The probability being computed is the probability that the attacker will catch up. We can call that event $A$ so we are computing $P(A).$
The way we are computing it is via the law of total probability. In other words, splitting it up into cases. The cases chose are how far along in their attack the attacker happens to be after $z$ blocks have been added. We let $B_k$ be the event that the attacker has made $k$ blocks of progress. We are assuming (more on this later) that $P(B_k) = \frac{1}{k!}\lambda^k e^{-\lambda}.$ We know that given that the attack has made $k$ blocks, the probability they catch up is (1) $1$ if they have already caught up (i.e. $k\ge z)$ (2) $(q/p)^{z-k}$ if they haven't caught up and are $z-k$ blocks behind.
Then we just decompose by the law of total probability: $$P(A) = \sum_k P(A|B_k)P(B_k) $$ where $P(B_k)=\frac{1}{k!}\lambda^k e^{-\lambda}$ and $P(A|B_k) = (q/p)^{z-k}$ for $k<z$ and $P(A|B_k) = 1$ for $k>z.$
So the probability they catch up is the probability they haven't got any blocks yet and they catch up, plus the probability that they've made one block and they catch up, plus the probability they've made two blocks and they catch up, etc. Note that as $k$ increases $P(A|B_k)$ (the probability the catch up given that they have made $k$ blocks) increases, eventually becoming $1,$ while $P(B_k)$ (the probability they've made $k$ blocks when the honest nodes have made $z$) decreases.
Now for the part about binomials. We can consider both the honest network and the attacker to be independent poisson processes, producing blocks at rate $\lambda_p$ and $\lambda_q$ so that the probability the next block is produced by honest network is $p=\frac{\lambda_p}{\lambda_p+\lambda_q}$ and similarly $q=1-p$ for the attacker. In this case the distribution of number of blocks $k$ the attacker has produced is negative binomial. This is because it is the number of successes for the attacker before $z$ failures (where a failure is the honest network producing a block). This seems to me a reasonable model of what's going on.
So we should instead have $$ P(B_k) = {k+z-1\choose k } q^k(1-q)^z.$$
Now, it happens sometimes that a negative binomial (or a binomial) can be approximated by a Poisson distribution. When is this true? Well, in general, it's when you have a lot of a attempts with a small probability. This would be the case when $z$ is large and $q$ is very small. This doesn't seem like a particularly natural limit for this problem as we'd imagine moderate sized $z$ and $q$ being of interest. An answerer Bob at the other site says something to the contrary but they're confusing the trials of the nonce for the trials to get the next block. The former process is actually a (not negative) binomial distribution which is a poisson process to a good approximation.
What Satoshi does is assume that the honest blocks take exactly their average time $$T_z = z/\lambda_p$$ to get the $z$ blocks. Then since we're still assuming the attacker generates blocks in a Poisson process, the number of blocks the attacker creates in that time is Poisson distributed with mean $$\lambda_q T_z = \frac{\lambda_qz}{\lambda_p} = \frac{q}{p}z.$$
But it seems a questionable approximation to suddenly have the honest nodes be deterministic while the attacker still has fluctuations. Of course, there will be a limit where this is effectively true, and it's in exactly the same limit as where the negative binomial above can be approximated as a Poisson.
It seems to me that questions (b) and (c) are both genuinely ambiguous, and both your solution and the one given in the textbook could be considered correct.
(b) What is the probability that Carl will have to wait more than 10 minutes for one of the others to show up?
I read this as asking for the probability that $\
\min(A, B)>10\ $, where $\ A\ $ is the time Carl has to wait for Alice to arrive and $\ B\ $ the time he has to wait for Alice to arrive. While this seems to me to be the most natural way of reading the question, I recognise that it can also be read as asking for the probability that at least one of $\ A\ $ and $\ B\ $ is greater than $10$—that is, the probability that $\ \max(A,B)>10\ $—and this may well seem to be the more natural reading to others.
(c) What is the probability that Carl will have to wait more than 10 minutes for both of the others to show up?
I initially read this as asking for the probability that Carl had to wait for more than $10$ minutes for both Alice and Bob to be present—that is, the probability that $\ \max(A,B)>10\ $. However, it seems to me just as natural to take it as asking for the probability that both Alice and Bob take more than $10$ minutes to arrive—that is, the probability that $\ \min(A,B)>10\ $.
Best Answer
A collision occurs if multiple stations choose the same slot. Equivalently, no collision occurs if they all choose a different slot. Assume that the four stations chose independently from one another. Let each number be represented by the uniformly distributed random variables $X_i, i=1\dots 4$.
The question is how you define the $i$'th event. I assume the question is asking for $1-P(X_1 \neq X_2 \neq X_3 \neq X_4)$ (meaning all different numbers). Defining $N=2^i$, this probability is equal to $1-\frac{N (N-1) (N-2) (N-3)}{N^4}$ since there are $N(N-1)(N-2)(N-3)$ ways of choosing four different numbers uniformly from a range of $N$. As a sanity check, if there were only two stations in the first collision, they would have to wait either 0 or 1 slot each. A collision would occur if they both chose 0 or both chose 1. This happens with probability 0.5 (2 out of the 4 possibilities). Our equation gives $1-\frac{2 \cdot 1}{2^2} = 0.5$
Corrections welcome.