Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 21 Oct 2013
Posts: 426

A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
13 Jan 2014, 11:26
Question Stats:
50% (01:50) correct 50% (01:35) wrong based on 455 sessions
HideShow timer Statistics
A jury pool consists of 6 men and w women. If 2 jurors are selected from the pool at random, is the probability that 2 men will be selected higher than the probability that 1 man and 1 woman will be selected? (1) w ≥ 3 (2) w < 6 OE Rephrase the question as Is 30 / (6 + w)(5 + w) > 12w / (6 + w)(5 + w)? (1): Always No Sufficient (2): w might be less than 3 but could be higher Insufficient Hi, Although I didn't put whole OE, I want to know how we can solve this question, please.
Official Answer and Stats are available only to registered users. Register/ Login.




Math Expert
Joined: 02 Sep 2009
Posts: 49300

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
11 Feb 2014, 01:36
joshlevin90 wrote: christianbze wrote: I guess you did not understand the solution.
First, you have to find the probability for 2 men being picked and for 1 man and 1 woman being picked.
2 men:  combinatorics: \(6 x 5\)  all possibilities: \((6+w)(6+w1)\)
1 man and 1 woman:  combinatorics: \(6 x w x 2\)  all possibilities: \((6+w)(6+w1)\) You can choose out of 6 men, w woman and you have two different places to set them.
So your inequality is: \(\frac{6x5}{(6+w)(6+w1)} > \frac{6 x w x 2}{(6+w)(6+w1)}\)
\(\frac{30}{(6+w)(5+w)} > \frac{12 w}{(6+w)(5+w)}\)
as both denominators are the same, we just check the numerators. So how long is \(30 > 12w\)? As long as w (which has to be an integer!) is smaller than 3. So 0, 1, and 2.
So look at your options. could you pls explain in detail how you got that equation? A jury pool consists of 6 men and w women. If 2 jurors are selected from the pool at random, is the probability that 2 men will be selected higher than the probability that 1 man and 1 woman will be selected?The total number of people in the jury = \(6+w\). The probability of selecting 2 men when selecting 2 jurors = \(\frac{6}{6+w}*\frac{5}{(6+w)1}=\frac{30}{(6+w)(5+w)}\); The probability of selecting 1 man and 1 woman when selecting 2 jurors = \(2*\frac{6}{6+w}*\frac{w}{(6+w)1}=\frac{12w}{(6+w)(5+w)}\): multiplying by 2 as MW can occur in two ways MW or WM; The question asks: is \(\frac{30}{(6+w)(5+w)}>\frac{12w}{(6+w)(5+w)}\)? > is \(30>12w\)? > is \(w<2.5\)? So, the question basically asks whether there are 2, 1, or 0 women in the jury. (1) w ≥ 3. Directly gives a NO answer to the question. Sufficient. (2) w < 6. Not sufficient. Answer: A. Similar questions to practice: ajarcontains8redmarblesandywhitemarblesifjoan101748.html (almost the same question from MGMAT) acertainjarcontainsonlybblackmarbleswwhitemarbles104924.htmlif2differentrepresentativesaretobeselectedatrandom128233.htmlacertainboxcontainsonlybluebgreengandredr153384.htmlHope this helps.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Intern
Status: Student
Joined: 06 Oct 2013
Posts: 26
Location: Germany
Concentration: Operations, General Management
GPA: 2.4
WE: Other (Consulting)

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
13 Jan 2014, 11:55
I guess you did not understand the solution. First, you have to find the probability for 2 men being picked and for 1 man and 1 woman being picked. 2 men:  combinatorics: \(6 x 5\)  all possibilities: \((6+w)(6+w1)\) 1 man and 1 woman:  combinatorics: \(6 x w x 2\)  all possibilities: \((6+w)(6+w1)\) You can choose out of 6 men, w woman and you have two different places to set them. So your inequality is: \(\frac{6x5}{(6+w)(6+w1)} > \frac{6 x w x 2}{(6+w)(6+w1)}\) \(\frac{30}{(6+w)(5+w)} > \frac{12 w}{(6+w)(5+w)}\) as both denominators are the same, we just check the numerators. So how long is \(30 > 12w\)? As long as w (which has to be an integer!) is smaller than 3. So 0, 1, and 2. So look at your options.
_________________
Thank You = 1 Kudos B.Sc., International Production Engineering and Management M.Sc. mult., European Master in Management Candidate
_______________________________________________________ #1 Official GMAT Prep 1: 530 (Q41 V21), 10/10/13 #2 Manhattan GMAT CAT 1: 600 (Q43 V30), 12/17/13 #3 Manhattan GMAT CAT 2: 640 (Q43 V34), 01/13/14 #4 Manhattan GMAT CAT 3: 660 (Q45 V35), 01/16/14 #5 Manhattan GMAT CAT 4: 650 (Q45 V34), 01/18/14 #6 Manhattan GMAT CAT 5: 660 (Q42 V38), 01/21/14 #7 Official GMAT Prep 2: 640 (Q48 V30), 01/26/14 GMAT 670 Q49 V34 AWA5 IR6  TOEFL ibt 110



