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If 2 different representatives are to be selected at random [#permalink]
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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/if2differe ... 68280.html
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Re: If 2 different representatives are to be selected at random [#permalink]
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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 [#permalink]
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08 Jan 2010, 15:49
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 [#permalink]
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23 Jun 2010, 12:40
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 [#permalink]
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23 Jun 2010, 12:54
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 [#permalink]
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11 Jul 2010, 04:53
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 [#permalink]
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11 Jul 2010, 06:44
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 [#permalink]
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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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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 > 1/2What 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) 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 > \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), 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 \(\frac{C^2_w}{C^2_{10}}>\frac{1}{2}\) > \(\frac{\frac{w(w1)}{2}}{45}>\frac{1}{2}\) > > \(w(w1)>45\) > \(w>7\)? (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 > \(C^2_{(10w)}\) # 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) \(w>5\) 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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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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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 }, 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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11 Aug 2012, 04:00
how to reach the final statement w> 6 from 2 we had (10w)(9w) < 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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11 Aug 2012, 04:11
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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17 Aug 2012, 04:10
PUNEETSCHDV wrote: how to reach the final statement
w> 6 from 2 we had (10w)(9w) < 9 , , why is w>6 If w=6 then (10w)(9w)=4*3=12 >9 and if w=7, then (10w)(9w)=3*2=6<9. When we increase w, from 7 to 10, (10w)(9w) 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 (10w)(9w) and then solve quadratic inequality, but number plugging for this particular case is better. Check this: ifxisanintegerwhatisthevalueofx1x24x94661.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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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*(X1)/2 / (10*9/2) > 1/2 X(X1)>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(M1)/10*9 < 1/10 M(M1) < 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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06 Nov 2012, 11:25
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 \(\frac{w}{10}*\frac{w1}{9}\) 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) 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 > \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\), 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 \(\frac{C^2_w}{C^2_{10}}>\frac{1}{2}\) > \(\frac{\frac{w(w1)}{2}}{45}>\frac{1}{2}\) > > \(w(w1)>45\) > \(w>7\)?
(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 > \(C^2_{(10w)}\) # 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) \(w>5\) 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: if2differentrepresentativesaretobeselectedatrandom128233.html#p1113622Hope it helps. Sorry , i am not understand how that formula stuff has been written(kind of lagging in that field) [color=#ff0000] \(\frac{10w}{10}*\frac{10w1}{9}<\frac{1}{10}\) > \((10w)(9w)<9\) > \(w>6\),



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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. 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 Pnew> 1/2 are women , then it can be say 1/2 (plus some point say .51) * .51 which is not sufficient 2. if Pmen < 1/10 then Pwomen 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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02 Jul 2014, 21:39
@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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