Bernoulli and Binomial Distributions: The Building Blocks of Probability

---

1. Introduction

Probability isn’t just about “chance”; it’s about patterns in outcomes.
To study these patterns, we use probability distributions — mathematical functions that describe how likely each outcome is.
Two of the simplest and most important distributions are the Bernoulli and Binomial distributions.

---

2. Bernoulli Distribution

A Bernoulli trial is a single experiment with exactly two possible outcomes:

• Success (usually coded as 1)
• Failure (usually coded as 0)

If the probability of success is \( p \), then:

\[ P(X = 1) = p, \quad P(X = 0) = 1 – p \]

Example: Tossing a fair coin:

\[ P(\text{Head}) = 0.5, \quad P(\text{Tail}) = 0.5 \]

Mean (Expected Value):

\[ \mu = p \]

Variance:

\[ \sigma^2 = p(1-p) \]

---

3. Binomial Distribution

The Binomial distribution models the number of successes in \( n \) independent Bernoulli trials.

If \( X \) is the number of successes, then:

\[ P(X = k) = \frac{n!}{k!(n-k)!} \, p^k (1-p)^{n-k} \]

where:

. \( n \) = number of trials
. \( k \) = number of successes
– \( p \) = probability of success in each trial

Example:

If you toss a fair coin 5 times, the probability of getting exactly 3 heads is:
\[ P(X = 3) = \frac{5!}{3!(5-3)!} (0.5)^3 (0.5)^2 = 10 \times 0.125 \times 0.25 = 0.3125 \]

Mean:

\[ \mu = np \]

Variance:

\[ \sigma^2 = np(1-p) \]

---

4. Mean & Variance Formula Sheet

---

5. Real-Life Applications

1. Sports – Predicting the number of goals a team might score.
2. Quality Control – Counting defective items in a batch.
3. Medical Studies – Estimating the number of patients who respond to treatment.

---

6. Quick Quiz

Q1. A coin has a probability \( p = 0.7 \) of landing heads. What is the mean and variance of one toss (Bernoulli)?

Q2. A die is rolled 6 times. What is the probability of getting exactly two “4”s?

---

Answers:

A1:

Mean: \( \mu = 0.7 \)
Variance: \( \sigma^2 = 0.7(0.3) = 0.21 \)

A2:

Here \( n = 6, k = 2, p = \frac{1}{6} \):
\[ P(X = 2) = \frac{6!}{2!(6-2)!} \left(\frac{1}{6}\right)^2 \left(\frac{5}{6}\right)^4 \]