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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!

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    Write it with a random variable: $\mathbb{P}(E = n | E > 1) = \frac{\mathbb{P}(E = n\ \cap \ E > 1)}{\mathbb{P}(E > 1)}$. – 2017-01-15
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    **Indeed.** $~$ A *lot* of confusion can be cleared up by ensuring that you *start* by writing a sensible phrase. $~$ Put into words, $\mathsf P(n\mid n\geq 2)$ is: "The probability of $n$ when given that $n$ is at least $2$". $~$ That makes no sense *and* is not what was required. $~$ What was required can be phrased as: "The probability that the *electron count during the interval* is $n$ when given that *this electron count* is at least $2$." $~$ Which may be expressed as $\mathsf P(E=n\mid E\geq 2)$, when we let $E$ represent *the electron count during the interval*. – 2017-01-15

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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)

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    first of all thank you so much that was really helpful. Secondly, in the next part of that question it asks about the average number of electrons, when i Sum the product n*P(N=n∣N≥2) I get $${λ\over 1-e^(-λ)-λe^(-λ)}$$ while the correct answer has λ(1-e^(-λ)) in the numerator – 2017-01-15
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    @MohamedAymanAly $$\begin{align}\sum_{n=2}^\infty \frac{n\lambda^n}{n!} &= \lambda \sum_{n-1=1}^\infty \dfrac{\lambda^{n-1}}{(n-1)!} \\[1ex] &= \lambda( \sum_{m=0}^\infty\frac{\lambda^m}{m!}-\frac{\lambda^0}{0!}) \\[1ex] &= \lambda(e^\lambda -1)\end{align}$$ – 2017-01-15
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    MohamedAymanAly, did the comment of @GrahamKemp help you? So $E[N\mid N\geq 2]= \lambda(e^\lambda - 1)/(e^\lambda -1-\lambda)$. – 2017-01-16
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    Also, I made a mistake in the final result, where I had an extra minus in the exponent of $e^\lambda$. It has been fixed. – 2017-01-16
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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.)

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