Learning Objectives
By the end of this chapter, you will be able to:
- Identify situations where binomial distribution applies
- Calculate binomial probabilities using the formula
- Find mean and variance of binomial distribution
- Apply binomial distribution to real-world problems
- Use binomial tables and understand cumulative probabilities
What is a Probability Distribution?
A probability distribution describes all possible values a random variable can take and their associated probabilities.
flowchart TD
A[Probability Distributions] --> B[Discrete]
A --> C[Continuous]
B --> D[Binomial]
B --> E[Poisson]
C --> F[Normal]
C --> G[t-distribution]
For this chapter, we focus on the Binomial Distribution - the most important discrete distribution.
Binomial Experiment
A binomial experiment has these characteristics:
The 4 Conditions (BINS)
flowchart TD
A[Binomial Experiment] --> B["B: Binary outcomes<br/>(Success or Failure)"]
A --> C["I: Independent trials"]
A --> D["N: Fixed number of trials"]
A --> E["S: Same probability<br/>for each trial"]
| Condition | Meaning | Example |
|---|---|---|
| Binary | Only two outcomes (Success/Failure) | Pass/Fail, Yes/No, Defective/Good |
| Independent | Each trial’s outcome doesn’t affect others | Coin flips, random selections |
| Fixed N | Number of trials is predetermined | Testing exactly 10 items |
| Same P | Success probability is constant | Each item has 5% defect rate |
Binomial Distribution Notation
If X follows a binomial distribution:
\[X \sim B(n, p)\]Where:
- $n$ = number of trials
- $p$ = probability of success on each trial
- $q = 1 - p$ = probability of failure
Binomial Probability Formula
The probability of getting exactly x successes in n trials:
\[P(X = x) = C(n, x) \times p^x \times q^{n-x}\]Or written as:
\[P(X = x) = \binom{n}{x} p^x (1-p)^{n-x}\]Where:
- $\binom{n}{x} = \frac{n!}{x!(n-x)!}$ is the number of ways to choose x successes from n trials
- $p^x$ is the probability of x successes
- $(1-p)^{n-x}$ is the probability of (n-x) failures
Step-by-Step Example 1: Basic Binomial
Problem: A fair coin is tossed 5 times. Find the probability of getting exactly 3 heads.
Solution:
Step 1: Verify binomial conditions
- Binary: Head or Tail ✓
- Independent: Each toss is independent ✓
- Fixed n: n = 5 tosses ✓
- Same p: p = 0.5 for each toss ✓
Step 2: Identify parameters
- n = 5 (number of tosses)
- x = 3 (successes wanted)
- p = 0.5 (probability of head)
- q = 0.5 (probability of tail)
Step 3: Calculate combination \(C(5, 3) = \frac{5!}{3! \times 2!} = \frac{5 \times 4}{2 \times 1} = 10\)
Step 4: Apply formula \(P(X = 3) = C(5,3) \times (0.5)^3 \times (0.5)^2\) \(= 10 \times 0.125 \times 0.25\) \(= 10 \times 0.03125 = 0.3125\)
Answer: P(3 heads) = 0.3125 or 31.25%
Step-by-Step Example 2: Quality Control
Problem: A factory produces items with a 10% defective rate. If 6 items are randomly selected, find the probability that: (a) Exactly 2 are defective (b) At most 1 is defective (c) At least 1 is defective
Solution:
Parameters: n = 6, p = 0.10, q = 0.90
(a) Exactly 2 defective:
\[P(X = 2) = C(6,2) \times (0.10)^2 \times (0.90)^4\] \[C(6,2) = \frac{6!}{2! \times 4!} = \frac{6 \times 5}{2} = 15\]\(P(X = 2) = 15 \times 0.01 \times 0.6561\) \(= 15 \times 0.006561 = 0.0984\)
(b) At most 1 defective: P(X ≤ 1) = P(X=0) + P(X=1)
\(P(X = 0) = C(6,0) \times (0.10)^0 \times (0.90)^6\) \(= 1 \times 1 \times 0.531441 = 0.5314\)
\(P(X = 1) = C(6,1) \times (0.10)^1 \times (0.90)^5\) \(= 6 \times 0.10 \times 0.59049 = 0.3543\)
\[P(X \leq 1) = 0.5314 + 0.3543 = 0.8857\](c) At least 1 defective:
\(P(X \geq 1) = 1 - P(X = 0)\) \(= 1 - 0.5314 = 0.4686\)
Step-by-Step Example 3: Survey Results
Problem: In a survey, 40% of citizens support a new policy. If 8 citizens are randomly selected, find: (a) P(exactly 4 support) (b) P(majority supports, i.e., more than 4)
Solution:
Parameters: n = 8, p = 0.40, q = 0.60
(a) Exactly 4 support:
\[C(8,4) = \frac{8!}{4! \times 4!} = \frac{8 \times 7 \times 6 \times 5}{24} = 70\]\(P(X = 4) = 70 \times (0.40)^4 \times (0.60)^4\) \(= 70 \times 0.0256 \times 0.1296\) \(= 70 \times 0.003318 = 0.2322\)
(b) Majority supports (X > 4):
\[P(X > 4) = P(X=5) + P(X=6) + P(X=7) + P(X=8)\]Let me calculate each:
\[P(X=5) = C(8,5)(0.4)^5(0.6)^3 = 56 \times 0.01024 \times 0.216 = 0.1239\] \[P(X=6) = C(8,6)(0.4)^6(0.6)^2 = 28 \times 0.004096 \times 0.36 = 0.0413\] \[P(X=7) = C(8,7)(0.4)^7(0.6)^1 = 8 \times 0.001638 \times 0.6 = 0.0079\] \[P(X=8) = C(8,8)(0.4)^8(0.6)^0 = 1 \times 0.000655 \times 1 = 0.0007\] \[P(X > 4) = 0.1239 + 0.0413 + 0.0079 + 0.0007 = 0.1738\]Answer: About 17.38% chance of majority support
Mean and Variance of Binomial Distribution
Mean (Expected Value)
\[\mu = E(X) = np\]Variance
\[\sigma^2 = npq = np(1-p)\]Standard Deviation
\[\sigma = \sqrt{npq}\]Example 4: Mean and Variance
Problem: In a town, 30% of households have solar panels. For a sample of 50 households, find the expected number with solar panels and the standard deviation.
