I am unsure if anyone has seen this problem anywhere, but I haven't. This is an original question, and later I realized, with my inability to answer it, I do have a fundamental gap in my understanding of probability theory. Conditional probability to be precise.
1. Say I were to give you an urn, and I say that there are either 5 black balls and 4 white balls in it OR there are 4 black balls and 5 white balls in it, but I'm not sure how many. What is the probability of drawing a white ball?
How different is the previous question from this one, which has a little more information.
2. There are 3 white balls and 2 black balls in the first box. There are 4 white and 4 black balls in the second box. A ball is randomly chosen from the first box and placed in the second box. A ball is then randomly chosen from the second box. What's the probability the chosen ball is white? The solution to the question 2, is obviously P(B) = P(A) *(P(B/A) + P(B/A')) = 23/45 with events A and B suitably defined (any adept mathematician, should be able to define A and B)
Would you do the same to question 1, by assuming a probability of 1/2 to each case and calculate as
1/2*(5/9 + 4/9) = 1/2