Is this a geometric distribution

Question

Let $X$ be $rv$ with the probability distribution, $$p(k) = (1 - x)x^k; k = 0, 1, 2,...$$ where $x \in (0, 1)$ is a constant.

Is this computing the number of failures with $Pr[\text{Success}] = 1-x$?

Answer 1

Yes, in general, geometric distribution takes the form of $$p(k) = (1-p)^k p, k=0,1,2\ldots, $$

Here $p=1-x$.

Note that failure or success depends on the definition of the user.

Answer 2

It depends on your definition of Geometric distribution.

If the geometric distribution is defined as the number of attempts until a success then $0$ does not belong to it's range. Under this intrpretation $ X = Z -1$ where $Z \sim \mathcal{G} (1-x)$.

Wikipedia