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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: gmatprep combs and probs ds [#permalink]
16 Mar 2008, 01:08

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

So we can't answer.

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

Thus ( ii ) answers.

Answer should be "B"

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

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 05 Sep 2008, 23:50, edited 1 time in total.

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.???

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, 11:13, edited 1 time in total.

Re: PS: Probability [#permalink]
16 Sep 2008, 14:47

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.

Re: PS: Probability [#permalink]
18 Sep 2008, 13:19

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

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

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.

Probability problems [#permalink]
19 Aug 2009, 03:09

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

Re: Probability problems [#permalink]
19 Aug 2009, 04:01

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.

Re: Probability problems [#permalink]
19 Aug 2009, 04:39

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

Thanks in advance

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

Re: Probability problems [#permalink]
19 Aug 2009, 04:50

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!

gmatclubot

Re: Probability problems
[#permalink]
19 Aug 2009, 04:50