I think the solution given, ie 2/3, is incorrect. The flaw in the reasoning is that the three events, W1, W2 or B remaining in the bag are given equal likelihood. This is not so. W1 and B are only half as likely as W2. When this is realised the correct answer is seen to be 75%.

A simpler way of seeing this is that after the counter is removed there are two equally probable states: that the unknown counter was removed from the bag or that it was the added white one. In the first of these cases the probability of drawing a white counter again is 100%; in the second it is 50%. Averaging these two we get 75%.

Sorry if I labour the obvious, but another way to see it is to think of doing the experiment 100 times. After the first draw the unknown counter would have been removed in 50 cases, the added white one in 50. So we would expect to draw the white one in 25 cases of the first 50, and all 50 of the second lot, giving 75 / 100.

Soler