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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?

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The number of customers that arrive in a specific time period has Poisson distribution. The waiting time between successive customer arrivals has exponential distribution. – David Mitra Jun 12 '12 at 18:21
This question is about a Poisson process rather than a Poisson distribution – Henry Jun 12 '12 at 23:28
Strongly related: Exponential distribution from Poisson – Jacob Akkerboom Mar 11 '14 at 10:38

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 $[T>1]$.

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