The answer is A.
From(1) -> W/B+W = W/(W/2 + W) = 2/3, so sufficient.
From (2), we can only infer that # of Black balls <= 1, but no info about White balls
Yes, we can assume from S2 that there are 0 or 1 black ball. because the probability is 0 can imply three things :
The probablity = B1/N * B2/(N-1)
B1 = # of black balls in container during first draw
B2 = # of black balls during second draw
N = Total # of balls
Because the probability of this even is zero, so:
1) First drawn ball is black but the second is not
2) Second drawn ball is black but the first is not
3) Both the balls are not black.
Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)
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