Finding a distribution for phone calls "A call centre receives a random number of calls per day, with a Poisson distribution with mean m. A given call concerns a particular newly released product with probability p, independently of all other calls and of the total number of calls. Compute the distribution (e.g. probability function) of the number of calls on a given day that concern the new product."
I guess it is reasonable to set the number of calls on a given day that concerns the new product to X, where we then want the distribution $P_{X|Y = k}$? Apparently the correct answer is $X \in Po(mp)$. I'm stuck.
 A: You are receiving calls at a constant average rate of $m$ per day, each call occurring independently of time of the previous event.   (The very definition of a Poisson distribution.)
Each call is also independently about the new product with an average 'success rate' of $p$.
Are not you receiving calls about the topic at a constant average rate of $mp$ per day, each such call occurring independently of the time of the previous such event?   Then, what type of distribution would the count of such calls have?
Or if you prefer: the conditional distribution of topical calls for a given amount of calls must be Binomially distributed, so we can calculate the marginal distribution:
$$\begin{align}\mathsf P(X=x) ~=~& \sum_{y=x}^\infty \mathsf P(Y=y)~\mathsf P(X=x\mid Y=y) \\ =~& \sum_{y=x}^\infty \dfrac{m^y\mathsf e^{-m}}{y!}\cdot\dfrac{y!~p^y(1-p)^{y-x}}{x!~(y-x)!}
\\\ddots& \end{align}$$
A: The call centre receives on average $m$ calls a day, and hence an average of $mp$ calls a day on the new product. Like all the calls, the calls on the new product arrive independently of each other, and hence have the Poisson distribution with mean $mp$, which is the given answer. No need to wrangle with named random variables.
This is a nice property of the Poisson distribution: it is infinitely divisible into Poisson distributions of smaller means, and those smaller means add up to the mean of the original distribution.
