For $i=1,2,\dots,l$, let $\mathbf{v}\_i =(v\_{i1},v\_{i2},\dots,v\_{in}) \in \mathbb{F}\_2^n$ be a sparse vector in GF(2) such that all $v\_{ij}$'s are independent for all $1 \le i \le l, 1 \le j \le n$ and $$\mathrm{Pr}[v\_{ij}=1] = \frac{\log(n)}{n}, \qquad 1 \le i \le l, 1 \le j \le n \\\mathrm{Pr}[v\_{ij}=0] = 1-\frac{\log(n)}{n}, \quad 1 \le i \le l, 1 \le j \le n$$ 
Let $\mathbf{e}\_1 = (1,0,0,\dots,0) \in \mathbb{F}^n\_2$ be a base vector in GF(2) in which the only non-zero element is the first element. What can we say about the probability that this base exists in the span of the vectors $\mathbf{v}\_1,\mathbf{v}\_2,\dots,\mathbf{v}\_l$?
$$\mathrm{Pr}[~\mathbf{e}\_1 \in \mathrm{Span}\{\mathbf{v}\_1,\mathbf{v}\_2,\dots,\mathbf{v}\_l\}~] = ~?$$
This probability quickly approaches 1 as $l$ approaches $n$ but is it possible to find a closed form for this probability in terms of $l$ and $n$? If not, can we find good upper and/or lower bounds for this probability in terms of $l$ and $n$? 



Probability of existence of a base in the span of sparse vectors in GF(2)
Let $V$ be the $n \times l$ matrix whose columns are your vectors.
For a given vector $x \in \mathbb F\_2^l$ with $k$ $1$'s, and $1 \le i \le n$, the number of 
$j$ for which $x\_j = 1$ and $v\_{ij} = 1$ 
 is a binomial random variable $W\_i(x)$ with parameters $(k,p)$, with $W\_i(x)$ for different $i$ independent. Thus $$\mathbb P((Vx)\_i = 1) = \mathbb P(W\_i(x) \ \text{odd}) = \dfrac{1}{2} (1 - (1-2p)^k)$$
so that
$$ \mathbb P(V x = e\_1) = 2^{-n} (1 - (1-2p)^k) (1 + (1-2p)^k)^{n-1}$$
The expected number of $x$ for which $V x = e\_1$ is then
$$ E(p,n,l) = \sum\_{k=0}^l {l \choose k} 2^{-n}(1 - (1-2p)^k) (1 + (1-2p)^k)^{n-1} $$ 
Now if $n \ge l$, $\ker V$ has dimension $\ge n-l$, so if there are
any such $x$ there are at least $2^{n-l}$ of them. Thus we can get an upper bound
$$\mathbb P(e\_1 \in \text{Span}(v\_1, \ldots, v\_l)) \le 2^{l-n} E(p,n,l)$$


