DS. In a bag there are R red balls, W white balls, and G

Question

DS. In a bag there are R red balls, W white balls, and G green balls. If a ball is chosen in random, is it more likely to be red or white? 
A. R/(W+G)>W/(R+G) 
B. G>R+W

damit · Answer

A? will explain in a bit if correct

ywilfred · Answer

From A, we can simply the inequality to:

R^2 + RG > W^2 + WG
R^2 - W^2 > G(R + W)
(R + W)(R - W) > G (R + W)
(R-W) > G

Since G must be positive, so R-W must be positive. For R-W to be positive, R must be greater than white. 
So From A, we know by probability, R is more likely.

Sufficient.

From B, G > R+W does not tell us anything. For all we know, R and W can be equal in number and so equally probably.
So not sufficient.

mallelac · Answer

I believe it is A.

Add 1 to both sides of A and take the inverse of the inequality. The resulting inequality clearly gives the answer that it is more likely white.

B is insufficient as we cannot ariive at any relation between R and W.

hardworker_indian · Answer

A.

A. R/(W+G)>W/(R+G) 
=> (2R)/(W+G+R) > (2W)/(W+G+R)
=> R>W
Actual ans: R is more probable than W
Sufficient

B. G>R+W
Green is more in number than combined R and W.
But we cannot conclude anything betweem R andW.
Insufficient.

DS. In a bag there are R red balls, W white balls, and G

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