 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 obtainSince $I_i{I}_j=I_{A_i\cap A_j}$, including $p_{ii}=p_i$, the variance isIn 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 givesEquivalently, the contribution of $N=0$ to $\mathbb{E}[(N-\mu {)}^2]$ is ${\mu }^2\mathbb{P}(N=0)$. 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 $0/0$.