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Giorgosaek
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A jury pool consists of 6 men and w women. If 2 jurors 28 May 2020, 21:58
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Bunuel
As per your solution, should we conclude that : if we should find the probability of x and y occurring at the same time (as per solution 1 man and 1 woman ) ,then if xy can also occur as yx -other things being equal- we should multiply by 2?
By that way the probability becomes almost double, kind of strange! Never seen that before.

Thank you in advance!
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shanks2020
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A jury pool consists of 6 men and w women. If 2 jurors 03 Jul 2020, 13:59
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Bunuel
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christianbze
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:
\(\frac{6x5}{(6+w)(6+w-1)} > \frac{6 x w x 2}{(6+w)(6+w-1)}\)

\(\frac{30}{(6+w)(5+w)} > \frac{12 w}{(6+w)(5+w)}\)

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:
https://gmatclub.com/forum/a-jar-contain ... 01748.html (almost the same question from MGMAT)
https://gmatclub.com/forum/a-certain-jar ... 04924.html
https://gmatclub.com/forum/if-2-differen ... 28233.html
https://gmatclub.com/forum/a-certain-box ... 53384.html

Hope this helps.
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Bunuel GMATNinja

Why do we not multiply when we use 6C1*wC1, but we do when using probability formulae?
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GDT
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A jury pool consists of 6 men and w women. If 2 jurors 04 Sep 2020, 06:46
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Bunuel
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Bunuel


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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VeritasKarishma

Can you pls explain with examples in what ques we have to consider both cases
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GDT
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A jury pool consists of 6 men and w women. If 2 jurors Updated on: 18 Sep 2020, 00:35
Originally posted by GDT on 17 Sep 2020, 06:43.
Last edited by GDT on 18 Sep 2020, 00:35, edited 1 time in total.
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VeritasKarishma

I remember reading a beautiful post by you that very well explains in what type of combination ques we have to consider both case that is selecting man first then woman and vice versa

Can you pls explain that or share the link of that post of "Quarter Wit Quarter wisdom "

I have searched all over the internet but can't find it

It will be of great help!
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A jury pool consists of 6 men and w women. If 2 jurors 17 Sep 2020, 18:23
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Hi Bunnel! everything except one thing, all I understood from explanation.

My point is how did you consider my multiplication of 2 in this context.
Here we are selecting one man from 6 men, and one woman from w women.
Now how does it matter who is first and who is second.Moreover, my understanding till date was that combination problem doesn't count the sequence of selection.
but yes if I go with fundamental then that 2 can pop up in expression.
one man from 6 men = 6C1
oNE women from W women= WC1
Now 2people from from 6+w people = 6+wC2= (6+W)(6+W-1)/2!

SO final expression is = 2*(6*w)/(6+w)(6+w-1)

This is my point. please correct if I am wrong.



Bunuel
joshlevin90
christianbze
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:
\(\frac{6x5}{(6+w)(6+w-1)} > \frac{6 x w x 2}{(6+w)(6+w-1)}\)

\(\frac{30}{(6+w)(5+w)} > \frac{12 w}{(6+w)(5+w)}\)

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:
https://gmatclub.com/forum/a-jar-contain ... 01748.html (almost the same question from MGMAT)
https://gmatclub.com/forum/a-certain-jar ... 04924.html
https://gmatclub.com/forum/if-2-differen ... 28233.html
https://gmatclub.com/forum/a-certain-box ... 53384.html

Hope this helps.
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A jury pool consists of 6 men and w women. If 2 jurors Updated on: 30 Nov 2023, 06:33
Originally posted by KarishmaB on 22 Sep 2020, 00:01.
Last edited by KarishmaB on 30 Nov 2023, 06:33, edited 1 time in total.
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GDT
VeritasKarishma

I remember reading a beautiful post by you that very well explains in what type of combination ques we have to consider both case that is selecting man first then woman and vice versa

Can you pls explain that or share the link of that post of "Quarter Wit Quarter wisdom "

I have searched all over the internet but can't find it

It will be of great help!
Show more

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pradeepgontla
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A jury pool consists of 6 men and w women. If 2 jurors 30 Sep 2020, 08:16
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Shobhit7
Ways of selecting 2 men: 6C2
Ways of selecting 1 man and 1 woman: 6C1*wC1
Total Ways: (6+w)C2

Rephrase Q stem:
Is 6C2 / (6+w)C2 > (6C1*wC1) / (6+w)C2 ?
Is 6C2 > 6C1*wC1 ?
Is 15>6w ?
Is w<15/6?
Is w<2.5?

St.1: No, w>=3
St.2: May be. w<6

Ans A
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I exactly used the same way all other explanations are way too lengthy and complex. This is the best possible solution.
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MS26
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A jury pool consists of 6 men and w women. If 2 jurors 25 Feb 2026, 19:28
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Men=6, women=w total=6+w
P(2Men)=6C2/6+wC2,=30/(6+w)*(5+w)
P(1M&1W)=6C1*wC1/6+wC2,=12w/(6+w)*(5+w)
now question asks us is P(2Men)>P(1M&W)?
equating this will give 30/(6+w)*(5+w)>12w/(6+w)*(5+w), i.e, 5/2>w,w<2.5
now going to the stem 1 w>=3 so answer will always be no
stem 2 w<6 , here answer will b eyes and answer is A
goodyear2013
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
Show Spoiler
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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