Probability of selecting $j$-th ball from urn is $\mathbb{P}(B_j\mid A_k)=\frac{k}{n}$ 
An urn contains $M$ balls, of which $M_1$ are white. A set of $n$ balls is obtained either
  by sampling balls with replacement (each ball is replaced in the urn after it is drawn), or
  without replacement (a ball once drawn remains out of the urn). 
Let $B_j$ be the event that
  the ball selected at the $j$-th step is white, and $A_k$ the event that a sample of size $n$ contains
  exactly $k$ white balls. Show that $$\mathbb{P}(B_j\mid A_k)=\frac{k}{n}$$both for sampling with replacement and for sampling without replacement.

It seems clear that $\mathbb{P}(B_1) =\frac{M_1}{M}$ and $\mathbb{P}(B_2) = \frac{M_1}{M} \frac{M-M1}{M-1} + \frac{M_1}{M} \frac{M_1-1}{M-1} = \frac{M_1}{M}$. We can probably suspect that $\mathbb{P}(B_j)=\frac{M_1}{M}$ as well, if $j \leq M_1$. 
However, it's not clear for me how to construct the probability $\mathbb{P}(B_j\mid A_k$) from here. 
If I'm not wrong, $$\mathbb{P}(A_k) = \frac{\left( \begin{array}{c} M_1\\k \end{array} \right)\left( \begin{array}{c} M - M_1\\ n -k \end{array} \right)}{\left( \begin{array}{c} M\\n \end{array} \right)} \tag{1}$$due to hypergeometric distribution. From here $$\mathbb{P}(B_j\mid A_k)=\frac{\mathbb{P}(B_j \cap A_k)}{\mathbb{P}(A_k)} $$.
But I'm not sure how to estimate the $\mathbb{P}(B_j \cap A_k)$. As far as I understand, instead of all possible events as in (1), event $\{ B_j \cap A_k \}$ should cover a part of it where the $B_j$ holds, but not sure how to get to the $\frac{k}{n}$ from here. 
Any hints or suggestions would be appreciated!
 A: To compute $P(B_j\cap A_k)$, you have to get your hands dirty with some combinatorics. Suppose that our sample space consists of all ordered selections of $n$ balls without replacement. There are $\frac{M!}{(M-n)!}$ such selections. We must count the number of sequences where there are $k$ white balls and where the $j^{th}$ ball is white. 
There are $M_1$ ways to choose the white ball that goes in the $j^{th}$ spot, then $\binom{M_1-1}{k-1}$ ways to choose the other $k-1$ balls, then $\binom{M-M_1}{n-k}$ ways to choose the black balls. Then, the $n-1$ balls (excepting the ball in spot $j$) can be ordered in $(n-1)!$ ways. Therefore, 
$$
P(B_j\cap A_k)=\frac{M_1 \binom{M_1-1}{k-1}\binom{M-M_1}{n-k}(n-1)!}{\frac{M!}{(M-n)!}}
$$
You need to redo all this work in the situation where you sample with replacement. 

Alternatively, you could argue that given $A_k$, all $\binom{n}k$ ways of choosing the order of black and white balls are equally likely, and there are $\binom{n-1}{k-1}$ of those which have a white ball in their $j^{th}$ spot, so $P(B_j|A_k)=\binom{n-1}{k-1}/\binom{n}k=k/n$.
