The geometric distribution models the number of trials needed to get the first success in a sequence of independent Bernoulli trials (each with success probability \(p\)).
It answers questions like: “How many times must we try before we succeed for the first time?”
Memoryless property: The probability of success in the next trial does not depend on how many failures have already occurred.
Two common formulations:
| Version | What it models | PMF | Support |
|---|---|---|---|
| Type 1 | Number of trials until first success (including the success) | \(P(X = k) = (1-p)^{k-1}p\) | \(k = 1, 2, 3, \dots\) |
| Type 2 | Number of failures before first success | \(P(Y = y) = (1-p)^y p\) | \(y = 0, 1, 2, \dots\) |
A quality control inspector tests items coming off an assembly line. Each item has a 5% chance of being defective. What is the probability that the first defect is found on the 5th item tested?
Here:
\(P(X = 5) = (1 - 0.05)^{5-1} \times 0.05\)
\(= (0.95)^4 \times 0.05 \approx 0.8145 \times 0.05 \approx 0.0407\)
So about 4.07%.
| Situation | Example |
|---|---|
| Quality control | Number of items tested until first defect |
| Sports | Number of shots until first goal |
| Customer behavior | Number of calls until first sale |
| Medicine | Number of patients until first responder |
| Engineering | Number of components until first failure |
library(ggplot2)
p <- 0.05
x <- 1:20
prob <- dgeom(x - 1, p) # Type 1: P(X = k) = (1-p)^(k-1) p
df <- data.frame(Trials = x, Probability = prob)
ggplot(df, aes(x = Trials, y = Probability)) +
geom_col(fill = "steelblue", alpha = 0.7) +
labs(title = paste("Geometric Distribution (p =", p, ")"),
subtitle = "Number of trials until first success",
x = "Number of trials", y = "Probability")