To answer this, we also need to know the rate with which these types of failures occur. The more often a complete failure occurs, the shorter the delay time should be. If the rate at which complete failures occur is much lower than the rate at at which single-event failures occur, the appropriate delay time may even be longer than the one for individual events: The appropriate delay time is a function both of the number of events and of the rate of failure, and that rate is probably different for the single-event failure type and the complete failure type.
For a quantitative treatment, let's say we have $n$ independent Poisson processes with identical parameter $\lambda$ (the events) and a further independent Poisson process with parameter $\mu$ (the complete failure). That is, on average there are $\lambda$ events of each event type per time unit and $\mu$ complete failures per time unit. Let's assume that a complete failure prevents all events from being registered until we notice and fix it, and let's say we want the alarm to be triggered when the probability of a complete failure having occurred reaches $p$.
Whenever we see an event, the probability of a complete failure having occurred is reset to $0$. If we then don't observe any events for a time $\tau$, then either no failure and no events occurred, or a failure occurred and no events occurred before the failure.
The time $t$ of the next occurrence for a Poisson process with parameter $\lambda$ has probability density function $\lambda\mathrm e^{-\lambda t}$, so the probability for no event to occur in the process by time $\tau$ is
$ \int_\tau^\infty\lambda\mathrm e^{-\lambda t}\,\mathrm dt=\mathrm e^{-\lambda \tau}\;. $
Thus, the probability for no failure to occur and no events to be observed by time $\tau$ is
$P(\text{no failure}\land\text{no events observed})=\mathrm e^{-\mu \tau}\left(\mathrm e^{-\lambda \tau}\right)^n=\mathrm e^{-(\mu+n\lambda) \tau}\;.$
The probability for a failure to occur and no events to be observed by time $\tau$ is a bit more involved. We have to integrate the probability $\mathrm e^{-n\lambda t}$ that no events have occurred by time $t$ with the probability density $\mu\mathrm e^{-\mu t}$ for a failure to occur at time $t$:
$ P(\text{failure}\land\text{no events observed})=\int_0^\tau\mu\mathrm e^{-\mu t}\mathrm e^{-n\lambda t}\,\mathrm dt=\frac{\mu}{\mu+n\lambda}\left(1-\mathrm e^{-(\mu+n\lambda) \tau}\right)\;. $
Thus the total probability for no events to be observed by time $\tau$ is
$ P(\text{no events observed})=\mathrm e^{-(\mu+n\lambda) \tau}+\frac{\mu}{\mu+n\lambda}\left(1-\mathrm e^{-(\mu+n\lambda) \tau}\right)=\frac{\mu+n\lambda\mathrm e^{-(\mu+n\lambda) \tau}}{\mu+n\lambda}\;. $
The conditional probability that no failure has occurred, given that no events were observed over time $\tau$, is the ratio of the probability of no failure and no events occurring by time $\tau$ and the total probability for no events to be observed:
$ \begin{align} P(\text{no failure}\mid\text{no events observed}) &= \frac{P(\text{no failure}\land\text{no events observed})}{P(\text{no events observed})} \\ &= \mathrm e^{-(\mu+n\lambda) \tau}\Big/\frac{\mu+n\lambda\mathrm e^{-(\mu+n\lambda) \tau}}{\mu+n\lambda} \\ &= \frac{\mu+n\lambda}{\mu\mathrm e^{(\mu+n\lambda) \tau}+n\lambda}\;. \end{align} $
As expected, this is $1$ for $\tau=0$ and goes to $0$ as $\tau\to\infty$.
To use this in your situation, you could measure the rate $\lambda$ at which events occur, the rate $\mu_1$ at which single-event failures occur and the rate $\mu_n$ at which complete failures occur. You're using a delay time $\tau_1=30s$ for single-event failures, and the probability $p_1$ of a single-event failure having occurred at which your currently implemented alarm goes off is
$ p_1=1-\frac{\mu_1+\lambda}{\mu_1\mathrm e^{(\mu_1+\lambda)\tau_1}+\lambda}\;. $
To get the alarm for complete failures to go off at the same failure probability $p_n=p_1$, you'd need to choose the delay time $\tau_n$ for complete failures to satisfy
$ \frac{\mu_n+n\lambda}{\mu_n\mathrm e^{(\mu_n+n\lambda)\tau_n}+n\lambda}=1-p_n=1-p_1=\frac{\mu_1+\lambda}{\mu_1\mathrm e^{(\mu_1+\lambda)\tau_1}+\lambda}\;. $
Solving for $\tau_n$ yields
$ \tau_n=\frac1{\mu_n+n\lambda}\log\left(\frac1{\mu_n}\left(\frac{\mu_n+n\lambda}{\mu_1+\lambda}\left(\mu_1\mathrm e^{(\mu_1+\lambda)\tau_1}+\lambda\right)-n\lambda\right)\right)\;. $