Find a chance that intersection of power set entries is an empty set We are given set $A = \{1,2, ...n\}$. $k$ entries picked from the power set of $A$.
 Task is to find probability that  $A_1 \cap A_2 \space \cap \space... \cap \space A_k = \emptyset$.
I came up with solution, but result looks overcomplicated and probably with better knowledge of combinatorics one could do better and find actual answer.
Solution
We are given: $card(A)=n$ and $X_i \in 2^A$. We will denote $card(X_i) = m_i$
Also, I derived following formulas:
Chance of appearing of specific element $a$ in the subset $X_i$: $\frac{C_{n-1}^{m_i-1}}{C_{n}^{m_i}} = \frac {m_i} n$.
Analogously, chance of appearing two specific elements in the subset $X_i$: $\frac{C_{n-2}^{m_i-2}}{C_{n}^{m_i}} = \frac {(m_i-1)m_i} {(n-1)n}$. 
Final idea is: probability of intersection $X_1,X_2...X_k$ being an empty set is:
 1 - P(intesection has card 1) - P(intesection has card 2) ... - P(intesection has card n).
I was able to write it as:
$$1 - \sum_{i=1}^n\prod_{j=1}^k\prod_{l=1}^i \frac{m_j - l +1}{n - l +1}$$
Could it be considered as valid solution? Also I would appreciate hints about mistakes in it. Thank you.
 A: This is not a complete answer (I think that it is quite a problem to find one). I am just solving it for special case $k=2$ (and $n>0$) in order to inspire, and to give an opportunity to compare with your own thinkings.
Denoting the cardinality of a set $B$ by $\left|B\right|$ we have:
$$P\left(X_{1}\cap X_{2}=\emptyset\right)=\sum_{r=0}^{n}P\left(X_{2}\subset A-X_{1}\mid\left|X_{1}\right|=r\right)P\left(\left|X_{1}\right|=r\right)$$
The sets $X_1$ and $X_2$ are supposed to be distinct and if $X_1=\emptyset$ then that is not a direct consequence of $X_1\cap X_2=\emptyset$. In that light we take the case $X_1=\emptyset$ apart and come to: $$P\left(X_{1}\cap X_{2}=\emptyset\right)=P\left(X_{1}=\emptyset\right)+\sum_{r=1}^{n}P\left(X_{2}\subset A-X_{1}\mid\left|X_{1}\right|=r\right)P\left(\left|X_{1}\right|=r\right)$$
$P\left(X_{1}=\emptyset\right)=2^{-n}$, $P\left(X_{2}\subset A-X_{1}\mid\left|X_{1}\right|=r\right)=\dfrac{2^{n-r}}{2^{n}-1}$
and $P\left(\left|X_{1}\right|=r\right)=\dbinom{n}{r}2^{-n}$.
Substitution of this then leads to: $$P\left(X_{1}\cap X_{2}=\emptyset\right)=\frac{3^{n}-1}{4^{n}-2^{n}}$$
A: A recursion: Let $T(n,k)$ be the number of family of subsets (say $\{A_1,A_2,\dots A_k\}$) of $[n]:=\{1,2,3,\dots ,n\}$ such that $\bigcap_{i=1}^{k}A_i=\emptyset$ . We define $T(n,k):=0$ if $2^n<k$ (in effect, we are not allowing repetitions among $A_k$'s). 
Let $B(n,k,r)$ be the number of family of subsets (say $\{B_1,B_2,\dots B_k\}$) of $[n]$ such  that $\mid \bigcap_{i=1}^{k}B_i\mid=r$. Let $I:=\bigcap_{i=1}^{k}B_i$. Then, $\bigcap_{i=1}^{k}\left(B_i\setminus I \right)=\emptyset$. Hence, $B(n,k,r)=\binom{n}{r}T(n-r,k)$. We have the recursion, $T(n,k)=\binom{2^n}{k}-\sum_{\text{over } r}B(n,k,r)$, leading to
$$T(n,k)=\binom{2^n}{k}-\sum_{r=1}^{\lfloor n-\log_2k \rfloor}\binom{n}{r}T(n-r,k)$$
I have no idea about solving this recursion from here. We can replace $\lfloor n-\log_2k \rfloor$ by $n$ or $k$ (as $T$ is $0$ after sometime), if that helps solving the recursion.
