FAD1015 L13 — Binomial Distribution

Lecture 13 of FAD1015 Mathematics III covering the binomial distribution as a discrete probability distribution. Source: (L13) FAD 1015 - Week 7_1 Binomial - std.pdf (34 slides).

Learning Objectives

Identify the binomial setting/characteristics
Apply the binomial probability formula
Use the binomial distribution table (cumulative probabilities)
Calculate the mean and standard deviation of a binomial distribution

1. The Binomial Setting

The binomial distribution applies to probability problems that have only two outcomes or can be reduced to two outcomes.

Natural two-outcome situations:

Baby born: male or female
Coin tossed: head or tail
Basketball game: win or lose

Situations reducible to two outcomes:

Medical treatment: effective or ineffective
Blood pressure: normal or abnormal
Multiple choice question: correct or incorrect

2. Four Characteristics of a Binomial Experiment

A binomial experiment must satisfy all four conditions:

Fixed number of trials — $n$ identical trials
Two possible outcomes — each trial is either success or failure
- $P(\text{success}) = p$
- $P(\text{failure}) = 1 - p = q$
Independent trials
Constant probability — the probability of success/failure remains the same for every trial

The binomial random variable $X$ is defined as: $$X = \text{number of successes observed when an experiment or trial is performed}$$

The probability distribution of $X$ is called the binomial probability distribution.

Checking for Binomial Distribution — Examples

Scenario	Fixed $n$	2 Outcomes	Independent	Constant $p$	Binomial?
Throwing dart till bullseye	No	Yes	Yes	Yes	No
Playing two football matches	Yes	No (win/draw/lose)	Yes	Yes	No
Playing two tennis games	Yes	Yes	No	Yes	No
Throw dice + toss coin for 6 and head	Yes	Yes	Yes	No	No
Throwing dice three times to get a 6	Yes	Yes	Yes	Yes	Yes

graph TD
    Q1["Fixed number of trials n?"] -->|No| NOT1["NOT Binomial"]
    Q1 -->|Yes| Q2["Exactly two outcomes<br/>per trial?"]
    Q2 -->|No| NOT2["NOT Binomial"]
    Q2 -->|Yes| Q3["Trials are independent?"]
    Q3 -->|No| NOT3["NOT Binomial"]
    Q3 -->|Yes| Q4["Constant probability p<br/>for every trial?"]
    Q4 -->|No| NOT4["NOT Binomial"]
    Q4 -->|Yes| YES["Binomial Distribution<br/>X ~ B(n,p)"]

    style Q1 fill:#e7f5ff,stroke:#1971c2
    style Q2 fill:#e7f5ff,stroke:#1971c2
    style Q3 fill:#e7f5ff,stroke:#1971c2
    style Q4 fill:#e7f5ff,stroke:#1971c2
    style NOT1 fill:#ffe3e3,stroke:#c92a2a
    style NOT2 fill:#ffe3e3,stroke:#c92a2a
    style NOT3 fill:#ffe3e3,stroke:#c92a2a
    style NOT4 fill:#ffe3e3,stroke:#c92a2a
    style YES fill:#d3f9d8,stroke:#2f9e44

3. The Binomial Probability Formula

If $X \sim \text{Bin}(n, p)$, then:

$$P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} = \frac{n!}{x!(n-x)!} \times p^x \times q^{n-x}$$

Where:

$n$ = total number of trials
$p$ = probability of success
$q = 1-p$ = probability of failure
$x$ = number of successes in $n$ trials
$n-x$ = number of failures in $n$ trials

Recall: $$^nC_r = \binom{n}{r} = \frac{n!}{r!(n-r)!}$$

4. Worked Examples — Using the Formula

graph LR
    subgraph identify["Step 1: Identify Parameters"]
        I1["n = number of trials"]
        I2["p = probability of success"]
        I3["x = desired successes"]
    end
    subgraph formula["Step 2: Apply Formula"]
        F1["P(X=x) = C(n,x) * p^x * q^(n-x)"]
    end
    subgraph interpret["Step 3: Interpret"]
        INT["State probability in context"]
    end

    I1 --> F1
    I2 --> F1
    I3 --> F1
    F1 --> INT

    style identify fill:#e7f5ff,stroke:#1971c2
    style formula fill:#ffe8cc,stroke:#d9480f
    style interpret fill:#d3f9d8,stroke:#2f9e44

Example 1: Exactly One Defective

Problem: Ten percent of all iPhones manufactured by Apple are defective. A quality control inspector randomly selects five iPhones from the production line. Is this a binomial experiment? What is the probability that exactly one of these iPhones is defective?

