am I splitting hairs?
"You pick a random bag and take out one marble. It is a white marble."
That is the end of the first action.
Question "what is the probability that the remaining marble from the same bag is also white?"
we are asked to take it that the first action removes bag B therefor it is logical to take it that which ever bag the white ball did not come from is nolonger in he picture. so we have one bag, containing one ball which is either black or white. so the probability of that one ball (which we know is either black or white) being white is 1/2.
Hold on I think I might be wrong.
If you ask What is the probability of you choosing the bag containing the two white balls? the answer is 1/3
It seems to me to be like heads or tails, the fact that it landed five times heads in a row the next time (if it's a fair coin) the odds are still 1/2
but the first action reduces the second action (that of determining the probable identity of the ball) to one bag (the same bag) which is either bag A or C. So that one bag which after the first action can only contain one ball which is either black or white.
"you know that you do not have Bag B (two black marble)"
but you also know you have one of two bags A and C.
If it is Bag A it contains one white ball and if it is Bag C it contains one black ball. There are not three balls to choose from, just the two. There is no "first" and "second" white ball because just as you have removed Bag B you have removed the so called "first" ball, it is the the ball you have in hand.
Why because had the balls in Bag A been numbered one and two and the balls in Bag C had one on the white and two on the black and then the ball removed from said bag been numbered one then you would know that which ever bag it came from (A or C) it must have left in the said bag a ball with number two on it.
Only by putting all three balls together in the said "same" bag can you have 2/3.
have I lost my marbles?