library(ggplot2)
n <- 50
p <- 0.02
df_qc <- data.frame(x = 0:8, pmf = dbinom(0:8, size = n, prob = p))
df_qc$highlight <- ifelse(df_qc$x <= 2, "Target (<=2)", "Other")
ggplot(df_qc, aes(x = factor(x), y = pmf, fill = highlight)) +
geom_col(width = 0.7) +
scale_fill_manual(values = c("Target (<=2)" = "darkgreen", "Other" = "grey70")) +
labs(title = "QC Defect Probability Matrix (n=50, p=0.02)",
x = "Number of Defective Chips (x)", y = "Probability", fill = "Region") +
theme_minimal()Binomial Distribution
Joel Sanqui (Appalachian State University)
Introduction
Let \(X\) be a random variable representing the total number of successes in \(n\) independent trials that satisfy the following:
- Each trial has two outcomes: Success (\(s\)) or Failure (\(f\)).
- The probability of success is \(p\).
- The probability of failure is \(q = 1-p\).
Finding the pmf
To find the total probability of exactly \(x\) successes, we split the problem into two distinct parts:
\[\begin{aligned} P(X = x) = &\quad \mathbf{\text{Combinations}} \\ &\times \mathbf{\text{Probability of a Specific Sequence}} \end{aligned}\]
1. Counting the Combinations
How many ways can we arrange exactly \(x\) successes across \(n\) total trials?
We use the binomial coefficient to choose \(x\) slots out of \(n\):
\[\binom{n}{x} = \frac{n!}{x!(n-x)!}\]
2. Probability of One Sequence
Since the trials are independent, we multiply their individual probabilities together.
For any specific sequence containing \(x\) successes and \(n-x\) failures:
\[\text{for } x = 0, 1, 2, \dots, n\]
\[\underbrace{p \cdot p \cdots p}_{x \text{ times}} \times \underbrace{(1-p) \cdot (1-p) \cdots (1-p)}_{n-x \text{ times}} = p^x (1-p)^{n-x}\]
3. The Final pmf
Combining the total number of possible arrangements with the probability of a single sequence yields the final formula:
\[f(x) = P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}\]
\[\text{for } x = 0, 1, 2, \dots, n\]
We write \(X \sim \text{Binomial}(n,p)\).
Mean of Binomial(n,p)
A Binomial variable can be written as the sum of \(n\) independent Bernoulli random variables: \(X = \sum_{i=1}^n Y_i\), where \(Y_i \sim \text{Bernoulli}(p)\).
First, find the expectation of a single Bernoulli trial: \[\mathbb{E}[Y_i] = (1 \cdot p) + (0 \cdot (1-p)) = p\]
By the linearity of expectation: \[\mathbb{E}[X] = \mathbb{E}\left[\sum_{i=1}^n Y_i\right] = \sum_{i=1}^n \mathbb{E}[Y_i] = \sum_{i=1}^n p = np\]
Variance of Binomial(n,p)
First, we find the variance of a single Bernoulli trial \(Y_i\): \[\mathbb{E}[Y_i^2] = (1^2 \cdot p) + (0^2 \cdot (1-p)) = p\] \[\text{Var}(Y_i) = \mathbb{E}[Y_i^2] - (\mathbb{E}[Y_i])^2 = p - p^2 = p(1-p)\]
Because the \(n\) Bernoulli trials are independent, the variance of their sum equals the sum of their variances: \[\text{Var}(X) = \text{Var}\left(\sum_{i=1}^n Y_i\right) = \sum_{i=1}^n \text{Var}(Y_i) = n p(1-p)\]
Example 1: Quality Control
Manufacturing plants use the binomial distribution to predict and manage product defects.
- Scenario: A factory tests a batch of microchips.
- Parameters: Suppose historically, the probability of any single chip being defective is \(p = 0.02\) and the factory tests \(n=50\) microchips. The Binomial parameters are then \(n=50\) and \(p=0.02\).
- Problem: Calculate the probability that \(x \le 2\) chips are defective and find the mean and variance of the number of defective chips.
Solutions
- Exact Probability \(P(X \le 2) = f(0)+f(1)+f(2)\) = 92.16%
- Mean or expected number of defective chips is (50)(.02) = 1
- Variance of the number of defective chips is (50)(.02)(.98) = .98.
