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Answer is E.

Since marbles are identical. Probability of drawing a white marble will always be 1/2 in any draw.

In any draw of one marble it will be either a black marble or a white marble.

I would like you to pay attention to the word "equiprobable".

Suppose we have two marbles in a bag: one black and one white. We pick a marble at random. What outcomes do we have?

Outcome 1: we picked the white marble.

Outcome 2: we picked the black marble.

"at random" means that picking each marble is equiprobable. So P1 = P2, where P1 and P2 are the probabilities of the outcomes.

We also know that at least one of the marbles is chosen, so the probabilities add up to 1.

P1 + P2 = 1

From these two formulas we deduce that the probability of picking the white marble is 1/2.

Suppose we have three marbles in a bag: one white marble and two black marbles. We pick a marble at random. What outcomes do we have?

Let's number the marbles in the way that #1 is the white marble, and #2, #3 are the black marbles. The outcomes we have:

Outcome 1: we picked the marble #1 (white).

Outcome 2: we picked the marble #2 (black).

Outcome 3: we picked the marble #2 (black).

These outcomes are equiporbable and one of them will definitely happen. So we have the same formulas P1 = P2 = P3 and P1 + P2 + P3 = 1.

From the formulas we deduce that P1 = 1/3 (the probability of picking the white marble is 1/3).

Could we consider outcomes as

Outcome 1: we picked the white marble.

Outcome 2: we picked a black marble.

?

Yes, we could. But these outcomes will NOT be equiprobable. So you can not say that the probability of each outcome is 1/2. In fact, the outcome "we picked a black marble" consists of "we picked the black marble #2" and "we picked the black marble #3". Its probability is 1/3 + 1/3 = 2/3 .