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

 It is currently 04 May 2016, 01:18

### 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

Author Message
Director
Joined: 03 Sep 2006
Posts: 878
Followers: 6

Kudos [?]: 526 [0], given: 33

Re: gmatprep combs and probs ds [#permalink]

### Show Tags

09 Mar 2008, 18:34
Case ( i )

Women can be = 6, 7, 8, 9, 10 and Men respectively will be 4,3,2,1,0.

Say, W = 6 and M = 4

Probability that both representatives selected will be
women, is p

p = 6C2/10C2 = 15/45 = 1/3

1/3 < 1/2

Say, W =10, M =0

p =1 > 1/2

Case ( ii )

Probability that both representatives selected will be
men, is P < 1/10

Probability that None are men = 1-P = Probability that both are women

p<1/10

1-p = 1-1/10 = 9/10 > 1/2

VP
Joined: 22 Nov 2007
Posts: 1092
Followers: 9

Kudos [?]: 374 [0], given: 0

Re: gmatprep combs and probs ds [#permalink]

### Show Tags

16 Mar 2008, 02:08
LM wrote:
Case ( i )

Women can be = 6, 7, 8, 9, 10 and Men respectively will be 4,3,2,1,0.

Say, W = 6 and M = 4

Probability that both representatives selected will be
women, is p

p = 6C2/10C2 = 15/45 = 1/3

1/3 < 1/2

Say, W =10, M =0

p =1 > 1/2

Case ( ii )

Probability that both representatives selected will be
men, is P < 1/10

Probability that None are men = 1-P = Probability that both are women

p<1/10

1-p = 1-1/10 = 9/10 > 1/2

you miss the case in which one is man and the other is woman.

this is the fastest way I have found to solve this one:

we must show whether wc2/10c2 >1/2

1. w>5
let's try with 6 or 8. with 6 we have 1/3, less than 1/2. with 8 we have 28/45, >1/2. not suff
2.mc2/10c2<1/10, mc2<4.5, m can be 3,2,1 or 0. if m=3, w=7, 7c2/10c2<1/2 ok. if m=2 and w=8, always 28/45>1/2 not suff

combining doesn't say much further. OA is E
Manager
Joined: 04 Jan 2008
Posts: 119
Followers: 2

Kudos [?]: 45 [0], given: 0

### Show Tags

06 Sep 2008, 00:21
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

Last edited by dancinggeometry on 06 Sep 2008, 00:50, edited 1 time in total.
Director
Joined: 23 Sep 2007
Posts: 790
Followers: 5

Kudos [?]: 147 [0], given: 0

### Show Tags

06 Sep 2008, 00:37
dancinggeometry 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 21 of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 1/10

(1) More than 21 of the 10 employees are women.???
Senior Manager
Joined: 10 Mar 2008
Posts: 371
Followers: 5

Kudos [?]: 176 [0], given: 0

### Show Tags

16 Sep 2008, 14:26
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

Last edited by vksunder on 17 Sep 2008, 12:13, edited 1 time in total.
Senior Manager
Joined: 31 Jul 2008
Posts: 306
Followers: 1

Kudos [?]: 32 [0], given: 0

### Show Tags

16 Sep 2008, 15:27
something wrong with the question (p can never be greater than 1)
SVP
Joined: 21 Jul 2006
Posts: 1538
Followers: 8

Kudos [?]: 524 [0], given: 1

### Show Tags

16 Sep 2008, 15:47
vksunder 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

w= # of women
m= # of men

Probability that both selected are women:

$$(w/10)*((w-1)/9) = (w^2-w)/90 > 1/2$$, so the question is asking us whether $$w^2-w>45$$--->because $$(1/2)*90 = 45$$

(1) So:
if $$w=6$$, then $$6^2-6 < 45$$ False
if $$w=9$$, then $$9^2-9 > 45$$ True --->NOT SUFF.

(2) $$(m/10)*((m-1)/9) = ((m^2-m)/90) < 10%*90$$ -----> $$(m^2-m/90) < (9/90)$$, so $$m^2-m < 9$$

pick numbers: if m=2 then w=8, and if m=3 then w=7.

so when $$w=8$$---> $$8^2- 8 >45$$ true
so when $$w=7$$---> $$7^2 - 7 <45$$ False ----> NOT Suff.

(Together)

$$W=8$$, then $$8^2-8 >45$$, and when $$w=7$$, then $$7^2-7 < 45$$

I just edited my calculations once you've corrected your typo.

Last edited by tarek99 on 17 Sep 2008, 14:46, edited 4 times in total.
Senior Manager
Joined: 10 Mar 2008
Posts: 371
Followers: 5

Kudos [?]: 176 [0], given: 0

### Show Tags

17 Sep 2008, 12:14
Sorry about the typo. I've replaced 21 with the correct value. That is, 1/2
SVP
Joined: 21 Jul 2006
Posts: 1538
Followers: 8

Kudos [?]: 524 [0], given: 1

### Show Tags

17 Sep 2008, 14:44
I just corrected my calculations after you've corrected your typo. Answer should be E. What's the OA?
VP
Joined: 30 Jun 2008
Posts: 1043
Followers: 13

Kudos [?]: 467 [0], given: 1

### Show Tags

18 Sep 2008, 11:06
tarek99 wrote:

$$(w/10)*((w-1)/9) = (w^2-w)/90 > 1/2$$, so the question is asking us whether $$w^2-w>45$$--->because $$(1/2)*90 = 45$$

tarek .. I do not get this step(which happens to be the basis for this problem)

