Unlike other discrete distributions I find that setting the range and solving for summation in negative binomial distributions to be quite tricky. Here's what I mean via a sample problem:
some biology students were checking the eye color of a large number of fruit flies. for an individual fly assume that the probability of white eyes is 0.25 and that of red eyes is 0.75. Treat these two observations as independent Bernoulli trials.
A) The probability that at least four flies have to be checked for eye color to observe a white eye fly is:
According to my text the answer is; $P(X\geq 4)= P(X>3)= 0.75^3$
I agree with the $P(X\geq 4)$ but am unsure why that is equal to P(X>3).Can you explain this?
B) The probability that at most four flies have to be checked for eye color to observe a white eyed fly.
My text says the answer is: $P(X\leq 4) = 1-(0.75)^4$
This I can understand quite well and it can be solved by taking the summation of 1 to 4 of $(1-0.75)^{x-1} (0.75)$, which in turn is 1-P(X>k).
Now, if the question asked P(X<4) how would you solve that? I feel like the first example is really confusing me.
Your question regarding the first part has a simple answer: the number of checks is a positive integer. Therefore, if your sample space is $\Omega$, and $X$ the random variable denoting the number of checks needed to find a white-eye, $X : \Omega \to \mathbb{R}$ takes values only in $\mathbb{N}$. So the set $\{\omega \in \Omega : X(\omega) > 3\}$ is equal to the set $\{\omega \in \Omega : X(\omega) \geq 4\}$, and so $P\{X \geq 4\} = P\{X > 3\}$.
So for the second part of the question: if the question asked for $P\{X<4\}$, that would simply be $1-P\{X\geq 4\}$, which is equal to $1-P\{X>3\}$. But since it's easy to check $P\{X<4\}$ directly, it would be $$P\{X<4\} = P\{X=1\} + P\{X=2\} + P\{X=3\} = 0.25 + 0.75 \cdot 0.25 + (0.75)^2 \cdot 0.25.$$