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

It is currently 30 Sep 2014, 07:53

Starting Soon:

Live Q&A Session With Kristen Egan - Assistant Director of Admissions at Darden Business School.


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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If 2 different representatives are to be selected at random

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
5 KUDOS received
Intern
Intern
User avatar
Joined: 20 Feb 2012
Posts: 41
Followers: 1

Kudos [?]: 64 [5] , given: 6

If 2 different representatives are to be selected at random [#permalink] New post 27 Feb 2012, 08:11
5
This post received
KUDOS
12
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

43% (02:29) correct 57% (01:23) wrong based on 428 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.
[Reveal] Spoiler: OA
Expert Post
23 KUDOS received
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23059
Followers: 3536

Kudos [?]: 27212 [23] , given: 2725

Re: If 2 different [#permalink] New post 27 Feb 2012, 08:15
23
This post received
KUDOS
Expert's post
3
This post was
BOOKMARKED
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{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) 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{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<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(w-1)}{2}}{45}>\frac{1}{2} --> --> w(w-1)>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_{(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 and w>6: w can be 7, answer NO or more than 7, answer YES. Not sufficient.

Answer E.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

1 KUDOS received
Manager
Manager
avatar
Joined: 14 Feb 2012
Posts: 228
Followers: 1

Kudos [?]: 70 [1] , given: 7

Re: If 2 different representatives are to be selected at random [#permalink] New post 08 May 2012, 00:41
1
This post received
KUDOS
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.
_________________

The Best Way to Keep me ON is to give Me KUDOS !!!
If you Like My posts please Consider giving Kudos

Shikhar

Manager
Manager
User avatar
Affiliations: Project Management Professional (PMP)
Joined: 30 Jun 2011
Posts: 213
Location: New Delhi, India
Followers: 2

Kudos [?]: 27 [0], given: 12

Re: If 2 different representatives are to be selected at random [#permalink] New post 15 Jun 2012, 02:24
shikhar wrote:
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.

I Made the same mistake Shikar :)

better to use the combination / equation method as explained before by bunuel
_________________

Best
Vaibhav

If you found my contribution helpful, please click the +1 Kudos button on the left, Thanks

2 KUDOS received
Intern
Intern
avatar
Joined: 18 May 2012
Posts: 2
Followers: 0

Kudos [?]: 7 [2] , given: 3

Re: If 2 different [#permalink] New post 06 Aug 2012, 09:32
2
This post received
KUDOS
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 :P}, 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 ]
Manager
Manager
User avatar
Joined: 31 Aug 2011
Posts: 209
Followers: 2

Kudos [?]: 62 [0], given: 44

CAT Tests
Re: If 2 different representatives are to be selected at random [#permalink] New post 11 Aug 2012, 04:00
how to reach the final statement


w> 6 from 2 we had (10-w)(9-w) < 9 , , why is w>6
_________________

If you found my contribution helpful, please click the +1 Kudos button on the left, I kinda need some =)

Manager
Manager
User avatar
Joined: 31 Aug 2011
Posts: 209
Followers: 2

Kudos [?]: 62 [0], given: 44

CAT Tests
Re: If 2 different representatives are to be selected at random [#permalink] New post 11 Aug 2012, 04:11
is plugging numbers only way to solve this quadratic inequality or do we have an algebric approach ?
_________________

If you found my contribution helpful, please click the +1 Kudos button on the left, I kinda need some =)

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23059
Followers: 3536

Kudos [?]: 27212 [0], given: 2725

Re: If 2 different representatives are to be selected at random [#permalink] New post 17 Aug 2012, 04:10
Expert's post
PUNEETSCHDV wrote:
how to reach the final statement


w> 6 from 2 we had (10-w)(9-w) < 9 , , why is w>6


If w=6 then (10-w)(9-w)=4*3=12>9 and if w=7, then (10-w)(9-w)=3*2=6<9. When we increase w, from 7 to 10, (10-w)(9-w) 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 (10-w)(9-w) and then solve quadratic inequality, but number plugging for this particular case is better.

Check this: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476 (solving quadratic inequalities)
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Joined: 27 Feb 2012
Posts: 138
Followers: 1

Kudos [?]: 17 [0], given: 22

Re: Probability of desired outcome [#permalink] New post 04 Nov 2012, 11:31
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*(X-1)/2 / (10*9/2) > 1/2
X(X-1)>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(M-1)/10*9 < 1/10
M(M-1) < 9
M can be 0,1,2,3
NS.

Combining also NS.
Ans E
_________________

---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Please +1 KUDO if my post helps. Thank you.

Manager
Manager
avatar
Status: K... M. G...
Joined: 22 Oct 2012
Posts: 51
Concentration: General Management, Leadership
GMAT Date: 08-27-2013
GPA: 3.8
Followers: 0

Kudos [?]: 6 [0], given: 118

Re: If 2 different [#permalink] New post 05 Nov 2012, 22:21
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{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) 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{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<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(w-1)}{2}}{45}>\frac{1}{2} --> --> w(w-1)>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_{(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 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)
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23059
Followers: 3536

Kudos [?]: 27212 [0], given: 2725

Re: If 2 different [#permalink] New post 06 Nov 2012, 04:55
Expert's post
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{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) 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{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<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(w-1)}{2}}{45}>\frac{1}{2} --> --> w(w-1)>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_{(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 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: if-2-different-representatives-are-to-be-selected-at-random-128233.html#p1113622

Hope it helps.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
avatar
Status: K... M. G...
Joined: 22 Oct 2012
Posts: 51
Concentration: General Management, Leadership
GMAT Date: 08-27-2013
GPA: 3.8
Followers: 0

Kudos [?]: 6 [0], given: 118

Re: If 2 different [#permalink] New post 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{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) 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{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<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(w-1)}{2}}{45}>\frac{1}{2} --> --> w(w-1)>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_{(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 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: if-2-different-representatives-are-to-be-selected-at-random-128233.html#p1113622

Hope it helps.




