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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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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 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? What is the probability of choosing 2 women out of 10 people \(\frac{w}{10}*\frac{w1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w1}{9}>\frac{1}{2}\) > \(w(w1)>45\) this is true only when \(w>7\). (w # of women \(<=10\)) So basically question asks is \(w>7\)? (1) \(w>5\) not sufficient. (2) \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), 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 from a [#permalink]
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08 Jan 2010, 16:49
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 \(\frac{C^2_w}{C^2_{10}}>\frac{1}{2}\) > \(\frac{\frac{w(w1)}{2}}{45}>\frac{1}{2}\) > > \(w(w1)>45\) > \(w>7\)? (1) \(w>5\), not sufficient. (2) \(C^2_{(10w)}\) # of selections of 2 men out of \(10w=m\) employees > \(\frac{C^2_{(10w)}}{C^2_{10}}<\frac{1}{10}\) > \(\frac{\frac{(10w)(10w1)}{2}}{45}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), 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 from a [#permalink]
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23 Jun 2010, 13:40
Bunuel wrote: sagarsabnis wrote: (2) \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), not sufficient
Answer E.
Hi Bunuel, How did you get w>6. I tried solving it by moving 9 on left side and expanding (w10) (w9)9<0 w^219w+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 from a [#permalink]
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23 Jun 2010, 13:54
sevenplus wrote: Bunuel wrote: sagarsabnis wrote: (2) \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), not sufficient
Answer E.
Hi Bunuel, How did you get w>6. I tried solving it by moving 9 on left side and expanding (w10) (w9)9<0 w^219w+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 \((10w)(9w)=(106)(96)=12>9\), but if \(w>6\), for instance 7, then \((10w)(9w)=(107)(97)=6<9\). 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? What is the probability of choosing 2 women out of 10 people \(\frac{w}{10}*\frac{w1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w1}{9}>\frac{1}{2}\) > \(w(w1)>45\) this is true only when \(w>7\). (w # of women \(<=10\)) So basically question asks is \(w>7\)? (1) \(w>5\) not sufficient. (2) \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), 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 from a [#permalink]
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11 Jul 2010, 07:44
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 \(\frac{w}{10}*\frac{w1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w1}{9}>\frac{1}{2}\) > \(w(w1)>45\) this is true only when \(w>7\). (w # of women \(<=10\)) So basically question asks is \(w>7\)? (1) \(w>5\) not sufficient. (2) \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), 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 BOA 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? What is the probability of choosing 2 women out of 10 people \(\frac{w}{10}*\frac{w1}{9}\) and this should be \(>1/2\). So we have \(\frac{w}{10}*\frac{w1}{9}>\frac{1}{2}\) > \(w(w1)>45\) this is true only when \(w>7\). (w # of women \(<=10\)) So basically question asks is \(w>7\)? (1) \(w>5\) not sufficient. (2) \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), not sufficient (1)+(2) \(w>5\), \(w>6\) not sufficient Answer E. 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, 22:39
@bunuel. Can't I do the following for second statement? (M/10) (M1)/9 < 1/10 M(M1) < 9 So, M < 3. Hence, W > 7. Sufficient. Pleas advise 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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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....




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