QC Visualization
QC Visualization
Example 2: A/B Testing
Digital marketers rely on the binomial distribution to see if website changes actually improve performance.
- Scenario: An e-commerce site routes users to a new checkout page design.
- Parameters: If the baseline checkout rate is known to be \(p = 0.08\) and the site routes \(n=500\) users to the new checkout page design, then the Binomial parameters are \(n=500\) and \(p=0.08\).
- Question: How likely it is to observe \(x \ge 50\) successful purchases by pure chance?
Solution
- Exact Binomial Probability \(P(X \ge 50) = 1 -P(X \le 49) = f(0) + f(1) + ...+f(49)\) = 6.22%
- Mean or expected number of purchases/checkouts is (500)(.08) = 40
- Variance of the number of purchases/checkouts is (500)(.08)(.92)
A/B Testing Visualization
library(ggplot2)
n <- 500
p <- 0.08
df_ab <- data.frame(x = 20:65, pmf = dbinom(20:65, size = n, prob = p))
df_ab$highlight <- ifelse(df_ab$x >= 50, "Target (>=50)", "Other")
ggplot(df_ab, aes(x = x, y = pmf, fill = highlight)) +
geom_col(width = 0.9) +
scale_fill_manual(values = c("Target (>=50)" = "firebrick", "Other" = "grey70")) +
labs(title = "A/B Testing (n=500, p=0.08)",
x = "Number of Checkouts/Purchases (x)", y = "Probability", fill = "Region") +
theme_minimal()A/B Testing Visualization
Binomial Distribution in R
In the above plots, we use the function dbinom to calculate the Binomial pmf. The probability in Example 1 is
To calculate the Binomial cdf, we use the pbinom function. The probability in Example 1 is also given by
To find quantiles of the Binomial distribution, we use the function qbinom.
To generate random numbers from a Binomial distribution, we use rbinom.
Example 3
What is the probability of getting exactly 50 heads if a fair coin is flipped 100 times?
Solution:
Let X = number of heads obtained after 100 flips. Then \(X\sim \text{Binomial}(100,0.5)\) and \(P(X = 50)\) is
Example 4
What is the probability of getting less than 45 heads?
Solution:
\(P(X < 45)\) is
Example 5
What is the probability of getting anywhere from 45 to 55 heads?
Solution:
\(P(45 \le X \le 55)\) is calculated as
Example 6
What is the 3rd quartile of the \(\text{Binomial}(100,0.5)\) distribution?
Solution:
The 3rd quartile is the 75th quantile and is calculated as
Empirical Simulation: Coin Flipping
What happens if we simulate a Binomial process empirically?
- Experiment: Simulate flipping a fair coin (\(p = 0.5\)) \(100\) times (\(n = 100\)).
- Repetitions: Repeat this whole process \(1,000\) times.
- Objective: Map out the generated values and approximate the distribution’s mean and variance from the simulation.
Simulation & Analysis
library(ggplot2)
set.seed(42) # Ensure reproducible results
# 100 trials, 1000 repetitions, p = 0.5
sim_trials <- rbinom(n = 1000, size = 100, prob = 0.5)
df_sim <- data.frame(heads = sim_trials)
# Calculate empirical values
sim_mean <- mean(sim_trials)
sim_var <- var(sim_trials)
# Dynamic Title with results
plot_title <- paste0("Empirical Coin Flips (1000 trials of n=100)\n",
"Simulated Mean: ", round(sim_mean, 2),
" | Simulated Variance: ", round(sim_var, 2))
# Plot histogram
ggplot(df_sim, aes(x = heads)) +
geom_histogram(binwidth = 1, fill = "purple", color = "white", alpha = 0.8) +
geom_vline(xintercept = sim_mean, color = "black", linetype = "dashed", size = 1) +
labs(title = plot_title, x = "Number of Heads (X)", y = "Frequency") +
theme_minimal()Note that the theoretical mean is 50 and the theoretical variance is 25 so the simulation is not bad.
Simulation & Analysis
References
History of the Distribution: Hald, A. (1998). A History of Mathematical Statistics from 1750 to 1930. New York, NY: John Wiley & Sons.
Properties of the Distribution Ross, S. M. (2014). A First Course in Probability (9th ed.). Pearson.
Computational Software: R Core Team (2026). R: A Language and Environment for Statistical Computing. R Foundation for Statistical Computing. Vienna, Austria. Available at https://www.R-project.org/.