Solution:

Let $X$ = number of defects in 5 trials
$X \sim B(5, 0.1)$
Verify conditions: fixed $n=5$ ✓, 2 outcomes ✓, independent ✓, constant $p=0.1$ ✓
$P(X = 1) = \binom{5}{1}(0.1)^1(0.9)^4 = 0.32805$

Example 2: Multiple Probability Questions (Same Scenario)

Using $X \sim B(5, 0.1)$:

Exactly three defective: $$P(X = 3) = \binom{5}{3}(0.1)^3(0.9)^2 = 0.0081$$
More than one defective: $$P(X > 1) = P(X=2) + P(X=3) + P(X=4) + P(X=5) = 0.08146$$
Less than two defective: $$P(X < 2) = P(X=1) + P(X=0) = 0.9185$$
At least one defective: $$P(X \geq 1) = 0.40951$$
At most two defective: $$P(X \leq 2) = P(X=2) + P(X=1) + P(X=0) = 0.99144$$

5. The Binomial Distribution Table

The provided binomial distribution table gives cumulative probabilities of the form:

$$P(X \geq r) = P(X=r) + P(X=r+1) + \dots + P(X=n)$$

The table lists $P(X \geq r)$ for common values of $n$ (number of trials) and $p$ (probability of success), with $r$ = number of successes.

Important: The table shows cumulative probabilities $P(X \geq r)$, not individual probabilities $P(X = r)$.

Example 3: Using the Binomial Table ($p \leq 0.5$)

Problem: Same iPhone scenario ($n=5, p=0.1$). Use the table to find:

$P(\text{exactly one})$ → read $P(X \geq 1)$ from table
$P(\text{exactly three})$ → $P(X \geq 3) - P(X \geq 4)$
$P(\text{more than one})$ → $P(X \geq 2)$
$P(\text{less than two})$ → $1 - P(X \geq 2)$
$P(\text{at least one})$ → $P(X \geq 1)$
$P(\text{at most two})$ → $1 - P(X \geq 3)$

Example 5: Using the Binomial Table When $p > 0.5$

Problem: Eighty percent of iPhones are defective. Inspector selects ten iPhones ($n=10, p=0.8$).

Trick — Flip success and failure:

Success ($p=0.8$) $X$	Failure ($p=0.2$) $Y$
0	10
1	9
2	8
3	7
4	6
5	5
6	4
7	3
8	2
9	1
10	0

If $X \sim B(10, 0.8)$, then $Y \sim B(10, 0.2)$ where $Y = 10 - X$.

Questions:

At most 3 defective → $P(X \leq 3) = P(Y \geq 7)$
More than 5 defective → $P(X > 5) = P(Y < 5) = 1 - P(Y \geq 5)$
Less than 3 defective → $P(X < 3) = P(Y > 7) = P(Y \geq 8)$
At least 7 defective → $P(X \geq 7) = P(Y \leq 3) = 1 - P(Y \geq 4)$
More than 1 but less than 5 → $P(1 < X < 5) = P(5 < Y < 9) = P(Y \geq 6) - P(Y \geq 9)$
Exactly 6 defective → $P(X = 6) = P(Y = 4) = P(Y \geq 4) - P(Y \geq 5)$

6. Reading the Table — Key Rules

From the NOTES slide:

Probability Wanted	How to Read from Table
$P(X \geq x)$	Read straight from table
$P(X \leq x)$	$1 - P(X \geq x+1)$
$P(X < x)$	$1 - P(X \geq x)$
$P(X = x)$	$P(X \geq x) - P(X \geq x+1)$
$P(X > x)$	$P(X \geq x+1)$

7. Mean and Standard Deviation

For $X \sim \text{Bin}(n, p)$:

Mean: $\mu = np$
Variance: $\sigma^2 = npq$
Standard Deviation: $\sigma = \sqrt{npq}$

Example 6: Mean, Variance and Standard Deviation

Problem: An egg supplier sends a daily shipment of 500 eggs. Previous records show 4% arrive damaged. Let $X$ be the number of damaged eggs. Find the mean and standard deviation.