Answer:

Step-by-step explanation:
A negative binomial random variable "is the number X of repeated trials to produce r successes in a negative binomial experiment. The probability distribution of a negative binomial random variable is called a negative binomial distribution, this distribution is known as the Pascal distribution".
And the probability mass function is given by:
Where r represent the number successes after the k failures and p is the probability of a success on any given trial.
Solution to the problem
For this case the likehoof function is given by:

If we replace the mass function we got:

When we take the derivate of the likehood function we got:
![l(p,x_i) = \sum_{i=1}^n [log (x_i +r-1 C k) + r log(p) + x_i log(1-p)]](https://tex.z-dn.net/?f=%20l%28p%2Cx_i%29%20%3D%20%5Csum_%7Bi%3D1%7D%5En%20%5Blog%20%28x_i%20%2Br-1%20C%20k%29%20%2B%20r%20log%28p%29%20%2B%20x_i%20log%281-p%29%5D)
And in order to estimate the likehood estimator for p we need to take the derivate from the last expression and we got:

And we can separete the sum and we got:

Now we need to find the critical point setting equal to zero this derivate and we got:


For the left and right part of the expression we just have this using the properties for a sum and taking in count that p is a fixed value:

Now we need to solve the value of
from the last equation like this:



![p[\sum_{i=1}^n x_i +nr]= nr](https://tex.z-dn.net/?f=%20p%5B%5Csum_%7Bi%3D1%7D%5En%20x_i%20%2Bnr%5D%3D%20nr)
And if we solve for
we got:

And if we divide numerator and denominator by n we got:

Since 