Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

23 Jan 2008, 07:12

4

This post received KUDOS

13

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

31% (02:35) correct
69% (01:44) wrong based on 528 sessions

HideShow timer Statistics

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?

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

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.

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..... _________________

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.

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.... thanks in advance

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

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.... thanks in advance

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

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

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%. _________________

------------------------------------------------------------- When you come to the end of your rope, tie a knot and hang on.

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.

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% _________________

------------------------------------------------------------- When you come to the end of your rope, tie a knot and hang on.

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.

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

gmatclubot

Re: Power Prep: Subcommittee
[#permalink]
22 Aug 2008, 11:56

Excellent posts dLo saw your blog too..!! Man .. you have got some writing skills. And Just to make an argument = You had such an amazing resume ; i am glad...

So Much $$$ Business school costs a lot. This is obvious, whether you are a full-ride scholarship student or are paying fully out-of-pocket. Aside from the (constantly rising)...

I barely remember taking decent rest in the last 60 hours. It’s been relentless with submissions, birthday celebration, exams, vacating the flat, meeting people before leaving and of...

Rishabh from Gyan one services, India had a one to one interview with me where I shared my experience at IMD till now. http://www.gyanone.com/blog/life-at-imd-interview-with-imd-mba/ ...