Suppose, we are interested in the event $x \leq 4$, where, $x$ is a random variable following $\mathcal{N}(0,1)$. We select an indicator function, where, $I(x \leq 4) = 1$. Does $I(x)$ can be thought to follow binomial distribution with probability of $\int_{-\infty}^{4}f(x)dx$, where $f(x)$ is the pdf correspondent to the normal distribution?
If true, can we calculate the expected value of the binomial event with probability of $\int_{-\infty}^{4}f(x)dx$?
If we repeat the said event $n$ number of times, can our expected value be simply, $n\int_{-\infty}^{4}f(x)dx$?
In general, take a probability space with whatever RVs you wish, then given an event $A$ (say $A=\{X\leq x\}$ or whatever) then the indicator function $\mathbb{1}_A$ is distributed like a Bernoulli RV with probability $p=\mathbb{P}(A)$. Then by repeating independent trials and summing such indicator fuctions $n$ times we get a binomial RV with distribution $\mathcal{B}(n,p)$.
A common example is to have some sampling of a normal, or some relevant RV, and count the number of observations that exceed some level $X>x$. Like acceptance sampling or stress testing depending on what $X$ models, etc.