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If 2 different representatives are to be selected at random

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If 2 different representatives are to be selected at random [#permalink] New post 08 Jan 2010, 13:18
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
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Re: If 2 different representatives [#permalink] New post 08 Jan 2010, 14:25
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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

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

Answer E.
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Re: If 2 different representatives [#permalink] New post 08 Jan 2010, 15:49
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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

Answer E.
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Re: If 2 different representatives [#permalink] New post 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.
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Re: If 2 different representatives [#permalink] New post 23 Jun 2010, 12:54
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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.

Hope it's clear.
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Re: Probability of 2 different representatives [#permalink] New post 23 Jun 2010, 12:59
Thanks Bunuel, Got it. +1 for you.
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Re: If 2 different representatives [#permalink] New post 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

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

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
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Re: If 2 different representatives [#permalink] New post 11 Jul 2010, 06:44
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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

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.
_________________

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?
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Re: If 2 different representatives [#permalink] New post 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

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

Answer E.



you rock buddy
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Re: If 2 different representatives   [#permalink] 12 Jul 2010, 10:47
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