 Solution

Let $X_i=1_{A_i}$ be the indicator random variable of $A_i$. Then $X={\sum }_i{X}_i$, so linearity of expectation givesUsing variance of a sum andwe obtain

For the circular array, let $X_i$ indicate that bulbs $i$ and $i+1$ are both on, with indices read modulo $n$. Then $\mathbb{E}{X}_i=p^2$. Non-neighbouring indicators are independent, whileThere are $n$ individual terms and $2n$ ordered neighbouring pairs in the double covariance sum, so

The event $B$ is exactly $\{X\ge 1\}$. If $p=n^{-0.6}$, Markov inequality givesIf $p=n^{-0.4}$, then $B^c=\{X=0\}$ implies $|X-\mathbb{E}X|\ge \mathbb{E}X$. By Chebyshev inequality,Therefore $\boxed{\mathbb{P}(B)\to 1}$ in the second regime.