We have a fisherman that goes fishing. The number of fishes that bite per hour follows a Poisson distribution with parameter $\lambda >0$. When the fish bites, it has a probability $p$ of going away, and $1-p$ of getting caught.
Let $Y$ be the number of fishes that get caught. Show that the random variable $Y \sim \mathcal{P}(\lambda (1-p))$
Since this problem is in the context of conditional expectation, I proceeded in the following way:
$$\Bbb{P}(Y=y)=\Bbb{E}(\Bbb{1}\{Y=y\})=\Bbb{E}(\Bbb{E}(\Bbb{1}\{Y=y\}|X))=\Bbb{E}(\Bbb{P}(Y=y|X))$$
So this is the step in which I get mixed up. Since $Y|X=x \sim Bi(x,(1-p))$ But here I have $Y=y|X$ So what is the real deal?
Let $Y$ and $Z$ be the no. of fish getting caught and the no. of fish escaping, respectively.
Then $Y+Z\in Poi(\lambda)$ (given).
Thus $$P(Y=k,Z=m)=P(Y=k,Z=m,Y+Z=m+k)=P(Y=k,Z=m|Y+Z=m+k)P(Y+Z=m+k)={m+k\choose m}(1-p)^kp^me^{-\lambda}\dfrac{\lambda^{m+k}}{(m+k)!}=\left[\dfrac{e^{-\lambda(1-p)}(\lambda(1-p))^k}{k!}\right]\left[\dfrac{e^{-\lambda p}(\lambda p)^m}{m!}\right]$$
showing $Y$ and $Z$ are independent and also showing $Y\in Poi(\lambda(1-p))$.