Re: Friday Puzzle -- Another from Lewis Carroll
Larry Barrett
I think 2/3. It seems like Mr. Bayes could come to the rescue here but I can not construct the events properly.
I look at it this way:
There are 4 possible sequences, all equally likely.
1. First ball in the bag is W, second ball in the bag is W, draw the first ball and observe color (W), color of remaining ball = W.
2. First ball in the bag is W, second ball in the bag is W, draw the second ball and observe color (W), color of remaining ball = W.
3. First ball in the bag is B, second ball in the bag is W, draw the first ball and observe color (B), color of remaining ball = W.
4. First ball in the bag is B, second ball in the bag is W, draw the second ball and observe color (W), color of remaining ball = B.
Sequence 3 must be eliminated because the color of the drqwn ball must be W (by the statement of the problem).
Of the remaining 3 sequences that satisfy the conditions of the problem, 2 or the 3 have a W ball remaining in the bag. So probability the remaining ball is W =2/3.