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Re: gmatprep combs and probs ds
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09 Mar 2008, 18:34
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 = 1P = Probability that both are women
p<1/10
1p = 11/10 = 9/10 > 1/2
Thus ( ii ) answers.
Answer should be "B"



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Re: gmatprep combs and probs ds
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16 Mar 2008, 02:08
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 = 1P = Probability that both are women
p<1/10
1p = 11/10 = 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 combining doesn't say much further. OA is E



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Zumit DS 009
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Updated on: 06 Sep 2008, 00:50
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: Zumit DS 009
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06 Sep 2008, 00:37
dancinggeometry 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 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.???



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PS: Probability
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Updated on: 17 Sep 2008, 12:13
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
Originally posted by vksunder on 16 Sep 2008, 14:26.
Last edited by vksunder on 17 Sep 2008, 12:13, edited 1 time in total.



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Re: PS: Probability
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16 Sep 2008, 15:27
something wrong with the question (p can never be greater than 1)



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Re: PS: Probability
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Updated on: 17 Sep 2008, 14:46
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: \((w/10)*((w1)/9) = (w^2w)/90 > 1/2\), so the question is asking us whether \(w^2w>45\)>because \((1/2)*90 = 45\) (1) So: if \(w=6\), then \(6^26 < 45\) False if \(w=9\), then \(9^29 > 45\) True >NOT SUFF. (2) \((m/10)*((m1)/9) = ((m^2m)/90) < 10%*90\) > \((m^2m/90) < (9/90)\), so \(m^2m < 9\) pick numbers: if m=2 then w=8, and if m=3 then w=7. so when \(w=8\)> \(8^2 8 >45\) true so when \(w=7\)> \(7^2  7 <45\) False > NOT Suff. (Together) \(W=8\), then \(8^28 >45\), and when \(w=7\), then \(7^27 < 45\) Answer E. I just edited my calculations once you've corrected your typo.
Originally posted by tarek99 on 16 Sep 2008, 15:47.
Last edited by tarek99 on 17 Sep 2008, 14:46, edited 4 times in total.



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Re: PS: Probability
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17 Sep 2008, 12:14
Sorry about the typo. I've replaced 21 with the correct value. That is, 1/2



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Re: PS: Probability
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17 Sep 2008, 14:44
I just corrected my calculations after you've corrected your typo. Answer should be E. What's the OA?



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Re: PS: Probability
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18 Sep 2008, 11:06
tarek99 wrote: \((w/10)*((w1)/9) = (w^2w)/90 > 1/2\), so the question is asking us whether \(w^2w>45\)>because \((1/2)*90 = 45\)
tarek .. I do not get this step(which happens to be the basis for this problem) Please elaborate ..
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Re: PS: Probability
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18 Sep 2008, 14:19
Quote: (w/10)*((w1)/9) = (w^2w)/90 > 1/2, so the question is asking us whether w^2w>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 x1/9), the total probability is 6/10 * 5/9 = 1/3 where (w/10)*((w1)/9) = (w^2w)/90 is equal to (6/10)*((61)/9) = 30/90



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Re: Zumit DS 009
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19 Oct 2008, 01:06
My answer is E
lets say there are 3 men and 7 woman
the chance for 2 men is 3/10*2/9=6/90 <1/10
the chance for 2 woman is 7/10*6/9 = 42/90=p.....p < 1/2
lets say thre are 2 men and 8 woman
the chance for 2 men is 2/10*1/9=2/90= <1/10
the chance for 2 woman is 8/10*7/9 = 56/90=p ..... p > 1/2
so must be E



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Re: Zumit DS 009
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19 Oct 2008, 01:33
dancinggeometry 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 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 IMO E
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Re: still prob
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23 Mar 2009, 13:39
marcodonzelli 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. 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*(n1)/2. total number of possible ways of selecting 2 employees out of 10 = 10C2 = 45. probability that two selected employees are women = n(n1)/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*(m1)/2. total number of ways = 45. p(both men) = m*(m1)/90 m*(m1)/90 < 1/10 m*(m1)<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. Hence E.



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Re: still prob
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24 Mar 2009, 15:46
Is this question correct? p is the probability? How can "p>21"? Should not p be between 0 and 1?



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Re: still prob
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24 Mar 2009, 22:21
matrix777 wrote: Is this question correct? p is the probability? How can "p>21"? Should not p be between 0 and 1? I'm not sure what question you are looking at. this question asks "is p>1/2"?



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Probability problems
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19 Aug 2009, 04:09
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
Thanks in advance



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Re: Probability problems
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19 Aug 2009, 05:01
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.
What is the OA?



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Re: Probability problems
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19 Aug 2009, 05:39
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(M1)<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 therefore answer is E
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Re: Probability problems
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19 Aug 2009, 05:50
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!




Re: Probability problems &nbs
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