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A jury pool consists of 6 men and w women. If 2 jurors
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13 Jan 2014, 10:26
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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?
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.
Re: A jury pool consists of 6 men and w women. If 2 jurors
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11 Feb 2014, 00:36
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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: \((6+w)(6+w-1)\)
1 man and 1 woman: - combinatorics: \(6 x w x 2\) - all possibilities: \((6+w)(6+w-1)\) You can choose out of 6 men, w woman and you have two different places to set them.
So your inequality is: \(\frac{6x5}{(6+w)(6+w-1)} > \frac{6 x w x 2}{(6+w)(6+w-1)}\)
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.
Re: A jury pool consists of 6 men and w women. If 2 jurors
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13 Jan 2014, 10:55
4
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: \((6+w)(6+w-1)\)
1 man and 1 woman: - combinatorics: \(6 x w x 2\) - all possibilities: \((6+w)(6+w-1)\) You can choose out of 6 men, w woman and you have two different places to set them.
So your inequality is: \(\frac{6x5}{(6+w)(6+w-1)} > \frac{6 x w x 2}{(6+w)(6+w-1)}\)
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
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14 Jan 2014, 04:38
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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?
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.
Re: A jury pool consists of 6 men and w women. If 2 jurors
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10 Feb 2014, 16:45
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
Re: A jury pool consists of 6 men and w women. If 2 jurors
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10 Feb 2014, 16:46
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: \((6+w)(6+w-1)\)
1 man and 1 woman: - combinatorics: \(6 x w x 2\) - all possibilities: \((6+w)(6+w-1)\) You can choose out of 6 men, w woman and you have two different places to set them.
So your inequality is: \(\frac{6x5}{(6+w)(6+w-1)} > \frac{6 x w x 2}{(6+w)(6+w-1)}\)
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?
Re: A jury pool consists of 6 men and w women. If 2 jurors
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10 Feb 2014, 21:26
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?
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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Bole So Nehal.. Sat Siri Akal.. Waheguru ji help me to get 700+ score !
Re: A jury pool consists of 6 men and w women. If 2 jurors
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31 Jul 2014, 06:53
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1
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.
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Re: A jury pool consists of 6 men and w women. If 2 jurors
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31 Jul 2014, 11:21
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.
Re: A jury pool consists of 6 men and w women. If 2 jurors
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02 Aug 2014, 11:51
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?
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.
A jury pool consists of 6 men and w women. If 2 jurors
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05 Aug 2014, 19:49
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.
A jury pool consists of 6 men and w women. If 2 jurors
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01 Jul 2015, 19:17
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+w-2)!].
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.
Re: A jury pool consists of 6 men and w women. If 2 jurors
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18 Oct 2016, 08:00
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?
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.
Re: A jury pool consists of 6 men and w women. If 2 jurors
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12 Mar 2018, 07:28
Top Contributor
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> 3 Let'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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Re: A jury pool consists of 6 men and w women. If 2 jurors
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