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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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A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

45% (03:40) correct
55% (01:56) wrong based on 82 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

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

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

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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Expert's post

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

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

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.

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

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, \(w>6\).

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

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

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

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. _________________

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

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.

you rock buddy

_________________

GGG (Gym / GMAT / Girl) -- Be Serious

Its your duty to post OA afterwards; some one must be waiting for that...

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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01 Nov 2015, 04:33

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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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....

gmatclubot

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

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