Customers arrive to a store such that the number of arriving customers in an hour has a Poisson distribution with mean $4$. A customer is male or female with equal probabilities. Let $X$ be the number of female customers in an hour and find $P(X = 0).$
So I have that $$X=\text{Number of female visitors.} \\Y=\text{Number of total visitors.} \ \ \ \\ Y\sim\text{Poi}(4) \quad \quad \quad \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$
And I'm looking for the probability that no females visited the store in a given hour.
If $X$ and $Y$ are discrete random variables and have joint pmf $p$, the conditional pmf of $Y$ and given $X=j$ is defined as
$$p_Y(Y=k|X=j)=\frac{p(X=j,Y=k)}{p_X(X=j)}. \tag{1}$$
Rewriting this I get
$$p_X(X=j)=\frac{p(X=j,Y=k)}{p_Y(Y=k|X=j)}.$$
I'm not sure but I believe that $p(x_j,y_k)=e^{-4}4^0/0!=1/e^4.$ To compute $p_X(x_j)$ for $j=0$ I also need $p_Y(y_k|x_j).$ How do I find this out?
Am I even doing this correctly to begin with?
There are two ways to approach this. First, for any number of customer arrivals you can compute the probability that none are women. Then you can use law of total probability to express the probability no women arrive as a sum. You appear to have chosen this approach and then fallen off the rails (seem to have forgotten there is a sum involved.) We have $P(X=0\mid Y=n)=1/2^n,$ so $$P(X=0)= \sum_n P(X=0\mid Y=n)P(Y=n)=\sum_n \frac{1}{2^n} \frac{4^ne^{-4}}{n!}$$
The second, slicker way is to use the decomposition property of the compound Poisson process, which tells you that the number of women that arrive is Poisson distributed with mean 2.