(Laplace example). We have $r + 1$ urn. In each urn, there is a $r$ of balls, with Urn number i contains both white and $r-i$ and black balls. The drawing takes place in the following way.
- We choose one of the urns with the same probability $\frac{1}{r+1} $
- We draw n + 1 times with the return from the previously selected urn. It is known that in n draws we chose only white balls. (a) Calculate the probability that in the next (n + 1)-drawing we will also draw white?
My solution: Let $X_1^{i} \ldots X_n^{i} $ be outcome from ith urn. We try compute probability $$ P(X_1^{i} = white \land \ldots \land X_n^{i} = white | X_{n+1}^{i}=white)= \frac{1}{r+1}(\frac{i}{r})^{n+1} $$ It's correct thinking? I suppose that we have independent trials.
If you are new to this kind of stuff, the best way to solve a problem like this is to write down each event clearly.
For example, you have $A = \{ \text{I drew $n$ white balls consecutively from an urn I picked} \}$. This alone leads you nowhere, but you can decompose it into disjoint events like $$ A_i = \{ \text{I drew $n$ white balls from urn #$i$ I picked}. \} $$ Then you will find you need $D_i=\{\text{I picked urn #$i$ in the first step}\}$, as well as the conditional event $A_i | D_i$. Now you can use the total probability formula to calculate $\Pr(A)$.
Next you need $C=\{ \text{I drew $n+1$ white balls consecutively from an urn I picked}\}.$ Please start from here.
With more experiences you no longer need to write down the specifics of all the events. Instead you will be able to plugin the numbers into whatever formula in one shot.