The waiting times for poisson distribution is an exponential distribution with parameter lambda. But I don't understand it. Poisson models the number of arrivals per unit of time for example. How is this related to exponential distribution? Lets say probability of k arrivals in a unit of time is P(k) (modeled by poisson) and probability of k+1 is P(k+1), how does exponential distribution model the waiting time between them?
I will use the following notation to be as consistent as possible with the wiki (in case you want to go back and forth between my answer and the wiki definitions for the poisson and exponential.)

$N_t$: the number of arrivals during time period $t$

$X_t$: the time it takes for one additional arrival to arrive assuming that someone arrived at time $t$

By definition, the following conditions are equivalent:

$ (X_t > x) \equiv (N_t = N_{t+x})$

The event on the left captures the event that no one has arrived in the time interval $[t,t+x]$ which implies that our count of the number of arrivals at time $t+x$ is identical to the count at time $t$ which is the event on the right.

By the complement rule, we also have:

$P(X_t \le x) = 1 - P(X_t > x)$

Using the equivalence of the two events that we described above, we can re-write the above as:

$P(X_t \le x) = 1 - P(N_{t+x} - N_t = 0)$

But,

$P(N_{t+x} - N_t = 0) = P(N_x = 0)$ 

Using the poisson pmf the above where $\lambda$ is the average number of arrivals per time unit and $x$ a quantity of time units, simplifies to:

$P(N_{t+x} - N_t = 0) = \frac{(\lambda x)^0}{0!}e^{-\lambda x}$

i.e.

$P(N_{t+x} - N_t = 0) = e^{-\lambda x}$

Substituting in our original eqn, we have:

$P(X_t \le x) = 1 - e^{-\lambda x}$

The above is the cdf of a exponential pdf.
