Binomial/Negative Binomial Distribution? Why not Poisson here? When I looked at the below problem, I thought of Poisson immediately. I converted the rate to making 9/10 shots. However the answer told me to use the binomial/negative binomial distribution for parts a,b respectively. Can anyone clarify why Poisson is not valid here? Isn't 90% a rate? 
A marksman scores a bull's eye on 90% of his shots.
a. What is the probability he gets at least eight bull's eyes if he shoots 10 times?
b. If he shoots until he gets eight bull's eyes, what is the probability he needs at
most ten shots?
The answer is 0.92981 for both parts, readily gotten by using binomial/negative binomial distribution. 
 A: The relevant distribution is binomial for part (a) because we are told in advance that the number of shots is fixed at $n = 10$.  The support of a Poisson random variable is all nonnegative integers, $0, 1, 2, \ldots.$  But if the marksman shoots only $10$ times, then it is impossible to observe, say, $11$ bull's eyes.  So a Poisson distribution would not be an appropriate model for the number of successes in a fixed number of trials.
For part (b), the relevant distribution is not binomial; it is negative binomial.  We want the probability that the total number of shots required in order to achieve $8$ successes is no more than $10$.  If $X \sim {\rm NegBinomial}(r = 8, p = 0.9)$, where $$\Pr[X = k] = \binom{k-1}{r-1} p^r (1-p)^{k-r}, \quad k = r, r+1, r+2, \ldots,$$ then $$\Pr[X \le 10] = \sum_{k=8}^{10} \binom{k-1}{7} (0.9)^8 (0.1)^{k-8} = \frac{1162261467}{1250000000}.$$ 
A: Without doing wither calculation, you can see that the two parts have the same answer, since if you start taking shots, then "hitting at least 8 of the first 10 shots" happens if and only if "your 8th hit happens on or before your 10th shot."
