Independence in Bernoulli scheme We've got probability space $(\Omega, F,P)$ and Bernoulli scheme with $n$ trials and with success probability equal to $p$. $A_k$ means exactly $k$ successes in $n$ trials. Prove that for any $B \in F$ and for any $k$, $P(B | A_k)$ doesn't depend on $p$.
So I thought of writing it this way: $P(B | A_k) = \frac{P(A_k \cap B)}{P(A_k)}$ and $P(A_k) =$$ {n}\choose{k}$$p^k q^{n-k}$, but what next?
 A: Call $X=(X_i)_{1\leqslant i\leqslant n}$ the $n$ results and $S_n=X_1+\cdots+X_n$, then $A_k=[S_n=k]$. It seems that, by $B\in F$, you mean that $B$ is measurable with respect to $X$, that is, that there exists some $C\subseteq\{0,1\}^n$ such that
$B=[X\in C]$. 
Call $C_k=\{c\in C\mid c_1+\cdots+c_n=k\}$, then $B\cap A_k=[X\in C_k]$ and $\mathbb P(X=c)=p^k(1-p)^{n-k}$ for every $c$ in $C_k$ hence $\mathbb P(B\cap A_k)=|C_k|\cdot p^k(1-p)^{n-k}$. 
Likewise, $\mathbb P(A_k)=\displaystyle{n\choose k}\cdot p^k(1-p)^{n-k}$ hence $\mathbb P(B\mid A_k)=\dfrac{|C_k|}{{n\choose k}}$, which does not depend on $p$.
A: This property is often referred to as the memorylessness property in probability and statistics. Let's rewrite $B$ in the same fashion as we have written $A_k$. If $A_k$ means exactly k successes in n trials, we can say $B_j$ means exactly $j$ successes in $m$ trials, where $j\geq k$, $m\geq n$. What we want to show is that:
$P(B_j |A_k)=C_{j-k}$, where $C_{j-k}$ means exactly $j-k$ successes in $m-n$ trials, using
$P(B_j | A_k) = \frac{P(A_k \cap B_j)}{P(A_k)}$, as you stated, where
$P(A_k) =$$ {n}\choose{k}$$p^k q^{n-k}$
$P(B_j) =$$ {m}\choose{j}$$p^k q^{n-k}$ and
$P(C_{j-k}) =$$ {m-n}\choose{j-k}$$p^{j-k}q^{m-n-(j-k)}$.
The next step is to apply our set theory definitions along with properties of $ {n}\choose{r}$ in order to reduce $P(B_j | A_k) = \frac{P(A_k \cap B_j)}{P(A_k)}$ to the form of $P(C_{j-k})$.
