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If 2 different representatives are to be selected at random

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If 2 different representatives are to be selected at random  [#permalink]

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New post 08 Jan 2010, 14:18
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If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10

OPEN DISCUSSION OF THIS QUESTION IS HERE: https://gmatclub.com/forum/if-2-differe ... 68280.html
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 27 Feb 2012, 09:15
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If 2 different representatives are to be selected at random from a group of 10 employees
and if p is the probability that both representatives selected will be women, is p > 1/2


What is the probability of choosing 2 women out of 10 people \(\frac{w}{10}*\frac{w-1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w-1}{9}>\frac{1}{2}\) --> \(w(w-1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))

So basically question asks is \(w>7\)?

(1) More than 1/2 of the 10 employees are women --> \(w>5\) not sufficient.

(2) The probability that both representatives selected will be men is less than 1/10 --> \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\) and \(w>6\): \(w\) can be 7, answer NO or more than 7, answer YES. Not sufficient.

Answer E.

You can use Combinations, to solve as well:

\(C^2_w\) # of selections of 2 women out of \(w\) employees;

\(C^2_{10}\) total # of selections of 2 representatives out of 10 employees.

Q is \(\frac{C^2_w}{C^2_{10}}>\frac{1}{2}\) --> \(\frac{\frac{w(w-1)}{2}}{45}>\frac{1}{2}\) --> --> \(w(w-1)>45\) --> \(w>7\)?

(1) More than 1/2 of the 10 employees are women --> \(w>5\), not sufficient.

(2) The probability that both representatives selected will be men is less than 1/10 --> \(C^2_{(10-w)}\) # of selections of 2 men out of \(10-w=m\) employees --> \(\frac{C^2_{(10-w)}}{C^2_{10}}<\frac{1}{10}\) --> \(\frac{\frac{(10-w)(10-w-1)}{2}}{45}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\) and \(w>6\): \(w\) can be 7, answer NO or more than 7, answer YES. Not sufficient.

Answer E.

Hope it's clear.
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 27 Feb 2012, 09:11
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If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2 ?

(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10.
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 08 Jan 2010, 15:25
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sagarsabnis wrote:
If 2 different representatives are to be selected at random from a group of 10 employees
and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10

Please can any one try it?


What is the probability of choosing 2 women out of 10 people \(\frac{w}{10}*\frac{w-1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w-1}{9}>\frac{1}{2}\) --> \(w(w-1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))

So basically question asks is \(w>7\)?

(1) \(w>5\) not sufficient.

(2) \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\), \(w>6\) not sufficient

Answer E.
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 08 Jan 2010, 16:49
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sagarsabnis wrote:
eehhhh!!! I didnt read the question properly and I was using 10C2 for calculation


You can use C, for solving as well:

\(C^2_w\) # of selections of 2 women out of \(w\) employees;

\(C^2_{10}\) total # of selections of 2 representatives out of 10 employees.

Q is \(\frac{C^2_w}{C^2_{10}}>\frac{1}{2}\) --> \(\frac{\frac{w(w-1)}{2}}{45}>\frac{1}{2}\) --> --> \(w(w-1)>45\) --> \(w>7\)?

(1) \(w>5\), not sufficient.

(2) \(C^2_{(10-w)}\) # of selections of 2 men out of \(10-w=m\) employees --> \(\frac{C^2_{(10-w)}}{C^2_{10}}<\frac{1}{10}\) --> \(\frac{\frac{(10-w)(10-w-1)}{2}}{45}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\), \(w>6\) not sufficient

Answer E.
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 23 Jun 2010, 13:40
Bunuel wrote:
sagarsabnis wrote:

(2) \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

Answer E.


Hi Bunuel,

How did you get w>6. I tried solving it by moving 9 on left side and expanding (w-10) (w-9)-9<0
w^2-19w+81<0
If we solve this quadratic we don't get a whole number for w.

Could you please elaborate? Your approach is very good as always.
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 23 Jun 2010, 13:54
2
sevenplus wrote:
Bunuel wrote:
sagarsabnis wrote:

(2) \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

Answer E.


Hi Bunuel,

How did you get w>6. I tried solving it by moving 9 on left side and expanding (w-10) (w-9)-9<0
w^2-19w+81<0
If we solve this quadratic we don't get a whole number for w.

Could you please elaborate? Your approach is very good as always.


As you correctly noted \(w\) must be an integer (as \(w\) represents # of women). Now, substituting: if \(w=6\), then \((10-w)(9-w)=(10-6)(9-6)=12>9\), but if \(w>6\), for instance 7, then \((10-w)(9-w)=(10-7)(9-7)=6<9\). So, \(w>6\).

Hope it's clear.
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 11 Jul 2010, 05:53
Bunuel wrote:
sagarsabnis wrote:
If 2 different representatives are to be selected at random from a group of 10 employees
and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10

Please can any one try it?