Sorry , i am not understand how that formula stuff has been written(kind of lagging in that field)

[color=#ff0000] \frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<9 --> w>6,
Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23059
Followers: 3536

Kudos [?]: 27212 [0], given: 2725

Re: If 2 different [#permalink] New post 07 Nov 2012, 04:44
Expert's post
breakit wrote:
Bunuel wrote:
breakit wrote:

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: if-2-different-representatives-are-to-be-selected-at-random-128233.html#p1113622

Hope it helps.




Sorry , i am not understand how that formula stuff has been written(kind of lagging in that field)

[color=#ff0000] \frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<9 --> w>6,


\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> cross-multiply (multiply both parts by 10*9): (10-w)(10-w-1)<9 --> (10-w)(9-w)<9.

Hope it's clear.
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Intern
Intern
avatar
Joined: 09 Sep 2012
Posts: 29
Schools: LBS '14, IMD '16
Followers: 0

Kudos [?]: 13 [0], given: 25

Re: If 2 different representatives are to be selected at random [#permalink] New post 13 Mar 2013, 15:44
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 P-new> 1/2 are women , then it can be say 1/2 (plus some point say .51) * .51 which is not sufficient

2. if P-men < 1/10 then P-women will be 9/10 ie there are many values between 1/4 and 9/10 which is not sufficient

Hence E
_________________

Your Kudos will motivate me :)

Manager
Manager
User avatar
Joined: 28 Apr 2013
Posts: 173
Location: India
GPA: 4
WE: Medicine and Health (Health Care)
Followers: 0

Kudos [?]: 26 [0], given: 84

Re: If 2 different [#permalink] New post 21 Nov 2013, 22:09
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{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) 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{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<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(w-1)}{2}}{45}>\frac{1}{2} --> --> w(w-1)>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_{(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 and w>6: w can be 7, answer NO or more than 7, answer YES. Not sufficient.

Answer E.

Hope it's clear.



How you figured it out (10-w)(9-w)<9 --> w>6 ??
Can you explain?
_________________

Thanks for Posting

LEARN TO ANALYSE

+1 kudos if you like

Expert Post
Math Expert
User avatar
Joined: 02 Sep 2009
Posts: 23059
Followers: 3536

Kudos [?]: 27212 [0], given: 2725

Re: If 2 different [#permalink] New post 22 Nov 2013, 01:18
Expert's post
rango 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{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) 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{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<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(w-1)}{2}}{45}>\frac{1}{2} --> --> w(w-1)>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_{(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 and w>6: w can be 7, answer NO or more than 7, answer YES. Not sufficient.

Answer E.

Hope it's clear.



How you figured it out (10-w)(9-w)<9 --> w>6 ??
Can you explain?


Please read the thread: if-2-different-representatives-are-to-be-selected-at-random-128233.html#p1113622
_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
25 extra-hard Quant Tests

Get the best GMAT Prep Resources with GMAT Club Premium Membership

Manager
Manager
User avatar
Joined: 28 Apr 2013
Posts: 173
Location: India
GPA: 4
WE: Medicine and Health (Health Care)
Followers: 0

Kudos [?]: 26 [0], given: 84

Re: If 2 different representatives are to be selected at random [#permalink] New post 22 Nov 2013, 04:24
Bunuel wrote:
PUNEETSCHDV wrote:
how to reach the final statement


w> 6 from 2 we had (10-w)(9-w) < 9 , , why is w>6


If w=6 then (10-w)(9-w)=4*3=12>9 and if w=7, then (10-w)(9-w)=3*2=6<9. When we increase w, from 7 to 10, (10-w)(9-w) 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 (10-w)(9-w) and then solve quadratic inequality, but number plugging for this particular case is better.

Check this: if-x-is-an-integer-what-is-the-value-of-x-1-x-2-4x-94661.html#p731476 (solving quadratic inequalities)


so that it is to put the random no.s; i tries to solve as inequalities and noticed that w(19-w) > 81, but could not reach out above mentioned value.

Thanks once again.
:banana
_________________

Thanks for Posting

LEARN TO ANALYSE

+1 kudos if you like

Re: If 2 different representatives are to be selected at random   [#permalink] 22 Nov 2013, 04:24
    Similar topics Author Replies Last post
Similar
Topics:
If 2 different representatives are to be selected at random vikramjit_01 7 18 May 2007, 13:25
If 2 different representatives are to be selected at random TeHCM 6 23 May 2006, 17:54
If 2 different representatives are to be selected at random sperumba 6 15 Jan 2006, 09:53
If 2 different representatives are to be selected at random nakib77 2 11 Nov 2005, 04:10
2 different representatives are to be selected at random AJB77 2 02 Jul 2005, 19:47
Display posts from previous: Sort by

If 2 different representatives are to be selected at random

  Question banks Downloads My Bookmarks Reviews Important topics  


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

Powered by phpBB © phpBB Group and phpBB SEO

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®.