The situation is that full rubbish bins are ready for collection at a rate of 2 per day in a given street. That is, the number of bins ready for collection is a Poisson process with rate 2. The bin lorry comes at day $T$, where $T$ ~ $U[6,8]$. I wish to find the probability that there are no bins when the bus arrives.
I see two possible approaches to this question. I can use the Poisson distribution $X$ ~ $Poi[2T]$ and find the probability $P(X = 0)$, or I can use the exponential distribution $Y$ ~ $Exp[2x]$ and find the probability $P(Y > T)$.
For the first option,
\begin{align} P(X=0)& = (2T)^0 e^{-2T}\\ & = e^{-2T} \end{align}
For with the second option,
\begin{align} P(Y > T)& = 1 - P(Y < T)\\ & = 1 - (1-e^{-2T})\\ & = e^{-2T} \end{align} So I end up with the same expression for both approaches, which is reassuring, but I am not sure how to solve it, assuming that this is correct. I am considering integrating over the range of t, that is
\begin{align} \int_{6}^{8}e^{-2T}dT \end{align}
Is this the correct way to solve the problem, or have I gone down completely the wrong track?
Conditioning on $T$, we may compute \begin{align} \mathbb P(N(T)=0) &= \int_0^\infty \mathbb P(N(T)=0\mid T=t)f_T(t)\,\mathsf dt\\ &= \int_0^\infty \mathbb P(N(t)=0)f_T(t)\,\mathsf dt\\ &= \int_6^8 \frac12 e^{-2t}\,\mathsf dt, \end{align} where $f_T(t) = \frac12\,\mathsf 1_{(6,8)}(t)$ is the density of $T$ and $\{N(t):t\in\mathbb R_+\}$ is a Poisson process with rate $2$.