What is the probability of choosing 2 women out of 10 people \(\frac{w}{10}*\frac{w-1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w-1}{9}>\frac{1}{2}\) --> \(w(w-1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))

So basically question asks is \(w>7\)?

(1) \(w>5\) not sufficient.

(2) \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\), \(w>6\) not sufficient

Answer E.



hi, since from basic data we find that w>7 and
2. statemant state that at least 2 men are there out of 10 ........ it is asking that . probability of choosing 2 men is 1/10 . ie. atleast 2 men are there. ....
so, now combining w>7 and two men the only option left is w=8.
hence ans is B
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 11 Jul 2010, 07:44
1
wrldcabhishek wrote:
Bunuel wrote:
sagarsabnis wrote:
If 2 different representatives are to be selected at random from a group of 10 employees
and if p is the probability that both representatives selected will be women, is p > 1/2

(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10

Please can any one try it?


What is the probability of choosing 2 women out of 10 people \(\frac{w}{10}*\frac{w-1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w-1}{9}>\frac{1}{2}\) --> \(w(w-1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))

So basically question asks is \(w>7\)?

(1) \(w>5\) not sufficient.

(2) \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\), \(w>6\) not sufficient

Answer E.



hi, since from basic data we find that w>7 and
2. statemant state that at least 2 men are there out of 10 ........ it is asking that . probability of choosing 2 men is 1/10 . ie. atleast 2 men are there. ....
so, now combining w>7 and two men the only option left is w=8.
hence ans is B


OA for this question is E. The problem with your solution is that we are not given that w>7, in fact we are asked to find whether this is true.
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 08 May 2012, 01:41
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I made a silly mistake ...which i thought is worth sharing .
My answer was D.
I thought that since number of women is greater than 5 so probability will be greater than ½ so A suff.
But A is insuff . coz the above statement will hold true only for a single event , but here 2 things are to be selected .
So better make equations and then derive condition.

Correct me if i am wrong.
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 06 Aug 2012, 10:32
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Guys,

The way i attacked this problem was that quesn asks me if p(W,W) >1/2 ? Therefore,

(1) gives me women as 6, 7, 8, 9 (can't be 10).
Now, for 6 women, the probab would be p(W,W) = 6/10 * 5/9 = 1/3 .....less than half
Now, for 7 women, the probab would be p(W,W) = 7/10 * 6/9 = 7/15 .....less than half
Now, for 8 women, the probab would be p(W,W) = 8/10 * 7/9 = 28/45 .....more than half
Now, for 9 women, the probab would be p(W,W) = 9/10 * 8/9 = 4/5...more than half
Clearly, (1) is insufficient to answer... [eliminating A & D]

(2) gives me p(M,M) <1/10. Now, this is insuff. as it tells nothing abt p(W,W) unless i verify the above finding of (1) and club both
[B also eliminated, now contention is between C & E]
For 4 men, p(M,M) = 4/10 * 3/9 = 2/15 (grtr than 1/10) .....not valid
For 3 men, p(M,M) = 3/10 * 2/9 = 1/15 (less than 1/10) ...valid
For 2 men, p(M,M) = 2/10 * 1/9 = 1/45 (less than 1/10) ...valid

Thus, for 7W3M => p(W,W)<1/2 & p(M,M)<1/15
And, for 8W2M => p(W,W)>1/2 & p(M,M)< 1/45

So, combining 2 stmts is still insufficient to answer the original quesn. Hence, E has to be correct answer.
[PS: Initially i chose C, as i couldn't understand Bunuel's explanation above {which is a rarity :P}, but as i was posting this query, i realized that while choosing C, i didn't considered the 2 men case & that's why i chose wrongly ]
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 11 Aug 2012, 05:00
how to reach the final statement


w> 6 from 2 we had (10-w)(9-w) < 9 , , why is w>6
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 11 Aug 2012, 05:11
is plugging numbers only way to solve this quadratic inequality or do we have an algebric approach ?
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 17 Aug 2012, 05:10
PUNEETSCHDV wrote:
how to reach the final statement


w> 6 from 2 we had (10-w)(9-w) < 9 , , why is w>6


If w=6 then (10-w)(9-w)=4*3=12>9 and if w=7, then (10-w)(9-w)=3*2=6<9. When we increase w, from 7 to 10, (10-w)(9-w) decreases so w can be 7, 8, 9 or 10.

PUNEETSCHDV wrote:
is plugging numbers only way to solve this quadratic inequality or do we have an algebric approach ?


We could expand (10-w)(9-w) and then solve quadratic inequality, but number plugging for this particular case is better.

Check this: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476 (solving quadratic inequalities)
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 04 Nov 2012, 12:31
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JayGriffith8 wrote:
If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2?
(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10.

