There are four balls in a hat: a blue one, a white one, and two red ones. Now I draw simultaneously two balls, look at them, and announce that at least one of them is red.
What is the chance that the other is red as well?
(In reply to re(2): Who's right
Why is anybody drawing balls saying anything about other colors? In this problem anybody drawing balls is going to announce that they have at least one red ball regardless if the red ball they see is first, second, or both.
There are four cases of seeing one red ball and making the announcement and one case of seeing two red balls and making the announcement. Therefore the answer to Ady's question is 1/5.
If the problem announced that the first ball is red (as opposed to at least one ball), then there are only three possibilities, making the probability 1/3.
That is the difference. "First ball" and "At least one ball" give different information as to the possible compositions of the pair of balls drawn.