Poisson conditional probability question So I have a question that asks the following:

The probability that $n$ electrons are emitted by a thermionic cathode in a time interval $t$ is given by a Poisson distribution with mean $\lambda$. What is the probability that $n$ electrons are emitted in a time interval $t$  in which at least 2 electrons are emitted?

I've tried to use conditional probability with the Poisson pmf given that $n$ is more than or equal to 2 but I can't seem to get anywhere and at this point any help would be really appreciated. Thanks in advance!
 A: Let $X$ be the number of electrons emitted in a time interval $t$. Note that $X\sim \text{Po}(\lambda)$. Then the probability is $$\mathbb P(X=n|X\ge 2)=\frac{\mathbb P(X=n\cap X\ge 2)}{\mathbb P(X\ge 2)}$$
It is easy to see that for $n=0,1;\mathbb P(X=n\cap X\ge 2)=0$. For $n\ge 2$, $$\frac{\mathbb P(X=n\cap X\ge 2)}{\mathbb P(X\ge 2)}=\frac{\mathbb P(X=n)}{1-\mathbb P(0\le X\le 1)}=\frac{e^{-\lambda}\cdot \frac{\lambda^n}{n!}}{1-\sum_{k=0}^1e^{-\lambda}\cdot \frac{\lambda^k}{k!}}$$
A: First of all, $X\sim \mathcal{P}(\lambda t)$, i.e., $\Pr\{X=n\}=\frac{(\lambda t)^n}{n!}e^{-\lambda t}$. The probability that $n$ electrons are emitted, given at least 2 are emitted is:
$$\Pr\{X=n|X\geq 2\} = \frac{\Pr\{X=n, X\geq 2\}}{\Pr\{X\geq2 \}}$$
Note that
$$\Pr\{X=n, X\geq2\} = \begin{cases}
\dfrac{(\lambda t)^n}{n!}e^{-\lambda t}, & n\geq 2\\
0, & \text{otherwise}
\end{cases}$$
and
$$\Pr\{X\geq2\}=\sum\limits_{k=2}^{+\infty}\frac{(\lambda t)^k}{k!}e^{-\lambda t}
=1-(1+\lambda t)e^{-\lambda t}$$
So,
$$\Pr\{X=n|X\geq2\}=\begin{cases}
\dfrac{(\lambda t)^n}{n!}\dfrac{e^{-\lambda t}}{1-(1+\lambda t)e^{-\lambda t}}, & n\geq 2\\
0, & \text{otherwise}
\end{cases}$$
