Question about Poisson process and arrival times Problem: On any given day you receive mail in mailbox with probability $p$. Assume whether mail is put in the mailbox or not is independent each day. 


*

*If the neighbor receive mail in his mailbox with prob $w$, independent each day, what is the probability the first mail (between you and neighbor) arrives after a week?

*Suppose your mail gets moldy with prob. $q$ when sitting in the mailbox independently of whether mail is actually received in the mailbox. What is the prob. the 1st received moldy mail will come within a week? What is the expected number of days you wait until you receive your 2nd moldy mail?
Attempt: for (1), so far I have
$$
\mathbb{P}[\min\{\text{me get 1st mail, neighbor get 1st mail\} after a week}]
   = \mathbb{P}[\text{I don't receive any mail in 1st week}] \times
     \mathbb{P}[\text{neighbor doesn't receive any mail in 1st week}],
$$
so my answer is $(1-p)^7(1-w)^7$, am I right?
(2) I am thinking
$$
\mathbb{P}[\text{1st received moldy mail will come within a week}] = q^7,
$$
not sure about the second part
 A: Note: This is not a Poisson Process. 
Let $X$ count the day you first get mail.  Let $Y$ count the day your neighbour first gets mail.   These are Geometric distributed random variables.
So for the first $\checkmark$:
$$\mathsf P(X\geq k) = (1-p)^{k-1}\quad \text{for }k\in \{1,2...\}\\ \mathsf P(Y\geq k) = (1-w)^{k-1}\quad \text{for }k\in \{1,2...\}
\\ \mathsf P((X\vee Y)\geq k) = \mathsf P(X\geq k, Y\geq k) = (1-p)^{k-1}(1-w)^{k-1}
\\ \mathsf P((X\vee Y)\geq 8) = (1-p)^7(1-w)^7$$

The probability that the first mouldy mail arrives within a week will be, the probability that any mail which arrives that week will be mouldy: $$1-(1-pq)^7$$
For the second part, what you are looking for is the expected number of days until moldy mail arrives.   If $\mathsf E(Z)$ is the expected number of days until the next mouldy mail arrives, then by linearity of Expectation, the expected number of days until the second mouldy mail arrives is: $2\mathsf E(Z)$.   To find $\mathsf E(Z)$ we use the Law of Iterated Expectation, partitioning on whether the next mail is mouldy or not.
$$\begin{align}
\mathsf E(Z) & = q \mathsf E(X) + (1-q)(\mathsf E(X)+\mathsf E(Z)) 
\\[2ex] \text{where}
\\[1ex] \mathsf E(X) & = 1/p & \text{is the expected days until the next mail arrives}
\\[3ex] \therefore 2\mathsf E(Z) & = ?
\end{align}$$
