Exponential Distribution and Poisson distribution I am trying to solve the following problem

In a shop on average 30 customers arrive in an hour. Assuming an exponential distribution, what is the probability that the time elapsed between two successive visits is 
  (i) more than 2 minutes 
  (ii) between 1 and 3 minutes

Is this distribution the same as a Poisson distribution where $λ=0.5$, because half a customer arrives every minute?
 A: The Poisson distribution with $\lambda=1/2$ is the discrete probability distribution of the number of customers arriving in one minute.  It takes values in the set $\{0,1,2,3,\ldots\}$.  The time between arrivals of successive customers is a continuous random variable, taking values in $(0,\infty)$.
Let $T$ be that random variable.  The probability $\Pr(T>t)$ is the same as the probability that the number of customer arriving before time $t$ is $0$.  That number of customers has a Poisson distribution with expected value $\lambda t = t/2$.  The probability that that is $0$ is therefore $\dfrac{(t/2)^0 e^{-t/2}}{0!}=e^{-t/2}$.  If you plug in $t=2$, you get the answer to your first question.
For your second question, the probability that it's more than one minute you get from plugging in $t=1$, and similarly for the probability that it's more than three minutes.  So now you need this:
$$
\Pr(1<T<3) = \Pr(T>1\  \&\  T\not>3).
$$
You need to show this last probability is
$$
\Pr(T>1)-\Pr(T>3).
$$
That's the same as showing
$$
\Pr(T>1\  \&\  T\not>3) + \Pr(T>3) = \Pr(T>1).
$$
So show that the two events $[T>1\  \&\  T\not>3]$ and $[T>3]$ are mutuallly exclusive and that if you put "or" between them, what you get is equivalent to the event $[T>1]$.