Math Expert
Joined: 02 Sep 2009
Posts: 49300

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
14 Jan 2014, 05:38



Intern
Joined: 22 Oct 2013
Posts: 4

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
10 Feb 2014, 17:45
i dont understand how you are getting your answer is there another way to explain it or if you can add in more steps as you go along solving it thank you



Intern
Joined: 22 Oct 2013
Posts: 4

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
10 Feb 2014, 17:46
christianbze wrote: I guess you did not understand the solution.
First, you have to find the probability for 2 men being picked and for 1 man and 1 woman being picked.
2 men:  combinatorics: \(6 x 5\)  all possibilities: \((6+w)(6+w1)\)
1 man and 1 woman:  combinatorics: \(6 x w x 2\)  all possibilities: \((6+w)(6+w1)\) You can choose out of 6 men, w woman and you have two different places to set them.
So your inequality is: \(\frac{6x5}{(6+w)(6+w1)} > \frac{6 x w x 2}{(6+w)(6+w1)}\)
\(\frac{30}{(6+w)(5+w)} > \frac{12 w}{(6+w)(5+w)}\)
as both denominators are the same, we just check the numerators. So how long is \(30 > 12w\)? As long as w (which has to be an integer!) is smaller than 3. So 0, 1, and 2.
So look at your options. could you pls explain in detail how you got that equation?



Senior Manager
Joined: 06 Aug 2011
Posts: 347

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
10 Feb 2014, 22:26
goodyear2013 wrote: A jury pool consists of 6 men and w women. If 2 jurors are selected from the pool at random, is the probability that 2 men will be selected higher than the probability that 1 man and 1 woman will be selected? (1) w ≥ 3 (2) w < 6 OE Rephrase the question as Is 30 / (6 + w)(5 + w) > 12w / (6 + w)(5 + w)? (1): Always No Sufficient (2): w might be less than 3 but could be higher Insufficient Hi, Although I didn't put whole OE, I want to know how we can solve this question, please. I tried... 2/6+w>1/6*1/w so i got 12w>6+W 11w>6 w>6/11?? so question becums is w>6/11? ans is A..S>=3 B is not sufficient ..because we cant say its greater than 6/11 or not.. My ans is correct.. Bt i think m not doing it rite !
_________________
Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !



Manager
Joined: 18 Jul 2013
Posts: 69
Location: Italy
GMAT 1: 600 Q42 V31 GMAT 2: 700 Q48 V38

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
28 Jul 2014, 13:29
Can someone tell me where i'm wrong?
1) Let's take the case when W=3. So there are 6 men and 3 women.
probability of taking 2 men (6C1*5C1)/9C2 = 30/72
probability of taking 1 man and 1 woman (6C1*3C1)/9C2 = 18/72
30/72 > 18/72, for 3 women. But as the number of women will rise, the probabilty will reverse. So insufficient.
Do i have to multiply 6C1*3C1 by two as we can take 1 man first, or 1 woman first?



Math Expert
Joined: 02 Sep 2009
Posts: 49300

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
30 Jul 2014, 09:19
oss198 wrote: Can someone tell me where i'm wrong?
1) Let's take the case when W=3. So there are 6 men and 3 women.
probability of taking 2 men (6C1*5C1)/9C2 = 30/72
probability of taking 1 man and 1 woman (6C1*3C1)/9C2 = 18/72
30/72 > 18/72, for 3 women. But as the number of women will rise, the probabilty will reverse. So insufficient.
Do i have to multiply 6C1*3C1 by two as we can take 1 man first, or 1 woman first? If there are 3 women, then: The probability of selecting 2 men is (6C2)/(9C2) = 15/36. The probability of selecting 1 men and woman is (6C1*3C1)/9C2 = 18/36. 15/36 < 18/36. If you increase the number of women the probability of the first case will always be less than the probability of the second. Hope it's clear.
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



Manager
Joined: 18 Jul 2013
Posts: 69
Location: Italy
GMAT 1: 600 Q42 V31 GMAT 2: 700 Q48 V38

A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
30 Jul 2014, 12:15
Bunuel wrote: If there are 3 women, then:
The probability of selecting 2 men is (6C2)/(9C2) = 15/36.
The probability of selecting 1 men and woman is (6C1*3C1)/9C2 = 18/36.
15/36 < 18/36. If you increase the number of women the probability of the first case will always be less than the probability of the second.
Hope it's clear.
Thank you for the answer but you mean 2*(6C1*3C1) don't you? we have to multiply by 2 because it can be WM or MW?



