Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 26 Sep 2016, 07:49

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 22 Jul 2009
Posts: 201
Location: Manchester UK
Followers: 2

Kudos [?]: 300 [0], given: 6

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

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
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 34861
Followers: 6483

Kudos [?]: 82666 [3] , given: 10115

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

### Show Tags

08 Jan 2010, 15:25
3
KUDOS
Expert's post
2
This post was
BOOKMARKED
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?
[Reveal] Spoiler:
E

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

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 34861
Followers: 6483

Kudos [?]: 82666 [1] , given: 10115

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

### Show Tags

08 Jan 2010, 16:49
1
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

(1)+(2) $$w>5$$, $$w>6$$ not sufficient

_________________
Manager
Joined: 26 Sep 2007
Posts: 65
Followers: 1

Kudos [?]: 15 [0], given: 5

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

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.

Math Expert
Joined: 02 Sep 2009
Posts: 34861
Followers: 6483

Kudos [?]: 82666 [2] , given: 10115

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

### Show Tags

23 Jun 2010, 13:54
2
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

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.

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, $$w>6$$.

Hope it's clear.
_________________
Manager
Joined: 26 Sep 2007
Posts: 65
Followers: 1

Kudos [?]: 15 [0], given: 5

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

### Show Tags

23 Jun 2010, 13:59
Thanks Bunuel, Got it. +1 for you.
Intern
Joined: 15 Mar 2010
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

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

Please can any one try it?
[Reveal] Spoiler:
E

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

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
Math Expert
Joined: 02 Sep 2009
Posts: 34861
Followers: 6483

Kudos [?]: 82666 [1] , given: 10115

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

### Show Tags

11 Jul 2010, 07:44
1
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

Please can any one try it?
[Reveal] Spoiler:
E

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

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.
_________________
Senior Manager
Joined: 25 Feb 2010
Posts: 481
Followers: 4

Kudos [?]: 79 [0], given: 10

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

Please can any one try it?
[Reveal] Spoiler:
E

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

you rock buddy

_________________

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

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

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11684
Followers: 527

Kudos [?]: 143 [0], given: 0

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.
_________________
Manager
Joined: 17 Nov 2013
Posts: 97
Concentration: Strategy, Healthcare
GMAT 1: 710 Q49 V38
GPA: 3.34
Followers: 4

Kudos [?]: 35 [0], given: 47

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

### Show Tags

02 Jul 2014, 22:39
@bunuel. Can't I do the following for second statement?

(M/10) (M-1)/9 < 1/10

M(M-1) < 9

So, M < 3. Hence, W > 7.

Sufficient.

Posted from GMAT ToolKit
_________________

Math Expert
Joined: 02 Sep 2009
Posts: 34861
Followers: 6483

Kudos [?]: 82666 [0], given: 10115

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

### Show Tags

03 Jul 2014, 06:17
smit29may wrote:
@bunuel. Can't I do the following for second statement?

(M/10) (M-1)/9 < 1/10

M(M-1) < 9

So, M < 3. Hence, W > 7.

Sufficient.

Posted from GMAT ToolKit

From M(M-1) < 9 --> $$M\leq{3}$$ (notice that M could be 3 here), thus $$W\geq{7}$$.

Hope it's clear.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11684
Followers: 527

Kudos [?]: 143 [0], given: 0

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.
_________________
SVP
Joined: 17 Jul 2014
Posts: 1713
Location: United States
Schools: Stanford '19
GMAT 1: 550 Q39 V27
GMAT 2: 560 Q42 V26
GMAT 3: 560 Q43 V24
GMAT 4: 650 Q49 V30
GPA: 3.56
WE: General Management (Transportation)
Followers: 14

Kudos [?]: 203 [0], given: 113

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....
Re: If 2 different representatives are to be selected at random from a   [#permalink] 09 Feb 2016, 19:26
Similar topics Replies Last post
Similar
Topics:
If 2 different representatives are to be selected at random from a... 2 24 Jul 2016, 05:38
44 If 2 different representatives are to be selected at random 15 04 Mar 2014, 00:17
86 If 2 different representatives are to be selected at random 21 27 Feb 2012, 09:11
If 2 different representatives are to be selected at random 1 12 Aug 2011, 13:08
12 If 2 different representatives are to be selected at random 16 01 Feb 2010, 00:19
Display posts from previous: Sort by