It is currently 25 Sep 2017, 01:19

Close

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
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.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If 2 different representatives are to be selected at random

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Senior Manager
Senior Manager
avatar
Joined: 02 Dec 2007
Posts: 449

Kudos [?]: 280 [0], given: 6

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

Show Tags

New post 05 Sep 2008, 11:03
1
This post was
BOOKMARKED
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 than1/10 .

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient

Last edited by Nihit on 05 Sep 2008, 11:15, edited 1 time in total.

Kudos [?]: 280 [0], given: 6

SVP
SVP
User avatar
Joined: 07 Nov 2007
Posts: 1795

Kudos [?]: 1037 [0], given: 5

Location: New York
Re: Probability [#permalink]

Show Tags

New post 05 Sep 2008, 12:06
Nihit 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 than1/10 .

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient


(1) More than 1/2 of the 10 employees are women.
Women count >6
assume 6W 4 M

P = 6C2/10C2 = 60/90=2/3 >1/2
so for W>=6 p>1/2
so for W=9
p= 9C2/10C2= 9*8/10*9 = 72/90 >1/2

not suffcient
(2) The probability that both representatives selected will be men is less than1/10
M 10-M women.

MC2/10C2 <1/10 --> M(M-1)/10*9<1/10
M(M-1)<9 this is possible only when M=4 or less.
So W>=6
same as (1)
not suffcient.

combined same result
E
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

Kudos [?]: 1037 [0], given: 5

Intern
Intern
avatar
Joined: 05 Sep 2008
Posts: 1

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

Re: Probability [#permalink]

Show Tags

New post 05 Sep 2008, 12:41
(1) More than 1/2 of the 10 employees are women.
Women count >6
assume 6W 4 M

P = 6C2/10C2 = 3/5 >1/2
so for W>=6 p>1/2
suffcient
(2) The probability that both representatives selected will be men is less than1/10
M 10-M women.

I DONT FOLLOW YOU HERE, PLEASE FORGIVE ME. ASSUMING 6 OF 10 ARE WOMEN, WOULDNT THAT PROBABILITY BE 6/10*5/9 OR 30/90 WHICH IS 3/9 (SO NOT ABOVE 1/2). ALTERNATIVELY, IT COULD BE ANOTHER EXTREME, 9 OF 10, WHICH WOULD MAKE PROBABILITY 9/10*8/9 WHICH IS 72/90 AND CLEARLY OVER 1/2. IT SEEMS TO ME THAT THIS IS NOT SUFFICIENT. PERHAPS I AM READING QUESTION WRONG? THANKS.


MC2/10C2 <1/10 --> M(M-1)/10<1/10
M(M-1)<1 this is possible only when M=1 or M=0
So W=9 or 10
p= 9C2/10C2 > 1/2
SUFFCIENT
I AGREE WITH YOU HERE, MAKES PERFECT SENSE.
D

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

SVP
SVP
User avatar
Joined: 07 Nov 2007
Posts: 1795

Kudos [?]: 1037 [0], given: 5

Location: New York
Re: Probability [#permalink]

Show Tags

New post 05 Sep 2008, 12:48
grover33 wrote:
(1) More than 1/2 of the 10 employees are women.
Women count >6
assume 6W 4 M

P = 6C2/10C2 = 3/5 >1/2
so for W>=6 p>1/2
suffcient
(2) The probability that both representatives selected will be men is less than1/10
M 10-M women.

I DONT FOLLOW YOU HERE, PLEASE FORGIVE ME. ASSUMING 6 OF 10 ARE WOMEN, WOULDNT THAT PROBABILITY BE 6/10*5/9 OR 30/90 WHICH IS 3/9 (SO NOT ABOVE 1/2). ALTERNATIVELY, IT COULD BE ANOTHER EXTREME, 9 OF 10, WHICH WOULD MAKE PROBABILITY 9/10*8/9 WHICH IS 72/90 AND CLEARLY OVER 1/2. IT SEEMS TO ME THAT THIS IS NOT SUFFICIENT. PERHAPS I AM READING QUESTION WRONG? THANKS.


MC2/10C2 <1/10 --> M(M-1)/10<1/10
M(M-1)<1 this is possible only when M=1 or M=0
So W=9 or 10
p= 9C2/10C2 > 1/2
SUFFCIENT
I AGREE WITH YOU HERE, MAKES PERFECT SENSE.
D


Thanks for pointing out.. thats silly mistake..
I updated the original post.
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

Kudos [?]: 1037 [0], given: 5

Current Student
User avatar
Joined: 04 Jan 2005
Posts: 283

Kudos [?]: 140 [0], given: 3

Location: Milan
Schools: Wharton, LBS, UChicago, Kellogg MMM (Donald Jacobs Scholarship), Stanford, HBS
Re: Probability [#permalink]

Show Tags

New post 05 Sep 2008, 13:43
I get E.

