Suppose $B_1, ..., B_k$ are dependent Bernoulli random variables with $$P(B_i = 1) \geq p$$ for each $1 \leq i \leq k$. Let Binom(k,p) be a binomial random variable with parameters $k$ and $p$.
Can we always say that $$P(B_1 + ... + B_k \geq w) \geq P(Binom(k,p) \geq w)$$ for each $w \geq 0$? (i.e. $B_1 + ... + B_k$ stochastically dominates $Binom(k,p)$). If not, what assumptions can we make so that the claim holds?
Let $U_1, \ldots, U_k$ be independent uniform random variables on $[0,1].$ Let $p_1, \ldots, p_k$ be numbers satisfying $p_i \geq p,$ where $p \in (0,1)$ is also a given number. Let $X_i = \mathbf{1}_{[0, p)}(U_i)$ and $Y_i = \mathbf{1}_{[p, p_i)}(U_i).$ Define $B_i = X_i + Y_i.$ Then, the $(X_i)$ are independent Bernoulli random variables with parameter $p$ and the $B_i$ are independent Bernoulli random variables with corresponding parameters $p_i = \mathbf{P}(B_i = 1) \geq p.$ Note that $\sum B_i \geq \sum X_i,$ so if $\sum X_i \geq w,$ then $\sum B_i \geq w,$ and this implies $$ \mathbf{P}(\mathsf{Bin}(k,p) \geq w) = \mathbf{P}\left( \sum X_i \geq w \right) \leq \mathbf{P} \left(\sum B_i \geq w \right). $$
Ammend. OP wanted the Bernoulli variables to be "dependent." We can take $B_1 = \ldots = B_k = B$ a Bernoulli variable of parameter $p + \delta.$ Then, for $w \in (0, k),$ $\mathbf{P}(B_1 + \ldots + B_k \geq w) = \mathbf{P}(B \geq w / k) = p + \delta.$ This does not depend on $w;$ if $w \in (0,1),$ then $\mathbf{P}(\mathsf{Bin}(k,p) \geq w) = 1 - \mathbf{P}(\mathsf{Bin}(k,p) = 0) = 1 - (1-p)^k;$ and if $w \in (k-1,k),$ $\mathbf{P}(\mathsf{Bin}(k,p) \geq w) = p^k.$ Therefore, for all large enough $k,$ $p^k < p + \delta < 1 - (1-p)^k.$ This obviously shows that you cannot have stochastic dominancy for all $p$ and $k$ and all types of dependency. If you want other types of dependency (e.g. something like $B_i = B + U_i,$ with the $U_i$ independent, so the $B_i$ only depend on the common $B$), you may be able to rescue the proof of the independent case but the algebra can get excruciating quickly.