Say I calculate the probability $P(A)$ that an event $A$ will happen, and obtain the associated confidence interval $z_1$. If I have $n$ trials, the expected number of trials that will have resulted in the event $A$ happening is $n\cdot P(A)$. But what then is the confidence interval $z_2$ for this expected number of $A$ events?
If the outcomes of each of the trials $n$ are completely correlated, I would imagine the confidence interval $z_2$ would be $z_1$ multiplied by $n$. However, it isn't clear what it would be for zero correlation or partial correlation. Is this a trivially simple problem?
I am not sure exactly what you are asking. Here is information that may be helpful.
If $X \sim \mathsf{Binom}(n, p),$ is the number of Successes observed in $n$ independent trials and $p$ is the probability of Success on any one trial, then $\hat p = X/n$ is the point estimate of $p.$ A Wald 95% confidence interval (CI) for $p$ is of the form $$\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}},$$
Multiplying through by $n$, you would have the following 95% CI for $E(X)$: $$X \pm 1.96\sqrt{n\hat p(1-\hat p)}.$$
Notes:
(1) As @Henry Comments, these formulas depend on a normal approximation to a binomial distribution. They work best when $n$ is large and $p$ is not too far from 1/2. A common approximate rule of thumb is that $np$ and $n(1-p)$ should both exceed 5.
(2) The Agresti CI for $p$ is more accurate, particularly when $n < 500$ or so: Define $\tilde n = n+4$ and $\tilde p = (p+2)/\tilde n.$ Then compute the CI for $p$ as:
$$\tilde p \pm 1.96\sqrt{\frac{\tilde p(1-\tilde p)}{\tilde n}},$$