Geometric Distribution

I've discovered the geometric distribution, it looks very handy.

$$E(X) = {1\over p}$$
$$V(X) = {1-p\over p^2}$$
$$P(Y=y) = {(1-p)}^{y-1}p$$

If I have a cheap laptop with an in built failure rate of 1% percent per month (per factory design),  I can find the probability of it failing in exactly 2 years time. Here, p=.01, and y=24

Therefore,
$$P(Y=24) = {(1-(.01))}^{(24)-1}(.01) = 0.008$$

If I want to know the chances of it failing before and up until that point, then I need the cumulative sum of P(Y=y) = 1,2,3,4,..,24
A shortcut to get this value is

$$P(Y<=y) = 1-(1-p)^y$$
$$P(Y<=24) = 1-(1-.01)^{24}=0.214$$

In R, I can plot this as follows

library(dplyr)
library(ggplot2)

dfGeom <- data.frame(x = 1:24, y = pgeom(1:24-1,.01))

ggplot(data=dfGeom, aes(x = x, y = y)) + 
  geom_bar(stat = "identity", position = "dodge") +
  geom_text(aes(y = y+.02, label = round(y,2), 
                x = as.numeric(x))) 
 

Meaning, chances of failure approaching year 2 and beyond is ~.2