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 wrote:
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:
\(\frac{6x5}{(6+w)(6+w-1)} > \frac{6 x w x 2}{(6+w)(6+w-1)}\)
\(\frac{30}{(6+w)(5+w)} > \frac{12 w}{(6+w)(5+w)}\)
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.htmlhttps://gmatclub.com/forum/if-2-differen ... 28233.htmlhttps://gmatclub.com/forum/a-certain-box ... 53384.htmlHope this helps.