Please give some comments and hints for the problem.
A binary linear code of dimension $d$ and length $n$ is a code over the alphabet $A=\mathbb{F}_2^d$ defined through a matrix $M \in \mathbb{F}_2^{n \times d}.$ For $\vec{a} \in A,$ the corresponding codeword is the vector $M \vec{a} .$ The minimum distance of the code is the minimum Hamming distance between two distinct codewords; that is, $\min \left\{\left\|M \vec{a}-M \vec{a}^{\prime}\right\|_1: \vec{a} \neq \vec{a}^{\prime} \in \mathbb{F}_2^d\right\}$.
Prove that if $2^d \,\mathbb{P}\left(\operatorname{Bin}\left(n, \frac{1}{2}\right)<k\right)<1,$ then there is a binary linear code of dimension $d$, length $n$ and minimum distance at least $k$.
My idea is a two-part proof.
Part 1: let $Z$ be a random variable counting the number of codes of dimension $d$, length $n$ and minimum distance at least $k$. I want to show that $\mathbb{E}[Z] \geq 1$ (or possibly just $\mathbb{E}[Z] > 0$, as $Z$ is discrete).
To this end I’m thinking of letting $\mathbb{E}[Z] = \sum_{j=1}^{2^{nd}} \mathbb{E}[Z_{M_j}]$ (by the linearity of expectation), where $Z_{M_j}$ is an indicator random variable signifying whether or not the matrix $M_j$ is a matrix associated with a code satisfying the condition. The number $2^{nd}$ is the number of matrices $M_j$.
Part 2: we let $\vec{a}$ and $\vec{a}’$ be two vectors in $A$ with entries chosen randomly and independently. Then for a given $M$, let $X_i$ be the indicator variable that $M\vec{a}$ and $M\vec{a}’$ differ at position $i$. Then again by linearity of expectation, the random variable $X$ counting the number of different positions, i.e. the Hamming distance between $M\vec{a}$ and $M\vec{a}’$, has expectation $\mathbb{E}[X] = \sum_{i=1}^{n} \mathbb{E}[X_i]$. I calculate that this number is $\frac{n}{2}$.
Still for the given code $M$, the probability that $X \geq k$ can be related to the assumption, i.e. $$\mathbb{P}(X \geq k) = 1 - \mathbb{P}(X < k) = 1 - \mathbb{P} \left(\frac{n}{2} < k\right) > 1 - \frac{1}{2^d}.$$
It is here that I got stuck. I don’t know how to relate the $Z_{M_j}$ in part 1 with the information I’ve obtained in part 2. I was thinking of using the union bound, as $Z_{M_j} = 1$ corresponds to the event that there’s a pair $M_j\vec{a}$ and $M_j\vec{a}’$ satisfying the condition, which is the union of the events that some pairs satisfying the condition. But this only gives me an upper bound on $\mathbb{E}[M_j]$, which is not what I need.
So here's a way to do it. Let $M$ have i.i.d. Bernoulli $1/2$ entries. Among all $a \neq a'$, there are only $2^d - 1$ possible vectors for $a - a'$ (remember we are in $\mathbb{F}_2$). For each $v$ nonzero vector in $\mathbb{F}_2^d$ note that $ Mv $ is a uniformly random vector of $\mathbb{F}_2^n$. In particular, $\| Mv\|_1 \sim \mathrm{Bin}(n,1/2)$. We may then bound $$\mathbb{P}( \exists~v \in \mathbb{F}_n^d \setminus \{0\} : \|Mv\|_1 < k) \leq (2^{d} - 1)\mathbb{P}(\mathrm{Bin}(n,1/2) < k) < 1\,.$$
In particular, this implies $$\mathbb{P}( \|Mv \|_1 \geq k \text{ for all }v \neq 0) > 0\,.$$
This means that there exists some $M$ so that $\|Mv\|_1 \geq k$ for all $v\neq 0$, which is indeed the code you want of minimum distance at least $k$.