ways of making "subcommittees that include Michael also include Anthony? "....Its just another way of putting where Michjael and Anthony are part of the same subcommittee.

Still stand by A..20%

Our universe consists of all subcommittees that include Michael (not all subcommittees). The question is how many of those subcommittees also include Anthony.
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25 Jan 2008, 11:11
ok C

Total Arrangements 6!=720

We figure out the number of possibilties when Mike sits in seat 1:

If he sits in seat 1, there are 5!=120 seats left. However Anthony can sit in only 2 seats out of the 5 so we multiply 2/5 * 120 = 48.

Since Michael has 6 different positions we get 48*6 = 288

288/720 = 40%
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22 Aug 2008, 10:56
walker wrote:
A

$$ratio=100%*\frac{C^4_1}{C^6_3}=100%*\frac{4*3*2}{6*5*4}=20%$$

Walker.. in the denominator you should be taking 6C3 / 2! ... I learnt this from one of your recent posts.
6C3 alone takes the order of the two committees into account. divide by 2! to eliminate that effect
Re: Power Prep: Subcommittee   [#permalink] 22 Aug 2008, 10:56

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