I've done part a and b, though a little stuck on identifying how to do the other questions. I don't really know where to look, any hints?
Patients arrive at a walk-in dental clinic according to a Poisson Process of rate 8 per hour.
(a) Find the probability that a time period of 20 minutes, exactly 4 patients arrive.
(b) Let T be the time (in hours) between the arrival of the first patient and that of the second patient. Name the distribution of T and state its mean.
Each patient requires one of two possible types of treatment (independently of all other patients and of the arrival process): with probability 0.6 he or she only needs a quick check up, whereas with probability 0.4 dental surgery is required.
(c) Find approximately the probability that amongst the first .50 patients, more than 25 will require dental surgery.
Suppose that the clinic is run by a single dentist, and that the time she takes to do a quick check-up is Exponentially distributed with a mean of 5 minutes, whereas the time she needs to perform dental surgery is Exponentially distributed with a mean of 15 minutes. Suppose that you arrive at the clinic just as the dentist calls in the patient before you and starts treating him, so you take a seat in the waiting room and await your turn.
(d) Calculate the probability that the dentist will take longer than 10 minutes treating the patient before you.
(e) It turns out that after you have waited for 10 minutes, the patient before you is still being treated. Given this information, what is the probability that he is undergoing dental surgery rather than a quick check-up?
For part (c) you don't need to know anything about the times and which the patients arrive and so you don't need to know anything about the Poisson process. It's just the probability of getting at least $25$ successes in $50$ independent trials with probability $0.4$ of success on each trial. Thus it involves a binomial distribution.
For part (d) you have \begin{align} & \Pr(\text{time} > 10) \\[10pt] = {} & \Pr(\text{check-up})\cdot\Pr(\text{time}>10 \mid {}\text{check-up)} + \Pr(\text{surgery}) \cdot\Pr(\text{time} > 10 \mid {} \text{surgery}) \end{align} For an exponential distribution with mean $5$, the probability of exceeding any number $t>0$ is $e^{-t/5},$ and similarly if the mean is $15$.
$$ \Pr(\text{surgery}\mid {}\text{time}>10) = \frac{\Pr(\text{surgery}\ \&\ \text{time}>10)}{\Pr(\text{time}>10)}. $$ The denominator was found in part (d).