Math Expert
Joined: 02 Sep 2009
Posts: 49300

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
31 Jul 2014, 07:53



Manager
Joined: 18 Jul 2013
Posts: 69
Location: Italy
GMAT 1: 600 Q42 V31 GMAT 2: 700 Q48 V38

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
31 Jul 2014, 12:21
Bunuel wrote: oss198 wrote: Bunuel wrote: If there are 3 women, then:
The probability of selecting 2 men is (6C2)/(9C2) = 15/36.
The probability of selecting 1 men and woman is (6C1*3C1)/9C2 = 18/36.
15/36 < 18/36. If you increase the number of women the probability of the first case will always be less than the probability of the second.
Hope it's clear.
Thank you for the answer but you mean 2*(6C1*3C1) don't you? we have to multiply by 2 because it can be WM or MW? No. When you do with combinations method you don't need to account for different ways of occurring because this is already taken care of with this method. Thank you very much for your patience



Intern
Joined: 28 Jan 2013
Posts: 29

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
02 Aug 2014, 12:51
goodyear2013 wrote: A jury pool consists of 6 men and w women. If 2 jurors are selected from the pool at random, is the probability that 2 men will be selected higher than the probability that 1 man and 1 woman will be selected? (1) w ≥ 3 (2) w < 6 OE Rephrase the question as Is 30 / (6 + w)(5 + w) > 12w / (6 + w)(5 + w)? (1): Always No Sufficient (2): w might be less than 3 but could be higher Insufficient Hi, Although I didn't put whole OE, I want to know how we can solve this question, please. The no. of ways in which you can select 2 men from 6 men = 6c2 = 6!/(2! * 4!) = 15 no. of ways in which you can select a man from 6 men & a women from w women = 6C1 * WC1 = 6W As per statement 1, W>=3 =>6W >=18 so always choosing a men & women has more chances than 2 men being selected from 6 As per statement 2, W<6 =>if w = 1 no. of ways you can choose a man & women is 6 which is less than 15(no. of ways for choosing 2 men from 6) but if W = 5 no. of ways you can choose a man & women is 6 * 5 = 30 which is greater than 15. Hence statement 2 alone cannot be used for solving the problem. So the answer is A.



Intern
Joined: 03 Oct 2011
Posts: 16

A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
05 Aug 2014, 20:49
Bunuel wrote: oss198 wrote: Bunuel wrote: If there are 3 women, then:
The probability of selecting 2 men is (6C2)/(9C2) = 15/36.
The probability of selecting 1 men and woman is (6C1*3C1)/9C2 = 18/36.
15/36 < 18/36. If you increase the number of women the probability of the first case will always be less than the probability of the second.
Hope it's clear.
Thank you for the answer but you mean 2*(6C1*3C1) don't you? we have to multiply by 2 because it can be WM or MW? No. When you do with combinations method you don't need to account for different ways of occurring because this is already taken care of with this method. Fantastic Bunuel!! I was struggling with the "multiply by 2" issue and trying to figure out when to do so and when not to. Problem solved! Thanks Much.



Manager
Joined: 08 Jun 2015
Posts: 110

A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
01 Jul 2015, 20:17
I did it a slightly different way...
Using Combinations....
The probability of 2 random jurors from the entire juror pool is (6+w)!/[(2!)(6+w2)!].
Looking at the first condition, (1) w≥3, just use 3 to represent women  2 men have the greatest odds of selection with the fewest number of women. If the probability of selecting 2 men is worse than the probability of 1 man and 1 woman when there are only 3 women, then selecting 2 men has even worse odds when we increase the number of women (e.g. w=4,5,6, etc.).
So, total combination of all in the juror pool is 9!/(2!*7!) = 9*8/2 = 36
The combination of 2 men is 6!/(4!*2!) = 6*5/2 = 15
The combination of 1 man and 1 woman is [6!/(5!*1!)]*[3!/(1!*2!)] = 6*3 = 18
2 men < 1 man, 1 woman 15/36 < 18/36
This answers as a no to condition (1) and thus it's sufficient.
Using a similar logic to condition (1), we test the fewest possible number of women for condition (2), which is 1 (or 0...). We already know that having 3 women will mean that selecting 2 men has a smaller probability. Then, will the min. number of women still result in the same outcome (less chance for two men...)? If so, then it satisfies. If not, then two men have a greater probability of selection and condition (2) is insufficient.
The total combination of all in the juror pool is 7!/(2!*5!) = 7*6/2 = 21
Combination of 1 man and 1 women with only 1 female juror in the pool is [6!/(5!*1!)]*[1!] = 6*1 = 6
Combination of 2 men is 6!/(4!*2!) = 6*5/2 = 15
1 man, 1 woman < 2 men 6/21 < 15/21
So 2 men have a greater probability. We already know that 3 women in the juror pool means a greater probability for the 1 man, 1 woman probability, so this condition is insufficient.



