exponential distribution involving customers 
Customers arrive at a certain shop according to a poisson process at
  the rate of $20$ customers per hour. what is the probability that the
  shop keeper will have to wait more than $5$ minutes (after opening)
  for the arrival of the first customers?

So i am trying to this problem by exponential, however im having some trouble with figuring out the pdf.
I was told $F(x)=20e^{-20x}$ if $x>0$, however the pdf on my paper says: $\frac{1}{\theta}e^{-\frac{x}{\theta}}$ so shouldnt it be: $\frac{1}{20}e^{-\frac{x}{20}}$.
I am just confused. Any explanations?

Let the discrete random variable $X=$ The number of minutes the shopkeeper will have to wait for the first customers to arrive.
$X\sim Po(\lambda)$ and working in minutes $\implies 3 \space$  customers arrive per minute, such that $\lambda=3$ so $$X\sim Po\left(3\right)$$ $$P\left(X \gt 5\right)=1-P\left(X \le 5\right) \approx \color{purple}{0.084}$$ 
