Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

08 Jan 2010, 14:18

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

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

HideShow timer Statistics

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

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

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]

Show Tags

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.

(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]

Show Tags

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

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]

Show Tags

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]

Show Tags

02 Jul 2014, 10:34

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

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).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

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
[#permalink]
09 Feb 2016, 19:26

Hey, guys, So, I’ve decided to run a contest in hopes of getting the word about the site out to as many applicants as possible this application season...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...

Whether you’re an entrepreneur, aspiring business leader, or you just think that you may want to learn more about business, the thought of getting your Masters in Business Administration...