Please let me know if my methodology is correct. Say there are 6 women wouldn't the probability be 6/10*5/9? Yielding 1/3? I understand the maths behind this but I need to know a simple fast way of deriving an answer in this case. I feel like I'd be going back and forth with scenarios and eating too much time.


No. of Women X
Probability of selecting 2 Women >1 /2
X*(X-1)/2 / (10*9/2) > 1/2
X(X-1)>45
so X should be greater than 7.
or No. of Men should be less than 3

Statement A: Women can be 6,7,8,9,10 NS
Statement B: M(M-1)/10*9 < 1/10
M(M-1) < 9
M can be 0,1,2,3
NS.

Combining also NS.
Ans E
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 06 Nov 2012, 12:25
Bunuel wrote:
breakit wrote:
Bunuel wrote:
If 2 different representatives are to be selected at random from a group of 10 employees
and if p is the probability that both representatives selected will be women, is p > 1/2


What is the probability of choosing 2 women out of 10 people \(\frac{w}{10}*\frac{w-1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w-1}{9}>\frac{1}{2}\) --> \(w(w-1)>45\) this is true only when \(w>7\). (w # of women \(<=10\))

So basically question asks is \(w>7\)?

(1) More than 1/2 of the 10 employees are women --> \(w>5\) not sufficient.

(2) The probability that both representatives selected will be men is less than 1/10 --> \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\) and \(w>6\): \(w\) can be 7, answer NO or more than 7, answer YES. Not sufficient.

Answer E.

You can use Combinations, to solve as well:

\(C^2_w\) # of selections of 2 women out of \(w\) employees;

\(C^2_{10}\) total # of selections of 2 representatives out of 10 employees.

Q is \(\frac{C^2_w}{C^2_{10}}>\frac{1}{2}\) --> \(\frac{\frac{w(w-1)}{2}}{45}>\frac{1}{2}\) --> --> \(w(w-1)>45\) --> \(w>7\)?

(1) More than 1/2 of the 10 employees are women --> \(w>5\), not sufficient.

(2) The probability that both representatives selected will be men is less than 1/10 --> \(C^2_{(10-w)}\) # of selections of 2 men out of \(10-w=m\) employees --> \(\frac{C^2_{(10-w)}}{C^2_{10}}<\frac{1}{10}\) --> \(\frac{\frac{(10-w)(10-w-1)}{2}}{45}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\), not sufficient

(1)+(2) \(w>5\) and \(w>6\): \(w\) can be 7, answer NO or more than 7, answer YES. Not sufficient.

Answer E.

Hope it's clear.


Hi I couldn't able to understand the write uo which i have marked in red . Please help here..(not sure how to mark those picture stuff)


Check here: if-2-different-representatives-are-to-be-selected-at-random-128233.html#p1113622

Hope it helps.




Sorry , i am not understand how that formula stuff has been written(kind of lagging in that field)

[color=#ff0000] \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\),
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 07 Nov 2012, 05:44
breakit wrote:
Bunuel wrote:
breakit wrote:

Hi I couldn't able to understand the write uo which i have marked in red . Please help here..(not sure how to mark those picture stuff)


Check here: if-2-different-representatives-are-to-be-selected-at-random-128233.html#p1113622

Hope it helps.




Sorry , i am not understand how that formula stuff has been written(kind of lagging in that field)

[color=#ff0000] \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) --> \(w>6\),


\(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> cross-multiply (multiply both parts by 10*9): \((10-w)(10-w-1)<9\) --> \((10-w)(9-w)<9\).

Hope it's clear.
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 13 Mar 2013, 16:44
If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2 ?

(1) More than 1/2 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10.

This is extreme value problem

for p > 1/2 , p1 * p2 > 1/2 ie p1 or p2 > 1/4

We move with this further .
1. if P-new> 1/2 are women , then it can be say 1/2 (plus some point say .51) * .51 which is not sufficient

2. if P-men < 1/10 then P-women will be 9/10 ie there are many values between 1/4 and 9/10 which is not sufficient

Hence E
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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 02 Jul 2014, 22:39
@bunuel. Can't I do the following for second statement?

(M/10) (M-1)/9 < 1/10

M(M-1) < 9

So, M < 3. Hence, W > 7.

Sufficient.

Pleas advise

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Re: If 2 different representatives are to be selected at random  [#permalink]

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New post 03 Jul 2014, 06:17
smit29may wrote:
@bunuel. Can't I do the following for second statement?

(M/10) (M-1)/9 < 1/10

M(M-1) < 9

So, M < 3. Hence, W > 7.

Sufficient.

Pleas advise

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From M(M-1) < 9 --> \(M\leq{3}\) (notice that M could be 3 here), thus \(W\geq{7}\).

Hope it's clear.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: If 2 different representatives are to be selected at random &nbs [#permalink] 03 Jul 2014, 06:17

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