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If 2 different representatives are to be selected at random [#permalink]
27 Feb 2012, 08:11

6

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19

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00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

43% (02:41) correct
57% (01:29) wrong based on 616 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.

Re: If 2 different [#permalink]
27 Feb 2012, 08:15

25

This post received KUDOS

Expert's post

9

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.

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

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

Re: If 2 different [#permalink]
06 Aug 2012, 09:32

3

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 }, 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 ]

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

Re: Probability of desired outcome [#permalink]
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.

Re: If 2 different [#permalink]
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)

Re: If 2 different [#permalink]
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)

Re: If 2 different [#permalink]
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)

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

Re: If 2 different [#permalink]
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? _________________

Re: If 2 different [#permalink]
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?

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

Re: If 2 different representatives are to be selected at random [#permalink]
08 Dec 2014, 13:31

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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 > 21 ?. (1) More than! of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 11.

Re: If 2 different representatives are to be selected at random [#permalink]
29 Jan 2015, 03:42

Expert's post

keshav11 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 > 21 ?. (1) More than! of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 11.

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...