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From a group of M employees, N will be selected, at random,

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From a group of M employees, N will be selected, at random, [#permalink] New post 07 Aug 2014, 13:58
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From a group of M employees, N will be selected, at random, to sit in a line of N chairs. There are absolutely no restrictions, either in the selection process nor in the order of seating — both are entirely random. What is the probability that the employee Andrew is seated somewhere to the right of employee Georgia?
Statement #1: N = 15
Statement #2: N = M



For a set of DS questions on Probability, as well as the OE of this particular question, see:
http://magoosh.com/gmat/2013/gmat-data- ... obability/

Mike :-)
[Reveal] Spoiler: OA

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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 07 Aug 2014, 19:36
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mikemcgarry wrote:
From a group of M employees, N will be selected, at random, to sit in a line of N chairs. There are absolutely no restrictions, either in the selection process nor in the order of seating — both are entirely random. What is the probability that the employee Andrew is seated somewhere to the right of employee Georgia?
Statement #1: N = 15
Statement #2: N = M



For a set of DS questions on Probability, as well as the OE of this particular question, see:
http://magoosh.com/gmat/2013/gmat-data- ... obability/

Mike :-)


Statement 1: insufficient since we need to know M to solve for the probability
Statement 2: N= M
This turns to sit N people into N chairs
For A is seated somewhere to the right of employee Georgia, we have (1+ 2+ 3+....+ (n-1)) possibilities
To illustrate this, lets say 4 people sit in 4 chairs
G in seat 1, A can be seated in 2,3, and 4 ( 3 possibilities)
G in seat 2, A can be seated in 3,4 ( 2 possibilities)
G in seat 3, A can be seated in 4 only (1 possibility)
Sum = 1+ 2+ 3
After G and A are seated, there are (n-2)! possible combination
( for example, in this case, after G and A are seated, there are two seats left, we have 2! possible combination)
The total possible combinations for N people in N chairs is N!
The possible combinations that meet the requirement is (1+2+3+...(N-1))* (N-2)! = [(N-1)*N/2] * (N-2)!
The probability = [(N-1)*N/2] * (N-2)!/N! = [(N-1)*N/2]/ [(N-1)*N] = 1/2
B is the answer.
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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 08 Aug 2014, 00:01
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Mmm:) That's easy.

(1) Not sufficient because we don't know M.
(2) Sufficient.
You need to arrange N people on N seats in such way that the employee Andrew is seated somewhere to the right of employee Georgia. The main idea that you have the same number of possibilities to put Andrew to the right of Georgia and to put him on the left. Just think that in each case when Andrew to the right of Georgia, you can switch their places. That's why, the probability is 1/2.

The correct answer is B.
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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 08 Aug 2014, 00:03
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You can apply the same idea in this problem if-bob-and-jen-are-two-of-5-participants-in-a-race-how-many-37225.html

Race can finish in 6!=120 ways, and exactly half of them Jen always finishes in front of Bob. The correct answer is 60.
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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 08 Aug 2014, 09:56
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Dear smyarga,
Good job! You took the very elegant solution to this. :-)

Dear vad3tha,
You found the answer via brute force calculations, but recognize that this approach can be costly, in terms of time & energy, on the real GMAT. If you want to achieve a truly elite GMAT score, part of what that requires is the ability to see the elegant solutions that involve few or no calculations. See:
http://magoosh.com/gmat/2013/how-to-do- ... th-faster/

Does all this make sense?
Mike :-)
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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 10 Aug 2014, 18:39
mikemcgarry wrote:
Dear smyarga,
Good job! You took the very elegant solution to this. :-)

Dear vad3tha,
You found the answer via brute force calculations, but recognize that this approach can be costly, in terms of time & energy, on the real GMAT. If you want to achieve a truly elite GMAT score, part of what that requires is the ability to see the elegant solutions that involve few or no calculations. See:
http://magoosh.com/gmat/2013/how-to-do- ... th-faster/

Does all this make sense?
Mike :-)


I agree with you, also. This is not what I did when I first saw the problem. I answered by trying out: 4 people in 4 chairs and 5 people in 5 chairs. I came up with 1/2.
I posted this solution afterward in case if ppl want to see the conceptual part the problem. Thanks for reminding me that.
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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 29 Oct 2014, 09:52
mikemcgarry wrote:
From a group of M employees, N will be selected, at random, to sit in a line of N chairs. There are absolutely no restrictions, either in the selection process nor in the order of seating — both are entirely random. What is the probability that the employee Andrew is seated somewhere to the right of employee Georgia?
Statement #1: N = 15
Statement #2: N = M



For a set of DS questions on Probability, as well as the OE of this particular question, see:
http://magoosh.com/gmat/2013/gmat-data- ... obability/

Mike :-)


Dear Mike,

If the value of M is known for first option ( just for example...M=20) then how to calculate probability for first statement?
Can you please help?
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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 29 Oct 2014, 13:08
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sach24x7 wrote:

Dear Mike,

If the value of M is known for first option ( just for example...M=20) then how to calculate probability for first statement?
Can you please help?