Intern
Joined: 12 Dec 2015
Posts: 18

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
03 May 2016, 10:32
Dear Bunuel, would you please confirm if this method is correct:
If there are 3 woman: 1 woman and 1 man: 6/9*3/8+3/9*6/8=36/72 2 man: 6/9*5/8=30/36 Probability for 1 woman and 1 man is higher than for 2 man
4 woman: 1 woman and 1 man: 6/10*4/9+4/10*6/9=48/90 2 man: 6/10*5/9=30/90 Probability for 1 woman and 1 man is higher than for 2 man
And so on, the probability of 1w to 1m will be higher as the number of woman increases.
Thank you!



Manager
Status: GMAT Coach
Joined: 05 Nov 2012
Posts: 135
Location: Peru
GPA: 3.98

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
18 Oct 2016, 09:00
goodyear2013 wrote: A jury pool consists of 6 men and w women. If 2 jurors are selected from the pool at random, is the probability that 2 men will be selected higher than the probability that 1 man and 1 woman will be selected? (1) w ≥ 3 (2) w < 6 OE Rephrase the question as Is 30 / (6 + w)(5 + w) > 12w / (6 + w)(5 + w)? (1): Always No Sufficient (2): w might be less than 3 but could be higher Insufficient Hi, Although I didn't put whole OE, I want to know how we can solve this question, please. P = Success/ Total possibilities P(2 men) = P (1st man) * P (2nd man) ={ 6/(6+w) } * { 5/(5+w) } = 30 / {(6+w) (5+w)} P(1 man + 1 woman ) = P (1st man) * P (2nd woman) + P (1st woman) * P (2nd man) = { 6/(6+w) } * { w/(5+w) } + { w/(6+w) } * { 6/(5+w) } = 12w / {(6+w) (5+w)} Question: is P(2 men) > P(1 man + 1 woman ) ? = is 30 / {(6+w) (5+w)} > = 12w / {(6+w) (5+w)} ? Divide both sides by 12 / {(6+w) (5+w)} => Question: is 2.5 > w ? (1) w ≥ 3 SUFFICIENT (2) w < 6 ; 2 = NO, 3 = YES ; INSUFFICIENT Answer is A
_________________
Clipper Ledgard GMAT Coach



CEO
Joined: 12 Sep 2015
Posts: 2871
Location: Canada

Re: A jury pool consists of 6 men and w women. If 2 jurors
[#permalink]
Show Tags
12 Mar 2018, 08:28
goodyear2013 wrote: A jury pool consists of 6 men and w women. If 2 jurors are selected from the pool at random, is the probability that 2 men will be selected higher than the probability that 1 man and 1 woman will be selected?
(1) w ≥ 3 (2) w < 6
Target question: Is P(2 men) greater than P(1 man and 1 woman)?Statement 1: w> 3Let's see what happens if w = 3 (note: this is the best chance that P(2 men) will be greater than P(1 man and 1 woman) P(2 men) = P(man selected 1st and man selected 2nd) = (6/9)(5/8) = 30/72 P(1 man and 1 woman) = P(man selected 1st and woman selected 2nd OR woman selected 1st and man selected 2nd) = P(man selected 1st and woman selected 2nd) + P(woman selected 1st and man selected 2nd) = (6/9)(3/8) + (3/9)(6/8) = 18/72 + 18/72 = 36/72 So, when w = 3, P(2 men) is not greater than P(1 man and 1 woman)IMPORTANT: Now that we've shown that P(2 men) is not greater than P(1 man and 1 woman) when w = 3, we can see that, as the value of w increases, the answer to the target question will always remain the same. As such, statement 1 is SUFFICIENT Statement 2: w < 6 Consider these two conflicting cases: Case a: w = 1 P(2 men) = (6/7)(5/6) = 30/42 P(1 man and 1 woman) = P(man selected 1st and woman selected 2nd OR woman selected 1st and man selected 2nd) = P(man selected 1st and woman selected 2nd) + P(woman selected 1st and man selected 2nd) = (6/7)(1/6) + (1/6)(6/7) = 6/42 + 6/42 = 12/42 So, when w = 1, P(2 men) is greater than P(1 man and 1 woman)Case b: w = 3 In statement 1, we already showed that, when w = 3, P(2 men) is not greater than P(1 man and 1 woman)Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT Answer = A Cheers, Brent
_________________
Brent Hanneson – GMATPrepNow.com
Sign up for our free Question of the Day emails




Re: A jury pool consists of 6 men and w women. If 2 jurors &nbs
[#permalink]
12 Mar 2018, 08:28






