$X$ be a geometric random variable, show that $P[X \geq n] = (1-p)^{n-1}$

Question

Question: $X$ be a geometric random variable, show that $P[X \geq n] = (1-p)^{n-1}$

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Attempted Solution:

Let $X$ be a geometric random variable, that is, $X$ counts the number of independent Bernoulli trials until the first success occurs.

We wish to find, given some number $n \in \Bbb{Z}_{\geq 0}$, $P[X \geq n]$, that is, the probability that $n$ or more attempts are required before the first success occurs.

Suppose that attempt results in success with probability $p$. Let $$[][][][][][][][][]....[]$$ denote a set of attempts. We know that $n-1$ attempts all result in failure, then the probability of those attempts are $1-p$. So the probability of having $n-1$ failures is $(1-p)^{n-1}$.

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Difficulty:

I am having trouble seeing that the event "having $n-1$ failures" equivalent to "need $n$ or more attempts"

Secondly, I know that for a geometric random variable, the PMF is $P[X = n] = (1-p)^{n-1}p$. Where does the $p$ go in this case?

Is there an alternative, axiomatic derivation of $P[X \geq n]$?

e.g. $P[X \geq n] = 1 - P[X < n]$ Don't know $P[X < n]$...

Much thanks!

Answer 1

$$P[X \geq x] = \sum\limits_{x = n}^\infty P[X = x] = \sum\limits_{x = n}^\infty (1-p)^{x-1}p = \dfrac{p}{1-p} \sum\limits_{x = n}^\infty (1-p)^x = \dfrac{p}{1-p} \dfrac{(1-p)^n}{p}$$

Establishes the result.