I have the following task. But first: The task is not a homework assignment, its just for me, I'm not a student yet.
Task :
The number of students attending a stochastic for computer scientists consultation is poisson-distributed with parameters $\lambda > 0$. These students each independently study computer science with probability $p$, where $p \in (0, 1)$.
Let $X$ be the number of computer science students attending office hours, and let $Y$ be the one Number of students who are office hours visit but not computer science students. Show that $X$ and $Y$ are independent. You can use, if necessary:
1: $\mathbb{P}(X=k | Y=n)=\left( \begin{array}{c}{n} \\ {k}\end{array}\right) p^{k}(1-p)^{n-k}$
2: $P(X=k)=\frac{\lambda^{k} p^{k}}{k !}$
My Idea:
The problem is simple but hart to prof formal(for me and thats why I need help).
The intersection of the event sets of X and Y is empty. Because there can not be a listener in the lecture who fulfills both results at the same time:
X: Attends the lecture and is a computer scientist.
Y: Attends the lecture and is not a computer scientist.
Could you help/show me how to make the a formal proof?
P.S. I hope I did not use the wrong tags. I could not find somthink better than Proof-writing . I'm so sorry.
Under the conditions described, as I understand them: \begin{eqnarray} \mathbb{P}\left(X=k, Y=j\right) &=& \frac{\lambda^{k+j}}{\left(k+j\right)!}e^{- \lambda}{k+j \choose k}p^k\left(1-p\right)^j\\ &=& \frac{\left( \lambda p\right)^k}{k!}e^{-\lambda p}\frac{\left( \lambda \left(1-p\right)\right)^{\,j}}{j!}e^{-\lambda \left(1-p\right)}\ . \end{eqnarray} Summing this equation separately over $\ j\ $ and over $\ k\ $ gives \begin{eqnarray} \mathbb{P}\left(X=k\right) &=& \frac{\left( \lambda p\right)^k}{k!}e^{-\lambda p}\ ,\mbox{ and}\\ \mathbb{P}\left(Y=j\right) &=& \frac{\left( \lambda \left(1-p\right)\right)^{\,j}}{j!}e^{-\lambda \left(1-p\right)}\ . \end{eqnarray} Combining these results gives $\ \mathbb{P}\left(X=k, Y=j\right)= $$\mathbb{P}\left(X=k\right)\mathbb{P}\left(Y=j\right)\ $ for all $\ k\ $ and $\ j\ $, which thus establishes the independence of $\ X\ $ and $\ Y\ $.