Question 867167
Re: TY 
Have no capability of going that far back in postings.  So many...more recent
for
C) It seems originally, I did use P(k) = p^k, only replaced the k  with x as
x is a variable normally used in Probability. P(k) = p^k more politically correct, given the question.

{{{P (x)= highlight_green(nCx)(p^x)(q)^(n-x) }}} 
a)n = 1,  {{{P (1)= highlight_green(1C1)(p^1)}}} = p^1,  1 trail, one success
b)n = 2, {{{P (2)= highlight_green(2C2)(p^2)}}} = p^2,   2 trials. two successes
c) n = k,  {{{P (k)= highlight_green(kCk)(p^k)}}} = p^k  just following the pattern n = x
If there was continual success in 100 trials, For ex: P(of that happening) = p^100
My take was the probability of ALL successes encompassing any number of trials.

Re: *TIP* that X can be written as X=I1+I2+....+In, 
where Ii are the random variables Bernoulli with probability of success p
Apply as You wish