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A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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13 Jan 2014, 11:26
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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.
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Re: A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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13 Jan 2014, 11:55
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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.
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Re: A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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14 Jan 2014, 05:38
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Re: A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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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



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Re: A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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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?



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Re: A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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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 !
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Re: A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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11 Feb 2014, 01:36
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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.
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Re: A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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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?



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Re: A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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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.
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A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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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?



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



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Re: A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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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.



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A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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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.



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A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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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.



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Re: A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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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!



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Re: A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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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
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Re: A jury pool consists of 6 men and w women. If 2 jurors [#permalink]
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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
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