Probability of at least two of $n$ independent events occurring subject to some conditions

This turned out stranger than I expected.

A good lower bound is $$\min\left(\frac{p^2}{4}, 1-\frac{4+2p-p^2}{4e^{\,p/2}}, 1-\frac{1+p}{e^{\,p}} \right)$$ where the components apply in the regions [0,0.923], [0.923,1.547] and [1.547,2].

For $n=3$, I got an explicit solution from Mathematica with these minimizing probabilities and arguments: \begin{align} \frac{p^2}{4} \ \ \ \ \ &\text{ via }\{\frac{p}{2}, \frac{p}{2}, 0\} \text{ if } p \in [0,1]\\ \frac{p^2(5-p)}{16} &\text{ via }\{\frac{p}{2}, \frac{p}{4}, \frac{p}{4}\} \text{ if } p \in [1,\frac{9}{5}]\\ \frac{p^2(9-2p)}{27} &\text{ via }\{\frac{p}{3}, \frac{p}{3}, \frac{p}{3}\} \text{ if } p \in [\frac{9}{5},2]\\ \end{align}

FullSimplify[Minimize[{q r + q s + r s - 2 q r s, p == q + r + s,
  0 <= q <= p/2, 0 <= r <= p/2, 0 <= s <= p/2}, {q, r, s}], Assumptions -> 0 < p < 2]

This and similar explicit results for $n=4$ suggest that for all $n$, the minimizing probabilities and arguments will be one of: \begin{align} \frac{p^2}{4} \ \ \ \ \ &\text{ via }\{\frac{p}{2}, \frac{p}{2}, 0, \ldots, 0\} \\ 1-\left(1+\frac{2p-p^2}{4-4q}\right) \left(1-q\right)^{n-1} &\text{ via }\{\frac{p}{2}, q, \ldots, q\} \\ 1-\left(1+p-\frac{p}{n}\right)\left(1-\frac{p}{n}\right)^{n-1} &\text{ via }\{\frac{p}{n}, \frac{p}{n}, \ldots, \frac{p}{n}\} \end{align} where $q=p/(2n-2)$. The formula at the top comes from taking the limit of this as $n$ goes to infinity.

The graph shows the blue line for $n=3$, the orange line for $n=4$, and the green line for the limit.