Geometric distribution

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It's crap, look at it

Geometric Distribution
$X\sim\text{Geo}(p)$ for $p$ the probability of each trials' success
$X\eq k$ means that the first failure occurred on the $k^\text{th}$ trial, $k\in\mathbb{N}_{\ge 1}$
Definition
Defined over	$X$ may take values in $\mathbb{N}_{\ge 0}\eq\{1,2,\ldots\}$
p.m.f	$\mathbb{P}[X\eq k]:\eq p^{k-1}(1-p)$
c.d.f / c.m.f	$\mathbb{P}[X\le k]\eq 1-p^k$
cor:	$\mathbb{P}[X\ge k]\eq p^{k-1}$
Properties
Expectation:	$$\mathbb{E}[X]\eq\frac{1}{p}$$
Variance:	$$\text{Var}(X)\eq\frac{1-p}{p^2}$$

Notes

during proof of $\mathbb{P}[X\le k]$ the result is obtained using a geometric series, however one has to align the sequences (not adjust the sum to start at zero, unless you adjust the $S_n$ formula too!)

Check the variance, I did part the proof, checked the MEI formula book and moved on, I didn't confirm interpretation.

Make a note that my Casio calculator uses $1-p$ as the parameter, giving $\mathbb{P}[X\eq k]:\eq (1-p)^{k-1}p$ along with the interpretation that allows 0