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If 2 different representatives are to be selected at random from a [#permalink]
08 Jan 2010, 13:18
00:00
A
B
C
D
E
Difficulty:
85% (hard)
Question Stats:
50% (03:42) correct
50% (02:06) wrong based on 42 sessions
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]
08 Jan 2010, 14:25
3
This post received KUDOS
Expert's post
2
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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]
08 Jan 2010, 15:49
1
This post received KUDOS
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]
23 Jun 2010, 12:40
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]
23 Jun 2010, 12:54
2
This post received KUDOS
Expert's post
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]
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
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]
11 Jul 2010, 06:44
1
This post received KUDOS
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]
12 Jul 2010, 10:47
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]
02 Jul 2014, 09:34
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Re: If 2 different representatives are to be selected at random from a [#permalink]
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