= Solution
The <indicator function> $I_A$ equals one when the <event> $A$ occurs and zero otherwise. Its <expected value> is $\mathbb E I_A=\mathbb P(A)$. By <linearity of expectation>, no <independence> assumption is needed to obtain
$$
\boxed{\mathbb E N=\sum_{i=1}^n p_i.}
$$
Since $I_iI_j=I_{A_i\cap A_j}$, including $p_{ii}=p_i$, the <variance> is
$$
\boxed{\operatorname{Var}N=\sum_{i,j=1}^n(p_{ij}-p_ip_j)=\sum_i p_i(1-p_i)+2\sum_{i<j}(p_{ij}-p_ip_j).}
$$
In particular, correlations between the <events> cannot simply be discarded.
Write $\mu=\mathbb E N$. If $\mu>0$, then $N=0$ implies $|N-\mu|\ge\mu$. Applying <Chebyshev's inequality> therefore gives
$$
\boxed{\mathbb P(N=0)\le\frac{\operatorname{Var}N}{\mu^2}.}
$$
Equivalently, the contribution of $N=0$ to $\mathbb E[(N-\mu)^2]$ is $\mu^2\mathbb P(N=0)$. \b[A positive mean is necessary for the displayed ratio to be defined.] If $\mu=0$, nonnegativity instead gives $N=0$ almost surely, and the printed ratio would be $0/0$.
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