1) As groover pointed out, for W=6

Pr=6/10 * 5/9 = 1/3 < 1/2 thus insuff (it's intuitive that if W6 M4 it's less than 1/2)

2) if M=3

Pr(2M)= 3/10 * 2/9 = 2/30 < 1/10

thus we could have 7 women

Pr(2W) = 7/10 * 6/9 = 7/15 < 1/2

In fact, for M=3, W=7, we have:

Pr(2W)= 7/15
Pr(2M)= 1/15
Pr(1W1M)= 3/10 * 7/9 + 7/10 * 3/9 = 7/15

so E

Where did I go wrong?

Kudos [?]: 140 [0], given: 3

SVP
SVP
User avatar
Joined: 07 Nov 2007
Posts: 1795

Kudos [?]: 1037 [0], given: 5

Location: New York
Re: Probability [#permalink]

Show Tags

New post 05 Sep 2008, 14:05
Paradosso wrote:
I get E.

1) As groover pointed out, for W=6

Pr=6/10 * 5/9 = 1/3 < 1/2 thus insuff (it's intuitive that if W6 M4 it's less than 1/2)

2) if M=3

Pr(2M)= 3/10 * 2/9 = 2/30 < 1/10

thus we could have 7 women

Pr(2W) = 7/10 * 6/9 = 7/15 < 1/2

In fact, for M=3, W=7, we have:

Pr(2W)= 7/15
Pr(2M)= 1/15
Pr(1W1M)= 3/10 * 7/9 + 7/10 * 3/9 = 7/15

so E

Where did I go wrong?


you are not wrong buddy!!!

Today is awful day for me!!! busy with work..and meetings.

E should be the answer.
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

Kudos [?]: 1037 [0], given: 5

VP
VP
User avatar
Joined: 05 Jul 2008
Posts: 1403

Kudos [?]: 423 [0], given: 1

Re: Probability [#permalink]

Show Tags

New post 05 Sep 2008, 14:22
x2suresh wrote:
Paradosso wrote:
I get E.

1) As groover pointed out, for W=6

Pr=6/10 * 5/9 = 1/3 < 1/2 thus insuff (it's intuitive that if W6 M4 it's less than 1/2)

2) if M=3

Pr(2M)= 3/10 * 2/9 = 2/30 < 1/10

thus we could have 7 women

Pr(2W) = 7/10 * 6/9 = 7/15 < 1/2

In fact, for M=3, W=7, we have:

Pr(2W)= 7/15
Pr(2M)= 1/15
Pr(1W1M)= 3/10 * 7/9 + 7/10 * 3/9 = 7/15

so E

Where did I go wrong?


you are not wrong buddy!!!

Today is awful day for me!!! busy with work..and meetings.

E should be the answer.


Arrived at E as well.

Used a slightly diff approach for (2)

The probability that both representatives selected will be men is less than1/10

means

The probability that both representatives selected will NOT be men is greater than or equal to 9/10.

this includes P(WW)+ P(MW) >= 9/10

We cant tell for sure how they will split out. Hence Insufficient.

Any thing wrong with this approach?

Kudos [?]: 423 [0], given: 1

Manager
Manager
User avatar
Joined: 09 Jul 2007
Posts: 245

Kudos [?]: 264 [0], given: 0

Re: Probability [#permalink]

Show Tags

New post 05 Sep 2008, 14:35
I agree with E.

But, this one is surely a tough question, Option 1 is easily ruled out, but needs a lot of heat and trial for the option 2.

Any easy approach ?

Kudos [?]: 264 [0], given: 0

VP
VP
User avatar
Joined: 05 Jul 2008
Posts: 1403

Kudos [?]: 423 [0], given: 1

Re: Probability [#permalink]

Show Tags

New post 05 Sep 2008, 14:47
ssandeepan wrote:
I agree with E.

But, this one is surely a tough question, Option 1 is easily ruled out, but needs a lot of heat and trial for the option 2.

Any easy approach ?


I agree. Did you look at what I did? Is it correct? I arrived at E

Kudos [?]: 423 [0], given: 1

Current Student
User avatar
Joined: 04 Jan 2005
Posts: 283

Kudos [?]: 140 [0], given: 3

Location: Milan
Schools: Wharton, LBS, UChicago, Kellogg MMM (Donald Jacobs Scholarship), Stanford, HBS
Re: Probability [#permalink]

Show Tags

New post 05 Sep 2008, 23:07
x2suresh wrote:
you are not wrong buddy!!!

Today is awful day for me!!! busy with work..and meetings.

E should be the answer.


Eheh, I will tell my nephews that I once corrected x2suresh in math! GMAC should award 20 more points to me for this (my exam is Wednesday)! :)

Kudos [?]: 140 [0], given: 3

Re: Probability   [#permalink] 05 Sep 2008, 23:07
Display posts from previous: Sort by

If 2 different representatives are to be selected at random

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.