Dear sach24x7
I'm happy to respond. :-)

My friend, that change would introduce an ambiguity into the question that doesn't exist in the original. If M = 20, and we are picking N = 15, do we mean
a) what is the probability that Andrew & Georgia are both among the 15, and that Andrew is to the right of Georgia?
or
b) given that Andrea & Georgia definitely are among the 15 selected, what is the probability that Andrew is to the right of Georgia?
The answer to (b) is simply 1/2, because of symmetry. That's an easy question. For the first question, question (a), we would need to use the techniques discussed here:
https://magoosh.com/gmat/2013/gmat-prob ... echniques/

Does all this make sense?
Mike :-)
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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 29 Oct 2014, 21:14
mikemcgarry wrote:
sach24x7 wrote:

Dear Mike,

If the value of M is known for first option ( just for example...M=20) then how to calculate probability for first statement?
Can you please help?

Dear sach24x7
I'm happy to respond. :-)

My friend, that change would introduce an ambiguity into the question that doesn't exist in the original. If M = 20, and we are picking N = 15, do we mean
a) what is the probability that Andrew & Georgia are both among the 15, and that Andrew is to the right of Georgia?
or
b) given that Andrea & Georgia definitely are among the 15 selected, what is the probability that Andrew is to the right of Georgia?
The answer to (b) is simply 1/2, because of symmetry. That's an easy question. For the first question, question (a), we would need to use the techniques discussed here:
https://magoosh.com/gmat/2013/gmat-prob ... echniques/

Does all this make sense?
Mike :-)


Thanks Mike!..

For option a) Will it be (20C2 / 20C15) * (1/2)

Please correct me if i;m wrong....
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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 29 Oct 2014, 22:17
mikemcgarry wrote:
From a group of M employees, N will be selected, at random, to sit in a line of N chairs. There are absolutely no restrictions, either in the selection process nor in the order of seating — both are entirely random. What is the probability that the employee Andrew is seated somewhere to the right of employee Georgia?
Statement #1: N = 15
Statement #2: N = M



For a set of DS questions on Probability, as well as the OE of this particular question, see:
http://magoosh.com/gmat/2013/gmat-data- ... obability/

Mike :-)


Since we have the constraint of Andrew being to the right of Georgia so it is clear that this means the answer divided by 2!

1) N = 15
ways of selection = C(M,N) = C(M,15)
M is unknown so insufficient.

2) N=M
ways of selection = C(M,N) = C(M,M) = 1
so, answer = 1/2! = 1/2
sufficient.
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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 30 Oct 2014, 10:38
Expert's post
sach24x7 wrote:
mikemcgarry wrote:
sach24x7 wrote:

Dear Mike,

If the value of M is known for first option ( just for example...M=20) then how to calculate probability for first statement?
Can you please help?

Dear sach24x7
I'm happy to respond. :-)

My friend, that change would introduce an ambiguity into the question that doesn't exist in the original. If M = 20, and we are picking N = 15, do we mean
a) what is the probability that Andrew & Georgia are both among the 15, and that Andrew is to the right of Georgia?
or
b) given that Andrew & Georgia definitely are among the 15 selected, what is the probability that Andrew is to the right of Georgia?
The answer to (b) is simply 1/2, because of symmetry. That's an easy question. For the first question, question (a), we would need to use the techniques discussed here:
https://magoosh.com/gmat/2013/gmat-prob ... echniques/

Does all this make sense?
Mike :-)


Thanks Mike!..

For option a) Will it be (20C2 / 20C15) * (1/2)

Please correct me if i;m wrong....

Dear sach24x7,
I'm happy to respond. :-)

Your denominator is right, but not your numerator.

Denominator = all possible group of 15 to be chosen from the 20 = 20C15

Numerator = all possible groups of 15 that include both Andrew & Georgia ---- well it's those two, and 13 others from the remaining 18, so that's 18C13.

P = [(18C13)/(20C15)]*0.5

You have to compare like to like. You can't compare all possible groups of 15 to all possible pairs, which is essentially what you did: that compares apples to oranges.

Does this make sense?

Mike :-)
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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 02 Nov 2014, 07:44
Hello Sir,(Magoosh)

Please help me clarify a doubt - what if A seat on 2nd last seat towards right of George n George seat in middle wont probability would be different in that scenario.
In Ds question if we have different ans - the ans will be not sufficient.
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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 02 Nov 2014, 07:59
Dear Mike,

Clear!!.....Thanks ! :-D
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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 03 Nov 2014, 10:16
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taleesh wrote:
Hello Sir,(Magoosh)

Please help me clarify a doubt - what if A seat on 2nd last seat towards right of George n George seat in middle wont probability would be different in that scenario.
In Ds question if we have different ans - the ans will be not sufficient.

Dear Taleesh,
I'm happy to respond. :-)
Believe it or not, where the two people sit doesn't matter. If there are N employees, and all N are selected to sit in some order in a row of N seats, then all permutations will be possible. For any scenario in which Georgia is to the right of Andrew, there's a matching scenario in which Georgia is to the left of Andrew.

Let's say N = 6, and the six employees are Andrew = A, Georgia = G, and four others {J, K, L, M}. We can create any scenario with A to the right, such as you describe:
M, K, G, J, A, L
and this is matched to a unique configuration with A to the left --- we simply switch the places of G & A
M, K, A, J, G, L
For each and every configuration with A to the right, we can find exactly one matching scenario with A to the left. That means, exactly half of the total scenarios have A to the right of G, and the other half, A to the left of G.

Does this make sense?
Mike :-)
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Re: From a group of M employees, N will be selected, at random, [#permalink] New post 03 Nov 2014, 11:42
Thanks alot (Magoosh). Got it well.
Re: From a group of M employees, N will be selected, at random,   [#permalink] 03 Nov 2014, 11:42
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