Answer should be A

Q asks "is the probability that the marble chosen will be red greater than the probability that the marble chosen will be white ? "

this is possible only if r>w

stm1 :
r/(b+w) > w/(b+r)

i.e. r(b + r) > w (b + w)

now since b is same(constant) in both expressions so r>w to make the expression true

hence SUFF

stmt 2: b-w >r
this would be possible for both the cases r>w & w>r

i.e b =20 ,w =14 ,r =5

& b=10, w =2 & r = 6

so Insuff

Thanks.
The Ans is A.

So is brute force substitution the thumb for most such problems ? I was pretty pressed for time, tried a few values and couldn't get it right and moved on..

-B

I did this way:
the prob of taking red ball out = r/b+w+r
for white ball = w/b+w+r
the question is whether r/b+w+r > w/b+w+r
or we have to find whether r > w
According to 1
r/(b+w) > w/(b+r)
Add 1 to both sides
this makes r+b+w/b+w>r+b+w/b+r
this givesb+r > b+w which means r > w SUFF
2.b - w > r
here we don't know if w>r or w<r
hence A

nice trick! Thanks for the explanation!

-B

one more shot,

we need to prove

r/(r+b+w) > b/(r+b+w)

cross multiplying

= r^2 + br+wr > wr+bw+w^2
= r^2+br > bw+ w^2
= r/(b+w) > w/(b+r)

which is the given first condition
hence A is suff.

b is not suff because, b>r+w is given, but it doesnt say anything about r and w.