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sportyrizwan
I will go with A

1. Since there are twice as many white balls as black balls.
Probability to extract a white ball will be 2/3

2. "If two balls are extracted from the bowl, the probability that both of them are black is 0"
This is only possible when either there is no black ball or either 1 black ball. So we cannot determine the probability of extracting a white ball...


Am I correct ?


Yup sporty. Just one clearification.

I assume that sufficiency of A is pretty clear.

As far as B is concern, black ball cant not be 0 since question states that
A bowl contains white and black balls

So it is clear that there is 1 black ball.
Why B is insufficient?

assume there are only 1 black and 1 white ball- probability to extract a white ball from the bowl= 1/2

Now assume there are 3 white balls and 1 black ball - probability to extract a white ball from the bowl is 3/4

B is insufficient. Answer is A.
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A
S1 gives 2/3- suffi
S2 gives, prob both black=0, which says that there is only one black ball in there. So, the no of white balls can be anything, so we cant find prob of white= s2 not suffi
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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.
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The stem can be rephrased as w/(w+b)?

1. w = 2b. Sufficient by substitution
2. This means that there is only one black ball. Let t = total, so black = 1 and white = t - 1. This implies (t-1)/[(t-1)+1] = (t-1)/t. We don't know the total, so insufficient

A.
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