Assuming I have a question like :
- Suppose that the amount of time one spends in a bank is exponentially distributed with mean ten minutes, that is, $λ = 1/10$. What is the probability that a customer will spend more than fifteen minutes in the bank? What is the probability that a customer will spend more than fifteen minutes in the bank given that she is still in the bank after ten minutes?
What I don't understand is the mean, if it's given that $λ=1/10$ which is $0.1$, does it mean we have $.1$ customers arriving every $10$ minutes ? It's a bit confusing to me and is my first solution correct?
$$\Pr[K>15] = e^{(-10)(0.1)}\frac{\big((10)(0.1)\big)^{15}}{15!}$$
You do not need to use Poisson processes here — this question is about the use of the exponential distribution, and its property of memorylessness. For $X\sim\operatorname{Exp}(\lambda)$, $\mathbb{E} X = \frac{1}{\lambda}$ (so indeed here the rate is $\lambda=\frac{1}{10}$), you have $$ \Pr[X > t ] = 1-F_X(t) = e^{-\lambda t} $$ and for $t \geq s \geq 0$ $$ \Pr[X > t \mid X \geq s] = \Pr[X > t-s ] = e^{-\lambda (t-s)}\ . $$ For $t=15$ and $s=10$, you get $$ \Pr[X > 15 ] = e^{-\frac{15}{10}} = e^{-\frac{3}{2}}, \qquad \Pr[X > 15 \mid X \geq 10] = e^{-\frac{5}{10}}= e^{-\frac{1}{2}}\ . $$
As for the parameter $\lambda$ in Poisson process, this is the rate (rate of "events") -- that is, in an interval of time of length $\tau$, if $N=(N_t)_{t>0}$ is a counting Poisson process with rate $\lambda$, the expected number of increments is going to be $\lambda\tau$: $\mathbb{E}[N_{t+\tau} - N_t]= \lambda\tau$ (there is "one increment every $1/\tau$ units of time").