Determine the distribution of $X$

Let $$N$$ a random variable such that $$N \sim \operatorname{Pois}(\lambda)$$. Furthermore, let $$X$$ a random variable such that $$\mathbb{P}( X = k\ | \ N=n)=\binom{n}{k}p^kq^{n-k},\ \ 0\leq k \leq n, \ \ p+q=1$$ Show that $$X \sim \operatorname{Pois}(\lambda p)$$

Honestly im hard stuck here. So i have to show that $$\mathbb{P}(X=k)=\dfrac{e^{-p\lambda}(p\lambda)^k}{k!}$$ and $$\operatorname{Supp}(X)=\mathbb{N}$$. Using the definition of conditional probability plus Bayes law i have $$\mathbb{P}( X = k\ | \ N=n)=\frac{\mathbb{P}( N = n\ | \ X=k)\mathbb{P}(X=k)}{\mathbb{P}(N=n)} \iff\binom{n}{k}p^kq^{n-k}\frac{e^{-\lambda}\lambda^n}{n!}=\mathbb{P}( N = n\ | \ X=k)\mathbb{P}(X=k)$$ So i need a expression for $$\mathbb{P}( N = n\ | \ X=k)$$ and i done. But i dont know how to get it.

• This is not a good approach as you are switching the conditioning in a scenario where the given conditioning is what you need. So Baye's rule is not useful here. Use the law of total probability (LoTP) to compute $P[X=k]$ by conditioning on $N=n$ for all $n\in\{0,1,2,...\}$. Everything in the resulting LoTP expression is directly given to you and waiting for you to use. Commented May 13 at 16:01
• It does not help here since you are looking for a proof, but $\mathbb{P}( N = n \mid X=k) = \mathbb{P}( N -X= n-k)$ as $X$ and $N-X$ are independent. $N-X \sim \operatorname{Pois}(\lambda q)$, matching $X \sim \operatorname{Pois}(\lambda p)$. Commented May 13 at 19:00

Recall the LoTP: Let $$\{B_n\}_{n \in I}$$ be disjoint events with union equal to the whole sample space (where $$I$$ is a finite or countably infinite set that contains all the $$n$$ subscripts). The law of total probability states that for any event $$A$$ we have $$P[A] = \sum_{n\in I} P[A|B_n]P[B_n]$$ where we remove any terms on the right-hand-side with $$P[B_n]=0$$.
Application: In your problem you fix $$k$$ as a nonnegative integer and define events $$A$$ and $$B_n$$ by $$A = \{X=k\}$$ $$B_n=\{N=n\} \quad \forall n \in \{0, 1, 2, ...\}$$ So the set of subscripts in this case is $$I=\{0,1,2,...\}$$.
Other problems: In other problems, say, when you partition on a dice roll, you might use $$I=\{1,2,3,4,5,6\}$$. Or if you partition on the result of two coin flips you might use $$I=\{HH,HT,TH,TT\}$$. Or if you partition on the value of a random variable $$G$$ that has a geometric distribution then $$I=\{1,2,3,...\}$$. \begin{align*} P[A]&=\sum_{i=1}^6P[A|\mbox{Roll is i}]P[\mbox{Roll is i}]\\ P[A]&= P[A|HH]P[HH]+P[A|HT]P[HT]\\ &\quad +P[A|TH]P[TH]+P[A|TT]P[TT]\\ P[A]&=\sum_{i=1}^{\infty}P[A|G=i]P[G=i] \end{align*}