Probabilistic interpretation of $\sum_{n \geq 0}\mathrm P_{n}(t)= 1$

Question

The following is a problem from Spiegel's Applied Differential Equations:

The probability $\mathrm{P}_{\mathrm n}(t)$ that a counter (such as a Geiger counter) will register exactly $\mathrm n$ nuclear particles in a time $t$ is determined by the system of differential equations:

$$\mathrm P_{\mathrm n}^\prime(t)=\lambda \left[\mathrm P_{\mathrm n-1}(t)-\mathrm P_{\mathrm n}(t) \right]$$

$$\mathrm P_{0}^\prime(t)=-\lambda\mathrm P_{0}(t)\text{ ; }\mathrm n\neq0 $$

Where $\lambda$ is a positive constant. Using appropriate conditions, find $\mathrm P_{\mathrm n}(t)$ for $\mathrm n =0,1,2,\dots$.

From you results, show that $\sum_{n \geq 0} P_n(t) ) 1$

¿What is the probabilistic interpretation of this?

Starting with $\mathrm n=0$ I get

$$ \mathrm P_{0}(t)= \exp(-\lambda t)$$

It not hard to show that

$$\mathrm P_{\mathrm n}(t) = {(\lambda t)^n \over n!}\exp(-\lambda t)$$

Then

$$\sum\limits_{n \geqslant 0} {{{\text{P}}_n}(t)} = \sum\limits_{n \geqslant 0} {\frac{{{{(\lambda t)}^n}}}{{n!}}\exp ( - \lambda t)} = \exp \left( {\lambda t} \right)\exp ( - \lambda t) = 1$$

¿What is the probabilistic interpretation of this?

Answer 1

It says that at any given time $t$ the counter will register exactly one number in $\{0,1,...\}$.

Each event is disjoint, and the union of events covers all the possibilities (ie, a number must be registered). Hence the probabilities sum to $1$.