How would I derive the CDF of P(T$\ge$t). I know the CDF of an exponential poisson distribution is F(t) = P(T$\le$t) = 1-P(T$>$t) = 1-P(X=0) = 1-$e^{(-\lambda)(t)}$, but in the case of (T$\ge$t) we cannot choose X to be 0.
I tried solving it by reversing what was done in the original Exponential Poisson distribution and got F(t) = P(T$\ge$t) = 1-P(T$<$t). In this case I think we cannot choose X to be 0 because there must already be an arrival by the time we hit time t because of the inequality P(T$<$t).
We have $$\Pr(T\ge t)=\Pr(T=t)+\Pr(T\gt t).\tag{1}$$
(i) Note that $\Pr(T\gt t)=1-\Pr(T\le t)=1-(1-e^{-\lambda t})=e^{-\lambda t}$.
(ii) Also, $\Pr(T=t)=0$. Indeed, if $Y$ is any random variable with continuous distribution, then for any real number $a$, we have $\Pr(Y=a)=0$.
Using Equation (1), and facts (i) and (ii), we get that $\Pr(T\ge t)=e^{-\lambda t}$.
Another way: $$\Pr(T\ge t)=\int_0^t \lambda e^{-\lambda x}\,dx=e^{-\lambda t}.$$
Remark: The fact that for a continuous random variable $Y$, we have $\Pr(Y=a)=0$, is useful. It means that we can casually replace $\lt$ by $\le$, or vice-versa, without affecting probabilities.