The rate at which customer arrive to the bookstore is $\lambda (t)=-(t-4)^2+16 $ where $t$ measured in hours. The customers can buy a book with probability $0.5$ and they can also buy a coffee with probability $0.5$ given they bought a book or $0.8$ given they bought no book. Find the probability of $k$ people buying coffee in a given day. A working day is 8 hours.
I am assuming that they buy a coffee with probability $0.5\times 0.5+0.5\times 0.8=0.25+0.4=0.65$ So on average every customer has a probability of $0.65$ to buy coffee.
What is the probability that exactly $k$ coffees are bought?
Is it the same as having $k/0.65$ customers walk in the shop?
$$P(N_8=k/0.65)=\frac{e^{-8\lambda(8)}8\lambda(8)^{(k/0.65)}}{(k/0.65)!}$$ and solve for $k$?. It seems very tedious.
Notes: I calculated the total expected #of arrivals using $\int \lambda (t)=85\frac{1}{3}$
First, it is not the same as the probability of having $k/0.65$ customers walk into the shop. For example, taking $k = 5$, it is really unlikely that exactly $2/0.65 \approx 7.692$ customers will walk into the shop. (Well, maybe less unlikely if the shop is located in King's Landing...)
What you have is a rate of customers walking in and buying coffee of $$ \lambda_c(t) = 0.65 \left(16-(t-4)^2 \right)= \frac{52}{5} - \frac{13(t-4)^2}{20} $$
The number of coffee buyers is Poisson distributed with a mean of $$ \int_0^8 \frac{52}{5} - \frac{13(t-4)^2}{20} \, dt = \frac{832}{15} $$
So $$P(k) = \frac{ \left(\frac{832}{15}\right)^k e^{-\frac{832}{15}} }{k!} $$