Poisson distributed radiation with a faulty counter There is this problem that I think I have solved. I need feedback if I have solved it correctly. I also have some questions regarding the intuition on the solution I have obtained.
Problem Statement: Radioactive decay 
Assume that a radioactive sample emits a random number of $\alpha$ particles in
any given hour, and that the number of $\alpha$ particles emitted in an hour is Poisson distributed with
parameter $\lambda$. Suppose that a faulty Geiger-Muller counter is used to count these particle emissions. In particular, the faulty counter fails to register an emission with probability $p$, independently of other emissions.
(a) What is the probability that the faulty counter will register exactly $k$ emissions in an hour?
(b) Given that the faulty counter registered $k$ emissions in an hour, what is the PMF of the actual number of emissions that happened from the source during that hour?
My Attempt: So with a discrete random variable having Poisson distribution, 
$$p_X(k) = \frac{e^{-\lambda}\lambda^{k}}{k!}$$
where $k \in \mathbb{N} \cup \{0\}$. In the given experiment, there are three random variables that are implicitly defined. Let $Y$ be the random variable which is the value that the counter registers after the first hour, let $Z$ be the random variable which denotes the number of failures in the counter during the hour, and let $X$ be the random variable which is the number of $\alpha$ particles emitted in one hour. 
(a) We need to find $\mathbb{P}(Y=k)$.
\begin{eqnarray*}
\mathbb{P}(Y=k) &=& \sum_{n=0}^{\infty}\mathbb{P}(Z=n) \times \mathbb{P}(X=n+k) \\ 
 &=& \sum_{n=0}^{\infty}{n+k \choose n} (1-p)^k p^n \frac{e^{-\lambda}\lambda^{n+k}}{(n+k)!} \\
&=& \frac{ \left \{ (1-p)\lambda \right \}^k e^{-\lambda(1-p)}}{k!}
\end{eqnarray*}
This is as if the Poisson distribution has changed with a new constant of $(1-p)\lambda$.
(b) Here, we need $$\mathbb{P}\left \{ (X=n+k)|(Y=k) \right \}= \frac{\mathbb{P} \left \{ (X=n+k)\cap(Y=k) \right \}}{\mathbb{P}(Y=k)}$$
and we know that $$\mathbb{P} \left \{ (X=n+k)\cap(Y=k) \right \}=\mathbb{P}(Z=n) \times \mathbb{P}(X=n+k)$$
Hence we finally get $$\mathbb{P}\left \{ (X=n+k)|(Y=k) \right \} = e^{-\lambda p} \frac{(\lambda p)^n}{n!}$$
We can replace $n+k$ with $N$ to get $$\mathbb{P}\left \{ (X=N)|(Y=k) \right \} = e^{-\lambda p} \frac{(\lambda p)^{N-k}}{(N-k)!}$$ if $N \geq k$ and $0$ otherwise.
I wish to know the following.


*

*Is my solution correct?

*If my solution is correct, what is the intuitive explanation to the fact that if $p=0$, answer to (b) is $0$? Specifically, when $p=0$, then the counter is not making any mistakes. Why then is the conditional probability $0$?

*If my solution is correct, what is the intuitive explanation to the fact that if $p \to 1$, answer to (b) is a Poisson process with coefficient $\lambda$? Specifically, when $p \to 1$, the counter is surely making a mistake each time. Why then is the conditional probability Poisson distributed when we know that the counter errs each time?

 A: Yes, that's correct.
The key observation here to makes sense of our intuition is that the observed decays and the unobserved decays are independent random variables.
Intuitively, as $p\to 0$, the detector becomes practically infallible, and we're virtually certain that the decays we observed are all of them. The conditional probability you get there in the limit isn't simply zero - it's the point mass $\delta(N-k)$, equal to $1$ if $N=k$ and zero otherwise.
As $p\to 1$, we lean on the independence. It doesn't matter how many decays we observed (although that isn't likely to be many) - the distribution of the unobserved decays always looks the same. Since, in the limit, all of the decays are unobserved, this approaches the Poisson distribution with parameter $\lambda$ that describes all of the decays.
The distribution of total decays given $k$ observed decays will be the appropriate Poisson distribution plus $k$. That's what the $N-k$ in your formulas is doing; it's shifting the argument. The answer to (b) is not a Poisson distribution but instead a translation of a Poisson distribution.
