What is the difference between a Poisson and an Exponential distribution? For a Poisson distribution:
$$\mathsf{P}(X=x)=\frac{e^{-\mu}\times \mu^x}{x!}$$
where $\mu$ is the mean number of occurrences.
For an Exponential distribution:
$$f(x;\lambda) =
\begin{cases}
\lambda e^{-\lambda x}  &  x \ge 0 \\
0 & x < 0
\end{cases}$$
where $\lambda$ is the rate parameter.
Apart from the fact that the formulas are obviously different, in layman's terms what is the difference between an Exponential and Poisson distribution? Or put in another way, why do we need them both? What does one of them do that the other doesn't? What is the difference between an Exponential distribution and an Exponential Density function? 
And yes, this is all very new to me as I was never taught distribution theory.
Thanks.
 A: Exponential is a continuous distribution while poisson is a discrete one.
A: As A.S.'s comment indicates, both distributions relate to the same kind of process (a Poisson process), but they govern different aspects: The Poisson distribution governs how many events happen in a given period of time, and the exponential distribution governs how much time elapses between consecutive events.
By way of analogy, suppose that we have a different process, in which events occur exactly every $10$ seconds.  Then the number of events that happen in a minute (i.e., $60$ seconds) is deterministically $6$, and the amount of time that elapses between consecutive events is, of course, deterministically $10$ seconds.
In contrast, in a Poisson process with a mean rate of one event every $10$ seconds (i.e., $\lambda = 1/10$), the number of events that happen in a minute is not deterministically $6$, but it has a mean of $6$.  The exact distribution is given by the Poisson distribution:
$$
P_k(t) = \frac{(\lambda t)^k}{k!} e^{-\lambda t}
$$
where $t = 60$ seconds is the time window.  Thus, the probability that no events occur in a minute is given by
$$
P_0(t) = \frac{(6)^0}{0!} e^{-6} = e^{-6} \doteq 0.0024788
$$
whereas the probability that $6$ events occur in a minute is given by
$$
P_6(t) = \frac{(6)^6}{6!} e^{-6} = \frac{46656}{720} e^{-6} \doteq 0.16062
$$
That is obviously much more likely, as you would expect.
Similarly, the time between events is also not deterministically $10$ seconds, but it has a mean of $10$ seconds.  The actual time distribution is the exponential distribution, which can be specified using its CDF (cumulative distribution function)
$$
F_T(t) \equiv P(T < t) = 1-e^{-\lambda t}
$$
The CDF provides essentially the same information as the PDF (probability density function), whose formulation you gave in your question; in fact, the derivative of the CDF is the PDF.  However, the CDF is sometimes easier to understand intuitively, so I'll explain using the CDF here.
In this case, the probability that the time between events is less than $10$ seconds is $F_T(10) = 1-e^{-1} \doteq 0.63212$, whereas the probability that the time between events is less than $60$ seconds is $F_T(60) = 1-e^{-6} \doteq 0.99752$.  The probability that it is greater than $60$ seconds is $1-F_T(60) = e^{-6} \doteq 0.0024788$, and you'll notice this is equal to the probability that no events occur in a given minute.  This is no coincidence; in a Poisson process, which is memoryless, the probability that the time between events is greater than a minute is naturally equal to the probability that no events occur in that minute!
