Normalize an infinite geometric probability law

Normalization gives $\sum_{x=0}^{\infty}c(4/5)^x=1$, hence $c\sum_{x=0}^{\infty}(4/5)^x=1$. Expand the source series: $c[(4/5)^0+(4/5)^1+(4/5)^2+\cdots]=1$. Recap the GP parameters from the Teacher note: first term $a=1$, ratio $r=4/5$, with $|r|<1$. Thus $c[1/(1-4/5)]=1$ and $c=1/5$.

$P(X\ge3)=\frac15(\frac45)^3+\frac15(\frac45)^4+\frac15(\frac45)^5+\cdots$. This tail has first term $a=\frac15(\frac45)^3$ and ratio $r=4/5$. Therefore $P(X\ge3)=\dfrac{\frac15(\frac45)^3}{1-\frac45}=64/125$.

$P(X\ge3)=1-P(X\le2)=1-[\frac15(\frac45)^0+\frac15(\frac45)^1+\frac15(\frac45)^2]=64/125$.