_________________

"You have to find it. No one else can find it for you." - Bjorn Borg

Manager
Joined: 28 Aug 2008
Posts: 101
Followers: 1

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

### Show Tags

18 Sep 2008, 14:19
Quote:
(w/10)*((w-1)/9) = (w^2-w)/90 > 1/2, so the question is asking us whether w^2-w>45--->because (1/2)*90 = 45
Quote:

Amitdgr... let me attempt to elaborate here:

Tarek's formula is just the words from the stem as an equation.. for example

From 1, say 6/10 are female... this means that when the first person is selceted the prob. is 6/10 (thus x/10) , when the next person is selceted, the prob. of Female becomes 5/9 (hence x-1/9), the total probability is 6/10 * 5/9 = 1/3 where

(w/10)*((w-1)/9) = (w^2-w)/90

is equal to (6/10)*((6-1)/9) = 30/90
Manager
Joined: 28 Apr 2008
Posts: 110
Followers: 1

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

### Show Tags

19 Oct 2008, 01:06

lets say there are 3 men and 7 woman-

the chance for 2 men is 3/10*2/9=6/90 <1/10

the chance for 2 woman is 7/10*6/9 = 42/90=p.....p < 1/2

lets say thre are 2 men and 8 woman-

the chance for 2 men is 2/10*1/9=2/90= <1/10

the chance for 2 woman is 8/10*7/9 = 56/90=p ..... p > 1/2

so must be E
VP
Joined: 17 Jun 2008
Posts: 1397
Followers: 8

Kudos [?]: 215 [0], given: 0

### Show Tags

19 Oct 2008, 01:33
dancinggeometry 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

IMO D
1)INSUFFI ,say women=5 then 5*4/10*9 <1/2

when women=8 then p>1/2

2)INSUFFI thouh im not able to put down he calculation

IMO E
_________________

cheers
Its Now Or Never

Senior Manager
Joined: 06 Jul 2007
Posts: 285
Followers: 3

Kudos [?]: 46 [0], given: 0

### Show Tags

23 Mar 2009, 13:39
marcodonzelli 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.

Fact 1: p(w) > 1/2

Let there are n women out of 10 employees. so n > 5.

number of ways of selecting 2 women out of n = nC2 = n*(n-1)/2.

total number of possible ways of selecting 2 employees out of 10 = 10C2 = 45.

probability that two selected employees are women = n(n-1)/90

n = 6, p=1/3
n = 7, p=42/90
n = 8, p=56/90.

1st is not sufficient.

Fact 2: let number of men in the sample is m. number of ways of selecting 2 men out of m = mC2 = m*(m-1)/2.

total number of ways = 45.

p(both men) = m*(m-1)/90

m*(m-1)/90 < 1/10

m*(m-1)<9
m = 3, 2, 1

second not sufficient.

combining both, we have two solutions 8 women and 2 men OR 7 women or 3 men.

Hence E.
Intern
Joined: 21 Jul 2006
Posts: 18
Followers: 0

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

### Show Tags

24 Mar 2009, 15:46
Is this question correct?
p is the probability?
How can "p>21"? Should not p be between 0 and 1?
Senior Manager
Joined: 06 Jul 2007
Posts: 285
Followers: 3

Kudos [?]: 46 [0], given: 0

### Show Tags

24 Mar 2009, 22:21
matrix777 wrote:
Is this question correct?
p is the probability?
How can "p>21"? Should not p be between 0 and 1?

I'm not sure what question you are looking at. this question asks "is p>1/2"?
Manager
Joined: 18 Jul 2009
Posts: 53
Followers: 3

Kudos [?]: 68 [0], given: 7

### Show Tags

19 Aug 2009, 04:09
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

Intern
Joined: 24 Jun 2009
Posts: 42
Followers: 0

Kudos [?]: 3 [0], given: 1

### Show Tags

19 Aug 2009, 05:01
E

A is insuff
B is insuff as it provides info for not two men i.e. it could either be two men or one man and one woman
together the prob of 2 women is between 1/3 and 9/10. therefore insuff.

What is the OA?
Manager
Joined: 14 Aug 2009
Posts: 123
Followers: 2

Kudos [?]: 96 [0], given: 13

### Show Tags

19 Aug 2009, 05:39
hrish88 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

for (1), suppose W=6, then both representatives are women would be 6C2, and total select ways are 10C2
p=6C2/10C2=(6*5)/(10*9)=1/3<1/2
therefore (1) is nsf.

for (2) suppose there are M mem,
then Pman=MC2/10C2<1/10, so, M(M-1)<9, M=0, 1, 2, 3
for M=3, then W=7,
p=7C2/10C2=(7*6)/(10*9)=42/90<45/90=1/2

_________________

Kudos me if my reply helps!

Intern
Joined: 24 Jun 2009
Posts: 42
Followers: 0

Kudos [?]: 3 [0], given: 1

### Show Tags

19 Aug 2009, 05:50
The condition II states that the prob of selecting two men is less than 1/10. This provides information for "prob of not selecting two men" and this means that either both reps are women OR one rep is a man and one rep is a woman. This is insuff coz it does not help calculate the prob of two women reps >1/2. hope this helps!
Re: Probability problems   [#permalink] 19 Aug 2009, 05:50

Go to page   Previous    1   2   3   4    Next  [ 67 posts ]

Display posts from previous: Sort by