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# If 2 different representatives are to be selected at random from a

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

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08 Jan 2010, 14:18
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Difficulty:

85% (hard)

Question Stats:

47% (02:34) correct 53% (01:54) wrong based on 90 sessions

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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
[Reveal] Spoiler: OA

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

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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?
[Reveal] Spoiler:
E

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

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

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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

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

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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

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.

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

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23 Jun 2010, 13:54
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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

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.

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 from a [#permalink]

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23 Jun 2010, 13:59
Thanks Bunuel, Got it. +1 for you.

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

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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?
[Reveal] Spoiler:
E

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

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 from a [#permalink]

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11 Jul 2010, 07:44
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Expert's post
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?
[Reveal] Spoiler:
E

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

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 from a [#permalink]

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12 Jul 2010, 11:47
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?
[Reveal] Spoiler:
E

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

you rock buddy

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

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02 Jul 2014, 10:34
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Re: If 2 different representatives are to be selected at random from a [#permalink]

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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.

Posted from GMAT ToolKit
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Re: If 2 different representatives are to be selected at random from a [#permalink]

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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.

Posted from GMAT ToolKit

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

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01 Nov 2015, 04:33
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Re: If 2 different representatives are to be selected at random from a [#permalink]

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09 Feb 2016, 19:26
oh damn..got myself screwed hard time on this one..and all because of a small mistake..
I took:
1) 6W so-> 6/10 * 5/9 = 2/3 but it should be 2/6 or 1/3 which is not sufficient.
2) tested few possibilities:
M=1 - works -> W=9 - yes
M=2 - 1/5*1/9 = 1/45 - works, W=8 - works
M=3 -> 3/10*2/9 = 1/15 - works, W=7 - works
M=4 -> 2/5*1/3 = 2/15 - works, W=6 -> again same mistake as in 1.
M=5 -> not work - so considered 2 sufficient.

because of a small calculation mistake, I got it wrong..god I hate myself....

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Re: If 2 different representatives are to be selected at random from a   [#permalink] 09 Feb 2016, 19:26
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