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If 2 different representatives are to be selected at random [#permalink]
27 Feb 2012, 08:11
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A
B
C
D
E
Difficulty:
95% (hard)
Question Stats:
44% (02:42) correct
56% (01:31) wrong based on 707 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
26
This post received KUDOS
Expert's post
13
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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 > 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.
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