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If 2 different representatives are to be selected at random [#permalink]
 27 Feb 2012, 09:11
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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 ? (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.
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Re: If 2 different [#permalink]
27 Feb 2012, 09:15
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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 > 1/2What 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.
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Re: If 2 different representatives are to be selected at random [#permalink]
08 May 2012, 01:41
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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.
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Re: If 2 different representatives are to be selected at random [#permalink]
15 Jun 2012, 03:24
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
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What is the probability that p > 1/2 ? [#permalink]
02 Aug 2012, 13:12
Q : 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.
My approach :
There are total 10 employees, and for P(2 women are selected) > 1/2, there should be atleast 6 women(W) in the group.
i.e, P(two women) = Total women / Total employees = W/10 = p --> W > 5.
From (1), we can clearly say that W > 5.
From (2), P(men) = M/10 = 1/10 --> M =1. Therefore W= 9.
Thus, both the statements are independently sufficient to answer this.
But the official answer says the option is E. How cum ? Can anyone explain ?
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Re: What is the probability that p > 1/2 ? [#permalink]
02 Aug 2012, 13:16
ankurjaini wrote: Q : 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.
My approach :
There are total 10 employees, and for P(2 women are selected) > 1/2, there should be atleast 6 women(W) in the group.
i.e, P(two women) = Total women / Total employees = W/10 = p --> W > 5.
From (1), we can clearly say that W > 5.
From (2), P(men) = M/10 = 1/10 --> M =1. Therefore W= 9.
Thus, both the statements are independently sufficient to answer this.
But the official answer says the option is E. How cum ? Can anyone explain ? Merging similar topics. Please refer to the solution above.
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Re: If 2 different [#permalink]
06 Aug 2012, 10:32
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 ]
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Re: If 2 different representatives are to be selected at random [#permalink]
11 Aug 2012, 05:00
how to reach the final statement w> 6 from 2 we had (10-w)(9-w) < 9 , , why is w>6
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Re: If 2 different representatives are to be selected at random [#permalink]
11 Aug 2012, 05:11
is plugging numbers only way to solve this quadratic inequality or do we have an algebric approach ?
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Re: If 2 different representatives are to be selected at random [#permalink]
17 Aug 2012, 05:10
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Probability of desired outcome [#permalink]
04 Nov 2012, 08:57
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.
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Re: Probability of desired outcome [#permalink]
04 Nov 2012, 12:31
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. Combining also NS. Ans E
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Re: Probability of desired outcome [#permalink]
04 Nov 2012, 15:16
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. Merging similar topics. Refer to the solutions above and ask if anything remains unclear. Also, please read carefully and follow: rules-for-posting-please-read-this-before-posting-133935.html Please pay attention to the rule #3. Thank you.
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Re: If 2 different [#permalink]
05 Nov 2012, 23:21
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)
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Re: If 2 different [#permalink]
06 Nov 2012, 05:55
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) Check here: if-2-different-representatives-are-to-be-selected-at-random-128233.html#p1113622Hope it helps.
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Re: If 2 different [#permalink]
06 Nov 2012, 12:25
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) Check here: if-2-different-representatives-are-to-be-selected-at-random-128233.html#p1113622Hope it helps. Sorry , i am not understand how that formula stuff has been written(kind of lagging in that field) [color=#ff0000] \frac{10-w}{10}*\frac{10-w-1}{9}<\frac{1}{10} --> (10-w)(9-w)<9 --> w>6,
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Re: If 2 different [#permalink]
07 Nov 2012, 05:44
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Re: If 2 different representatives are to be selected at random [#permalink]
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 Hence E
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Re: If 2 different representatives are to be selected at random
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13 Mar 2013, 16:44
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