---
title: "Bernoulli and Binomial Distributions"
subtitle: "SD05 — Discrete Distributions"
author: "Samer El Kababji"
output:
  html_document:
    toc: true
format:
  html:
    code-links:
      - text: "Download R Markdown"
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        href: SD05-01-bernoulli-and-binomial-distributions.Rmd?download=1
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE, message = FALSE, warning = TRUE)
```

**Course:** Statistics and Probability for Data Science (DS14270).  
**Source deck:** Samer El Kababji, *SD05 — Discrete Distributions*.

This lesson develops Bernoulli trials and their binomial sum. Each formula is
followed by executable base R, including the deck's coin and credit-card
examples. Simulations are labeled as extensions.

## One Bernoulli trial

A Bernoulli random variable records failure or success:

$$
X\in\{0,1\},\qquad P(X=1)=p,\qquad P(X=0)=1-p.
$$

Its moments are

$$
E(X)=p,\qquad \operatorname{Var}(X)=p(1-p).
$$

```{r bernoulli-pmf-and-moments}
p <- 0.20
x <- 0:1
probability <- c(1 - p, p)

mean_X <- sum(x * probability)
variance_X <- sum((x - mean_X)^2 * probability)

data.frame(x, probability)
c(mean = mean_X, variance = variance_X)
```

For example, if a visitor buys a desktop computer with probability $0.20$,
then the purchase indicator has mean $0.20$ and variance $0.16$.

## From Bernoulli trials to a binomial count

Suppose there are $n$ trials, each trial has two outcomes, the trials are
independent, and the success probability $p$ is constant. The total number of
successes

$$
X=X_1+\cdots+X_n
$$

has a binomial distribution, written $X\sim\operatorname{Bin}(n,p)$, with PMF

$$
P(X=k)=\binom nkp^k(1-p)^{n-k},\qquad k=0,1,\ldots,n.
$$

```{r binomial-pmf}
n <- 5
p <- 0.5
k <- 0:n
pmf <- choose(n, k) * p^k * (1 - p)^(n - k)

data.frame(k, formula = pmf, R = dbinom(k, size = n, prob = p))
sum(pmf)
```

For five fair coin tosses, the probability of exactly three heads is

$$
P(X=3)=\binom53(0.5)^3(0.5)^2=0.3125.
$$

```{r five-toss-example}
choose(5, 3) * 0.5^3 * 0.5^2
dbinom(3, size = 5, prob = 0.5)
```

## Cumulative probabilities and quantiles

The binomial CDF adds the masses up to $k$:

$$
F(k)=P(X\leq k)=\sum_{j=0}^{k}\binom njp^j(1-p)^{n-j}.
$$

```{r binomial-cdf-and-quantile}
# At most 3 heads in 5 fair tosses
pbinom(3, size = 5, prob = 0.5)

# Smallest k for which P(X <= k) is at least 0.90
qbinom(0.90, size = 5, prob = 0.5)
```

## Mean and variance

Because a binomial variable is a sum of independent Bernoulli variables,

$$
E(X)=np,\qquad
\operatorname{Var}(X)=np(1-p),\qquad
\operatorname{SD}(X)=\sqrt{np(1-p)}.
$$

In the deck's credit-card example, 10 customers independently pay by credit
card with probability $0.75$. Thus

$$
E(X)=7.5,\qquad \operatorname{Var}(X)=1.875,
\qquad \operatorname{SD}(X)\approx1.37.
$$

The integer counts within one standard deviation of the mean are 7 and 8.

```{r credit-card-example}
n <- 10
p <- 0.75
mu <- n * p
variance <- n * p * (1 - p)
sigma <- sqrt(variance)
within_one_sd <- (0:n)[abs((0:n) - mu) <= sigma]

c(mean = mu, variance = variance, sd = sigma)
within_one_sd
sum(dbinom(within_one_sd, size = n, prob = p))
```

## Simulation check (fictional extension)

```{r binomial-simulation}
set.seed(14270)
simulated_counts <- rbinom(100000, size = 10, prob = 0.75)

c(
  simulated_mean = mean(simulated_counts),
  theoretical_mean = 10 * 0.75,
  simulated_variance = var(simulated_counts),
  theoretical_variance = 10 * 0.75 * 0.25
)
```

## Practice questions

1. For $X\sim\operatorname{Bin}(12,0.30)$, find $P(X=4)$.
2. For the same variable, find $P(X\leq4)$ and $P(X\geq1)$.
3. Find its mean and standard deviation.
4. Which binomial assumption fails if later trials become easier after every
   success?

::: {.callout-note collapse="true"}
### Check your answers

1. `dbinom(4, 12, 0.30)` gives approximately $0.2311$.
2. `pbinom(4, 12, 0.30)` gives approximately $0.7237$, and
   $P(X\geq1)=1-P(X=0)\approx0.9862$.
3. $E(X)=3.6$ and $\operatorname{SD}(X)=\sqrt{2.52}\approx1.587$.
4. The success probability is no longer constant; independence may also be
   questionable.
:::