A patient arrives at a doctor's office. With probability $\frac{1}{5}$ he receives service immediately, while with probability $\frac{4}{5}$ his service is deferred an hour. After an hour's wait again with probability $\frac{1}{5}$ his needs are serviced instantly or another hour of delay is imposed and so on. What is the distribution of the number of patients who receive service over an $8$-hr period assuming the same procedure is followed for every arrival and the arrival pattern is that of a Poisson process with parameter $1$.
My initial guess was that the distribution would be like a negative binomial. But I am getting stuck while writing the probability of $[N(t)=n]$ conditioned on a suitable event. It seems as if the events I should condition on are too complex.
Any suggestions?
Let $N(t)$ be the number of arrivals by hour $t$. Let $X$ be the number of patients serviced by hour $8$, and $X_i$ the number of arrivals in hour $i$ that are serviced by hour $8$, so that $X=X_1+\cdots+X_8$.
\begin{align} P(X_i=n) &= \sum_{k=n}^{\infty} P(N(i)-N(i-1)=k)P(X_i\mid N(i)-N(i-1)=k) \\ &= \sum_{k=n}^{\infty} \dfrac{1}{k!}e^{-1} \binom{k}{n}p_i^n(1-p_i)^{k-n} \\ &\qquad\qquad\qquad\text{where $p_i=$ prob. of a patient being serviced within $(8-i)$ hours } \\ &= \dfrac{p_i^n}{n!} e^{-1} \sum_{k=n}^{\infty} \dfrac{(1-p_i)^{n-k}}{(n-k)!} \\ &= \dfrac{p_i^n}{n!} e^{-p_i}. \end{align}
So $X_i\sim Pois(p_i)$. Also, the $X_i$s are independent, so $X\sim Pois(p_1+\cdots +p_8)$.
Finally, we find $p_i$:
\begin{align} p_i &= \dfrac{1}{5} \sum_{k=0}^{8-i} \left(\dfrac{4}{5}\right)^k \\ &= \dfrac{1}{5} \dfrac{1-(4/5)^{9-i}}{1-4/5} \\ &= 1-(4/5)^{9-i}. \end{align}