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:

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

Show Tags

27 Feb 2012, 09:11

4

This post received KUDOS

53

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

44% (01:39) correct
56% (01:31) wrong based on 1279 sessions

HideShow timer Statistics

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.

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.

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

Show Tags

08 May 2012, 01:41

2

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

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 }, 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 ]

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.

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.

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)

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

Show Tags

13 Mar 2013, 16:44

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

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

Show Tags

08 Apr 2017, 10:47

1

This post received KUDOS

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 ? w* (w - 1) > 45 w >= 8?

(1) More than 1/2 of the 10 employees are women. NS. w can be 6 through 10.

(2) The probability that both representatives selected will be men is less than 1/10. m(m-1) < 9 m <=3, w can be 7 through 9. NS.

Combined, m w 3 7 - does not satisfy. 2 8 - satisfies

NS.

E. _________________

In case you find my posts helpful, give me Kudos. Thank you.

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

Show Tags

28 Aug 2017, 19:22

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.

Bunuel, how do you go from \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) thank you. it would be nice if someone can explain this

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.

Bunuel, how do you go from \(\frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10}\) --> \((10-w)(9-w)<9\) thank you. it would be nice if someone can explain this

It's done by multiplying both sides by 90.
_________________

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

Show Tags

28 Aug 2017, 23:56

BANON 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.

An easy way to solve statement 2 might be:

Probability(both men)<1/10. Probability(neither men)= Probability(both women)< 1-1/10; Probability(both women) <9/10. (is it more than half? can be or cannot be)... Insufficient.

Version 8.1 of the WordPress for Android app is now available, with some great enhancements to publishing: background media uploading. Adding images to a post or page? Now...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...

“Keep your head down, and work hard. Don’t attract any attention. You should be grateful to be here.” Why do we keep quiet? Being an immigrant is a constant...