Poisson Distribution - Optimistaion 
A store offers a new seasonal product featured. Let $N$ be the random variable
  which means the number of clients who come to the store during the season, where
  $N \sim \operatorname{Poisson}(19)$. It is estimated that the probability that a customer buys the new product is $0.38$
  regardless from a customer to another.
a) It is assumed here that the store has an unlimited supply of product. Random variables $X$ and $Y$ are such that
$X$: The number of customers who purchase the product ;
$Y$: The number of customers who do not buy the product.
Are the variables $X$ and $Y$ independent ? Justify .

My answer: I think they are dependent, because there is a fixed number of costumers, and when the number of buyers changes, the number of non buyers changes too.

b) The store has a profit of 63 million for each unit sold. Each unsold unit
  should be stored for next year at a cost of 38 million.
Determine what the number of units stored should be to
  maximize its average profit.

My answer: Let's start with the pdf of Poisson distribution.
$p_x(k)=\displaystyle\frac{e^{-\alpha}\alpha^k}{k!}$ , for k=0,1,...
$\alpha=19$
$p_x(k)=\displaystyle\frac{e^{-19}19^k}{k!}$ , for k=0,1,...
Now I don't know how to relate my 0.38  to the equation I just came up with...
Since 38% of the people buy, I need to first estimate how many costumers I'll have with the Poisson distribution, then assume that 38% only will  buy. So to maximize my profits, let's say there is 100 costumers estimated, I'll only put 38, so I'll sell every one of them. But I don't know how to apply this logic to the problem.
Your help is really appreciated! :)
 A: For Part A
Given: We have customers arriving at a constant average rate of $19$ per season, each at a time independent of the arrival of the previous customer, and each customer has an independent probability for being a purchaser of $0.38$.
Note: Although the seasonal counts, of purchasers and non-purchasers, are conditionally dependent, for a given total seasonal count of customers, that is not what you were asked. 
Hint: A Poisson distribution is, by definition, that of the count of events occurring in an interval (of time and/or space) if these events occur with a constant average rate and independently of the time since the last event.
Put them together to identify the distributions of the seasonal counts of purchasers and non-purchasers, and assess whether they depend on each other or not?
Part B should then fall into place.

Formula you should know:
The total count is Poisson distributed with rate $19$ per season.   So $N\sim\mathcal {Poiss}(19)$, which implies that: $$\mathsf P(N=x+y) = \dfrac{19^{(x+y)}~\mathsf e^{-19}}{(x+y)!} \quad\Big[x\in\Bbb N, y\in\Bbb N\Big]$$
For a given total seasonal count, the count of purchases will be conditionally binomial distributed with rate $0.38$ (can you justify?).   So $X\mid N\sim \mathcal {Bin}(N, 0.38)$, which implies that: $$\mathsf P(X=x\mid N=x+y) = \dfrac{(x+y)!~{(0.38)^x~(0.62)^y}}{x!~y!}\quad\Big[x\in\Bbb N, y\in\Bbb N\Big]$$
Now


*

*$\mathsf P(X=x) = \sum_{\psi=0}^\infty \mathsf P(X=x\mid N=x+\psi)~\mathsf P(N=x+\psi)$

*$\mathsf P(Y=y) = \sum_{\chi=0}^\infty \mathsf P(Y=y\mid N=\chi+y)~\mathsf P(N=\chi+y)$

*$\mathsf P(X=x, Y=y) = \mathsf P(X=x\mid N=x+y)~\mathsf P(N=x+y)$

