small rate and probability Is it true that when a rate is small, it can be considered as a probability of an event happening in a small time interval? For example, if a rate is Q [1/min], the probability of the event happening in a short time interval $dt$ is $Qdt$ (or is it just $Q$?). I want to know why it is so, how small the rate has to be to be considered small, and how short the duration has to be for the probability definition. Under this condition, the interpretation of 
$\int_{0}^{T} Qdt$
is the probability of an event happening in a duration $T$? If so, why is it so? If not, what is the interpretation?
 A: If the rate at which an event occurs is $\lambda$, then the probability of an event occurring within a short period of time $dt$ is equal to $dP = \lambda dt$.  On other words, $dP/dt = \lambda$, from which it may be easier to see the intuition behind calling it a rate.
For instance, if the rate is $0.01$ events per second, then over a period of $3$ seconds, the probability of an event occurring is approximately $(0.01 \mbox{ events/second})(3 \mbox{ seconds}) = 0.03 \mbox{ events}$.
It's not so much that the rate has to be small, so much as the product of the rate and the time interval has to be small ($\ll 1$).  The estimate is only exact in the limit as the product approaches zero.
The actual probability of an event occurring in a positive interval of time can be determined as follows.  Suppose that the probability that no event has occurred in the interval $[0, t]$ is $G(t)$.  What is the probability that no event occurs in the interval $[0, t+\Delta t]$?  It would be (under the conditions of independence assumed in the use of the term "rate") $G(t)(1-\lambda \Delta t)$.  Another way to say that is that $\Delta G(t) = G(t+\Delta t)-G(t) = -\lambda G(t) \Delta t)$.  In the limit, as $\Delta t \to 0$, we would have
$$
\frac{d}{dt} G(t) = -\lambda G(t)
$$
and the solution to that differential equation, with the initial condition $G(0) = 1$ (since it is certain that no event occurs in the time interval $[0, 0]$) is $e^{-\lambda t}$.  Therefore, the probability that at least one event does occur in time $t$ is $F(t) = 1-G(t) = 1-e^{-\lambda t}$.  You will note that, as we said above, $\lambda t$ is a good approximation for $F(t)$ so long as $\lambda t \ll 1$, but is not a good approximation for $F(t)$ otherwise.  For example, if $t = 1/\lambda$ (the "time scale" of the event), then $\lambda t = 1$, but $F(t) = 1-e^{-1} = 1-1/e \doteq 0.63212$.
The reason that the expression isn't simply $\lambda t$ (or $\int_{t'=0}^t Qt' \, dt' = Qt$ in your notation) is that that expression merely counts the expected number of events in time $t$.  If there were guaranteed to be only one event in that time, then that expected number would also be the probability of an event, but since there could also be $2, 3, 4, \ldots$ events in that time frame, your expression overestimates the probability of an event.  As long as multiple events are unlikely to occur (that is, when $\lambda t \ll 1$), the discrepancy is not large, but for larger $t$, the inaccuracy is substantial.  Note, for instance, that $\lambda t > 1$ for $t > 1/\lambda$, and probabilities cannot be greater than $1$, of course.
