Poisson Conditional probability Problem My exam is tomorrow and my lecturer is not replying so any help would be appreciated here. I am struggling a bit with applying the conditional probability rule in poisson; the question asks: 
Find the probability that n electrons are emitted in a time interval in which at least 2 electrons are emitted given the processes is poisson distributed:
I Interpret that as $$P(n|n\ge2) = {P(n \& n\ge2)\over P(n\ge2)}$$
but I don't know how to find the term in numerator.
Thanks!
 A: Let $N\sim \operatorname{Pois}(\lambda)$. We use the definition of a conditional probability,
$$
P(N=n\mid N\geq 2) 
=
\frac{P(N = n,N \geq 2)}{P(N\geq 2)}
$$
Then split it up in the two cases: $n<2$ and $n\geq 2$. For $n<2$ the right hand side is zero. For $n\geq 2$ then of course $N\geq 2$ and so
$$
P(N=n,N\geq 2) = P(N = n) = \frac{\lambda^n}{n!}e^{-\lambda}.
$$
For the denominator, use that
\begin{align*}
P(N\geq 2) &= 1-P(N\leq 1) 
\\&= 
1-P(N = 0) - P(N = 1) 
\\&=
1-\frac{\lambda^0}{0!}e^{-\lambda} -\frac{\lambda^1}{1!}e^{-\lambda} 
\\&=
1-e^{-\lambda}(1+\lambda). 
\end{align*}
Finally, collecting things we get
$$
P(N = n\mid N\geq 2) = \begin{cases}
0,& n<2\\
\dfrac{\lambda^n}{n!(e^{\lambda}-1-\lambda)}, & n\geq 2.
\end{cases}
$$
(for the case $n\geq2$, multiply the numerator and denominator by $e^\lambda$ to get the result)
A: I will illustrate a solution to this problem for the case $\lambda = 5$ by simulation
in R statistical software with 10 million realizations of $N \sim Pois(\lambda).$
Subsequently, we disregard cases in which $N = 0$ or $1$ to obtain the 
conditional distribution of $M.$ The results agree with the conditional
distribution derived in the Answer by @Therkel (+1). (I will leave a discussion
of $E(M)$ until after the simulation.) 
Of course, the PDF of $N$ is given by 
$P(N = i) = f_i = e^{-\lambda} \frac{\lambda^i}{i!},$ for $i = 0, 1, 2, \dots.$
The PDF of $M$ is given by
$P(N = i|N \ge 2) = P(M = i) = f_i^\prime = f_i/P(N \ge 2),$ for $i = 2, 3, 4, \dots.$ This PDF is denoted as pdf.c in the simulation code.
lam = 5
N = rpois(10^7, lam)
mean(N)
## 5.000718
M = N[N >=2]              # conditioning
mean(M)
## 5.176375               # aprx E(M) from simulation
i=0:30;  den = 1-ppois(1,lam)
pdf.c = dpois(i,lam)/den  # cond'l PDF ... 
pdf.c[i<2]=0              # ... O's for values 0 and 1
sum(i*pdf.c)
## 5.175546               # nearly exact E(M) sum first terms of series

To get the conditional mean: $E(M) = \sum_{i=2}^\infty if_i^\prime.$
Enough terms are summed at the end of the code that I'm sure $E(M) \approx 5.18$
is correct for $\lambda = 5.$
I am not exactly sure whether this answer is equivalent to values suggested in Comments
by @Mohammad or @Graham Kemp. [One has two answers and faulty typesetting and the other
seems to have problems with signs ($e^5 -1 \approx  147$ and $e^{-5} -1$ is negative)
 and does not contain the denominator.] 
Below is a histogram of the simulated values (tan bars) compared with
values of the conditional CDF $f_i^\prime$ (blue dots). Agreement is very good. (With 10 million iterations
for $N$ and nearly 9,600,000 realizations of $M,$ results should be
accurate to three places, and the resolution of the graph is less than three places.)
 
