Find the probability that the company will receive exactly k claims from A policies tomorrow. Consider a large insurance company with two types of
policies: policy A and policy B. Suppose that the number of claims the
company sees in a given day has a Poisson distribution with parameter $\lambda$. Suppose further that a randomly selected claim is from a type A policy with probability $p$.
Find the probability that the company will receive exactly $k$ claims from A policies tomorrow. 
Hint: Multinomial theorem and ${}_nC_k=0$ if $k>n$.
My work: 

I am so confused about how this is related to the multinomial theorem and setting up.
 A: Your work so far is basically correct---and you used the multinomial theorem in computing the probability that exactly $k$ of $n$ cases are of type $A$---and it just remains to do the following:
(1) You use the parameters $p_A$ and $p_B$ for the respective probabilities that the two policy types are chosen. But we're given that $p = p_A$, and since each claim is either of type $A$ or $B$, $p_B = 1 - p_a = 1 - p$, so we can substitute and write the probability in terms of the parameter $p$.
(2) It remains to sum over all of the possible values of $n$. The fact that ${n \choose k} = 0$ for $k > n$ corresponds to the fact that if the number $n$ of claims that come in is less than $k$, it's not possible for $k$ of those claims to have type $A$.

 Substituting in your expression per (1), simplifying, and add over all $n \geq k$ gives  that the desired probability is \begin{align}\sum_{n \geq k} \frac{\lambda^n e^{-\lambda} p^k (1 - p)^{n - k}}{k! (n - k)!} &= \frac{e^{-\lambda} \lambda^k p^k}{k!} \sum_{n \geq k} \frac{\lambda^{n - k} (1 - p)^{n - k}}{(n - k)!} \\ &= \frac{e^{-\lambda} \lambda^k p^k}{k!} \sum_{m \geq 0} \frac{[\lambda (1 - p)]^m}{m!} \\ &= \frac{e^{-\lambda} \lambda^k p^k}{k!} e^{\lambda (1 - p)} \\ &= \color{#df0000}{\boxed{\frac{(\lambda p)^k}{k! e^{\lambda p}}}} .\end{align} Alternatively, since each case event is independent, the number of claims of type A submitted each day itself follows a Poisson distribution. The mean number of cases of type A is $(\textrm{average number of cases per day}) \cdot (\textrm{probability that a given case is of type A})$, that is, $\lambda p$, and substituting in the formula for the Poisson distribution immediately recovers the above answer.

