Number of successes when the successes are positively correlated In $n$ trials, each with success-probability $p$, what is the probability of at least $k$ successes, $P[n,k]$ ?
The answer depends on the dependence between the trials:
A. If the trials are independent, then the number of successes is a Binomial variable, so:
$$
P^{\text{ind}}[n,k] = \sum_{i=k}^n {n\choose k} p^i (1-p)^{n-1}
$$
B. If the trials can be arbitrarily dependent, then $P[n,k]$ might be 0. E.g, take $p=1/2$, $n=2,k=2$, and assume $X_2 = 1 - X_1$ (trial 2 fails iff 1 succeeds; from Henry's comment).
C. Now suppose that the variables are dependent "in the good direction":
$$
\forall i_1,\ldots,i_j: \Pr[X_{i_1}=1\cap\cdots\cap X_{i_j}=1] \geq \Pr[X_{i_1}=1]\cdots\Pr[X_{i_j}=1] = p^j
$$
that is, if we already had some successes, the probability of another success weakly increases. 
Here, $P[n,k]$ cannot be zero. A trivial upper bound is $P[n,k] \geq p^k$ (as commented by muzzlator), since even if we consider only trials $1,\ldots,k$, the assumption implies that the probability  all of them succeed is at least $p^k$
Is there a better lower bound for $P[n,k]$ in this case?
Initially I thought that $P[n,k]\geq P^{\text{ind}}[n,k]$, since intuitively, the dependence between the variables can only increase the number of successes. However, this is not true. As a simple example (from muzzlator's comment), take $k=1, n\to\infty$ and assume that all variables are the same: $X_i=X_1$ for all $i$. Then, $P^{\text{ind}}[n,1]\to 1$ while $P[n,1]=p$.
So, what is a correct lower bound on $P[n,k]$?
 A: Trying a different approach here. Once again, I'm very sorry for how rough it is, and I'll try improve on it over time, but I'm stretched for time atm so I'll leave it here as something you might consider rather than as a full-fledged answer
Define the $i$-intersection as the region where exactly $i$ of the variables are true.
We want to get the most bang for buck by placing as much probability in the $(k-1)$-intersections as possible. That way, the probability budget here is spent in the most efficient way possible since the area is shared among $k-1$ things. There's no harm for our criteria putting our probability here since we're independent in the "good" direction.
The rest of our probability budget we want to spend in the $n$-intersection. This is because we use the least probability while knocking off all the variables here. That is to say, let's maximise our chance of exactly $k-1$ successes or otherwise, always have $n$ successes.
If the $n$-intersection were to have probability $p^k$, then we'd have to make sure every variable gets $p - p^k$ probability added to it. We can distribute these across the $(k-1)$-intersections. Each variable is associated with ${n-1} \choose {k-2}$ $(k-1)$-intersections so we need to make sure each $k-1$ intersection gets $(p - p^k)/{{n-1}\choose{k-2}}$. We then sum these up, there are total of $n \choose {k-1}$ of these so we are left with:
$$ \frac{{n}\choose{k-1}}{{n-1}\choose {k-2}} (p - p^k) $$
Simplifying, this is:
$$ \frac{n}{k-1} (p - p^k)$$
The total area is therefore:
$$ \frac{n}{k-1} (p - p^k) + p^k $$
If this probability is less than or equal to $1$, we're good.
In general, we want to know when:
$$ \frac{n}{k-1} (p - A) + A \leq 1 $$
$$ \frac{k - 1 - n}{k-1} A \leq \frac{k - 1 - np}{k-1}$$
$$ A \geq \frac{np - k + 1}{n - k + 1} $$
So I'm going to say choice for lower bound maximum of $p^k$ and $\frac{np - k + 1}{n - k + 1}$.
Handwavey reasoning that the constraint is true: we have less area per variable in the ($k$ and above)-intersections (what I call the center) and therefore more in the lower intersections (the outside). Moreover, more of this is shared with all the other variables. Moreover, even if we use more than $p^k$ in the center, we're never going to add to it more than how much we'd have for $k$ or more successes with independent Bernoulli trials. <-- Edit: the flaw in this reasoning is that we count the ($k$ and above)-intersections area towards any intersection of fewer variables, and so having more area-per-variable on the outside doesn't mean much if the area in the center has decreased by too much.
Simple example of this:
In the case where $n=3, k=2, p=0.9$, $p^2 = 0.81$ and $\frac{np - k + 1}{n - k + 1} = 0.85$ and so the max is $0.85$. We can choose a distribution where probability of a single success is $0.05$ for each variable ($0.15$ total) and probability of all three succeeding is $0.85$
If $n=3, k=2, p=0.1$, then $p^2 = 0.01$ and $\frac{np - k + 1}{n - k + 1} = -0.35$ and so the maximum is $0.01$. We then share $0.01$ for all successes and have $0.09$ for each variable being the only success.
One more example:
If $n = 20, k = 10, p = 0.5$, then maximum is $\frac{np - k + 1}{n - k + 1} = \frac{1}{11}$. We now have $20 \choose 9$ places to put $(\frac{1}{2} - \frac{1}{11})/{19 \choose 8}$. That is, probability of exactly $9$ successes is $(\frac{1}{2} - \frac{1}{11})/{19 \choose 8}$ for each combination of $9$ things, and probability of all $20$ things succeeding is $\frac{1}{11}$. We'll see if the constraint is met here in the case of $P(X_1 \cap X_2) \geq 0.25$: We expect $X_1$ and $X_2$ to be in $18\choose 7$ of those $9$-intersections. That is a total probability of ${18\choose 7}(\frac{1}{2} - \frac{1}{11})/{19 \choose 8} = \frac{36}{209}$, add to that $\frac{1}{11}$ and we have $\frac{5}{19} > \frac{1}{4}$
To summarize: Consider the following distribution, where $A=\max\big(p^k, {np-k+1\over n-k+1}\big)$:


*

*with probability $A$, all $n$ trials succeed. 

*with probability $(p-A) {{n \choose {k - 1}} \over {{n-1}\choose{k-2}}}$, a $k-1$-tuple of trials is selected at random, these $k-1$ trials succeed and the rest fail. 
Then:


*

*The success probability of each single variable is (by construction):


$$
A + (p-A) {{{n-1}\choose{k-2}}\over{{n-1}\choose{k-2}}}
= p
$$


*

*The success probability of at least $k$ trials is exactly $A$.


It remains to prove that the success probability of each $j$-tuple of variables is at least $p^j$, i.e:
$$
A + (p-A){ {n-j \choose k-j-1} \over {n-1\choose k-2} }
\geq
p^j
$$
This is obvious for $j\geq k$ since $A\geq p^k$. We proved it above for $j=1$. It remains to prove (or disprove) this for $1<j<k$.
Edit: See g g's examples below, he proves that the second condition is false. It's interesting that it always appears to be the $k-1$-intersection that causes the constraint, that is, can we use:
$$A + (p-A){ 1 \over {n-1\choose k-2} } \geq p^{k-1} $$
as the only additional constraint?
