The exercise says:
Let $X$ ~ $G (p)$. Prove that $P (X ≥ k) = (1 - p)^k$, with $k ∈ N$.
My try to demonstrate it:
- $P (X ≥ k) = 1-P(X<k)$
- $P (X+1 ≥ k+1) = 1-P(X+1<k+1) = 1-P(X⩽k)$
- $P (X+1 ≥ k+1) = 1-((1-p)^kp)$
- $P (X+1-1 ≥ k+1-1) = 1-(((1-p)^kp)-1)$
- $P (X ≥ k) = 1-(((1-p)^kp)-1)$
$\bbox[0.5ex]{X\sim\mathcal{Geo}_0(p)}$ means $X$ is a count of failures before a success in an indefinite sequence of independent Bernoulli trials with identical success rate $p$.
$\bbox[0.5ex]{\mathsf P(X\geq k)\quad [k\in\Bbb N]}$ is then the probability for the first success occurring after at least $k$ consecutive failures. Which is simply the probability for obtaining $k$ consecutive failures.
$$\mathsf P(X\geq k)=(1-p)^k$$
Of course you can show this by using $\bbox[0.5ex]{\mathsf P(X=k)~=~(1-p)^kp}$ (but why is that so)?
$$\begin{align}\mathsf P(X\geq k) &= \sum_{j=k}^\infty (1-p)^j p \\ & = (1-p)^kp\sum_{j=k}^\infty (1-p)^{j-k} \\ & \vdots \end{align}$$