Re: Solutions to exercises
In problem 1.1 the bags and the suggested use of Bayes Theorem is a bit of a red Herring. You have 4 balls 3 of which are black. You pick one ball which is black. There are now 3 balls 2 of which are black. The chance that you pick a black one is 2 out of 3. It doesn't matter how many bags there are (as long as there is at least one ball in each bag) or from which bag you choose the second ball (as long as there is a ball in it).
So for example if I have 10 balls spread over 3 bags (e.g. 4,4,2) and one is white and the first ball I pick is black. Then the chance I pick another black ball from any particular bag is 8/9. Try solving this using the Bayesian Theorem :^).
You can think of the bags as groupings on the balls on a table in the dark. If you select a ball from one group it doesn't matter which group you select the second ball from.
