TITLE: Calculating probability from a summation of the Negative Binomial Distribution
QUESTION [0 upvotes]: I am trying to understand an example of calculating $P(X>75)$ for a random variable $X$ when you have the formula (from the Negative Binomial Distribution for this example):
$$f(x)= P(X=x)= {x+4 \choose x}(0.08)^5(0.92)^x \hspace{20pt} \text{for } x=0,1,2...$$
I know the complement is easier to calculate and so you can do $P(X>75)=1-P(X\leq75)$. Still, the computation for every $X$ would be very time-consuming. My textbook lists the solution using a summation instead:
\begin{align}
P(X>75) &= 1-P(X\leq75) \\
&= 1-\sum_{x=0}^{75}f(x) \\
&= 0.2235
\end{align}
How is the value of the sum $\sum_{x=0}^{75}f(x)$ calculated? I have attempted to manipulate it to look like the binomial theorem.
\begin{align}
\sum_{x=0}^{75}{{x+4 \choose x}(0.08)^5(0.92)^x} &= (0.08)^5 \sum_{x=0}^{75}{{x+4 \choose x}(0.92)^x}
\end{align}
I am unsure of how to continue.

REPLY [1 votes]: Most likely, they computed it numerically.Here is a piece of R code that computes this probability.
f=function(x){(0.08^5)(0.92^x)(factorial(x+4))/(factorial(x)*24)}
a=seq(0,75,1)
a=as.matrix(a)
pi=apply(a,1,f)
prob=1-sum(pi)  #the final probability

prob

[1] 0.2234999
{"set_name": "stack_exchange", "score": 0, "question_id": 1529650}