Question about Poisson process and arrival times

Question

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.

1. 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?

2. 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

Answer 1

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.

$$\begin{align} \mathsf E(Z) & = q \mathsf E(X) + (1-q)(\mathsf E(X)+\mathsf E(Z)) \\[2ex] \text{where} \\[1ex] \mathsf E(X) & = 1/p & \text{is the expected days until the next mail arrives} \\[3ex] \therefore 2\mathsf E(Z) & = ? \end{align}$$