Probability that two subspaces are complementary $p$ - a prime number
$Gr_{n,k}(F_p)$ - set of all k-dimensional subspaces of the vector space $F_p^n$
$F_p$ - finite field with p elements
For $1\leq k\leq n-1$ fixed, the subspaces $V$ and $W$ are drawn independently and randomly from $Gr_{n,k}(F_p)$  and  $Gr_{n,n-k}(F_p)$ resp. $A_p$ is the event that  $V$ and $W$ are complementary: $F_p^n=V \oplus W$. How to show that $$\lim_{p\to \infty}P(A_p)=1$$ holds? Thanks!
 A: Think in terms of basis vectors.  You have a linearly independent set $B_V=\{\vec v_1,\dots, \vec v_k\}$ of basis vectors for $V$, and a linearly independent set $B_W=\{\vec w_1,\dots,\vec w_{n-k}\}$ of basis vectors for $W$.  We are interested in the probability that if we put these two sets together we get a linearly independent set of vectors (hence a basis for $F_p^n$).
Now let's look at something less likely. Namely randomly drawing $n$ vectors (with replacement) from $F_p^n$ and getting a basis.  (Less likely, because we don't have the guarantee of linear independence of certain subsets.)  We call this event $C_p$.
As we subsequently draw each vector, the probability that the new draw is linearly independent of all previous vectors goes down. 
The probability that the first vector is lienarly independent (i.e., nonzero) is $\frac{p^n-1}{p^n}$.  Then the conditional probability that the next vector is linearly independent (given that the first is linearly independent) is $\frac{p^n-p}{p^n}$ (since the first linearly independent vector defines a subspace of dimension $1$, and hence with $p$ vectors).  The conditional probability that next vector is linearly independent (given that the first two vectors were) is $\frac{p^n-p^2}{p^n}$. And so on.  
The probability, $P(C_p)$ that the $n$ randomly drawn vectors are all linearly independent is given by 
$$P(C_p)=\frac{p^n-1}{p^n}\cdot \frac{p^n-p}{p^n}\cdot \frac{p^n-p^2}{p^n}\cdots\frac{p^n-p^{n-1}}{p^n}=\left(1-\frac{1}{p^n}\right) \left(1-\frac{1}{p^{n-1}}\right)\cdots \left(1-\frac{1}{p}\right) $$
We notice that the last factor is the least of the factors. 
So $P(C_p)\ge \left(1-\frac1p \right)^n\to 1$  as $p\to \infty$. So $P(C_p)\to 1$ also.
However for any $p$, we have $P(A_p)\ge P(C_p)$.
Hence also $P(A_p)\to 1$ as $p\to\infty$.
