A bag contains $n$ balls of unknown colours. The first ball taken from the bag was white. What is the probability of second ball is also white, when we assume, that the number of white balls in bag is equally likely?
I guess the key part here is to make use of Bayes' theorem. My idea was to consider such 'disjoint' events.
$A_1$ when bag contains exactly $1$ white ball.
$A_2$ when bag contains exactly $2$ white ball.
$\dots$
$A_n$ when bag contains exactly $n$ white ball.
But I cannot see what is $B$ here, and this idea lead me to dead end.
I have also made a simple c++ program, which simulates the situation and produces the answer nearly equal to $2/3$. I know this is somewhat useless, but knowing the answer is helpful.
Your prior distribution is that there is $\frac 1{n+1}$ chance of each number of white balls from $0$ to $n$. The drawing of a white ball says the probability of $A_0$ is zero. The chance of $A_i$ is now proportional to $i$ as you had $\frac in$ chance of drawing a white ball if you started with $i$ of them. The sum of $\frac in$ is $\frac {n(n+1)}{2n}=\frac {n+1}2$ so the probability of $A_i$ is now $\frac {2i}{n(n+1)}$ The chance we get a white ball assuming we started with $i$ white balls is $\frac {i-1}{n-1}$ so the chance we get a white ball overall is $$\sum_{i=1}^n \frac {2i}{n(n+1)}\cdot\frac {i-1}{n-1}=\frac 1{n(n^2-1)}\sum_{i=1}^n(2i^2-2i)\\=\frac 2{n(n^2-1)}\left(\frac {n(n+1)(2n+1)}6-\frac {n(n+1)}2\right)\\=\frac 2{n(n^2-1)}\cdot\frac {n(n+1)(2n-2)}6\\=\frac 23$$