Solution:
n = 50, p = 0.30, q = 0.70
Mean: \(\mu = np = 50 \times 0.30 = 15\)
Variance: \(\sigma^2 = npq = 50 \times 0.30 \times 0.70 = 10.5\)
Standard Deviation: \(\sigma = \sqrt{10.5} = 3.24\)
Interpretation: On average, 15 households have solar panels, with a standard deviation of about 3.24.
Binomial Probability Table
For common values, you can use binomial tables. Here’s a partial table for n = 5:
| x | p=0.1 | p=0.2 | p=0.3 | p=0.4 | p=0.5 |
|---|---|---|---|---|---|
| 0 | 0.5905 | 0.3277 | 0.1681 | 0.0778 | 0.0313 |
| 1 | 0.3281 | 0.4096 | 0.3602 | 0.2592 | 0.1563 |
| 2 | 0.0729 | 0.2048 | 0.3087 | 0.3456 | 0.3125 |
| 3 | 0.0081 | 0.0512 | 0.1323 | 0.2304 | 0.3125 |
| 4 | 0.0005 | 0.0064 | 0.0284 | 0.0768 | 0.1563 |
| 5 | 0.0000 | 0.0003 | 0.0024 | 0.0102 | 0.0313 |
Using the Table
Example: For n = 5, p = 0.3, find P(X = 2)
From table: P(X = 2) = 0.3087
Step-by-Step Example 5: Exam-Style Problem
Problem: An examination has 10 multiple-choice questions, each with 4 options. A student guesses all answers randomly. Find: (a) The probability of getting exactly 5 correct (b) The probability of passing (at least 4 correct) (c) The expected number of correct answers
Solution:
Parameters:
- n = 10 questions
- p = 1/4 = 0.25 (probability of guessing correctly)
- q = 0.75
(a) Exactly 5 correct:
\[C(10,5) = \frac{10!}{5! \times 5!} = 252\]\(P(X = 5) = 252 \times (0.25)^5 \times (0.75)^5\) \(= 252 \times 0.000977 \times 0.2373\) \(= 252 \times 0.000232 = 0.0584\)
(b) Passing (X ≥ 4):
This requires calculating P(X=4) + P(X=5) + … + P(X=10)
Or easier: Use complement if X ≥ 4 is passing.
P(X ≥ 4) = 1 - P(X ≤ 3) = 1 - [P(0) + P(1) + P(2) + P(3)]
\(P(X=0) = (0.75)^{10} = 0.0563\) \(P(X=1) = 10(0.25)(0.75)^9 = 0.1877\) \(P(X=2) = 45(0.25)^2(0.75)^8 = 0.2816\) \(P(X=3) = 120(0.25)^3(0.75)^7 = 0.2503\)
\[P(X \leq 3) = 0.0563 + 0.1877 + 0.2816 + 0.2503 = 0.7759\] \[P(X \geq 4) = 1 - 0.7759 = 0.2241\](c) Expected correct answers:
\[E(X) = np = 10 \times 0.25 = 2.5\]Answer: By random guessing, the student expects to get only 2.5 correct on average, with about 22.4% chance of passing.
Shape of Binomial Distribution
flowchart TD
A[Shape depends on p] --> B["p < 0.5: Right-skewed"]
A --> C["p = 0.5: Symmetric"]
A --> D["p > 0.5: Left-skewed"]
E["As n increases, becomes more symmetric<br/>(approaches normal distribution)"]
When p = 0.5
- Distribution is perfectly symmetric
- Mean is at the center
When p < 0.5
- Distribution is skewed right
- More probability on lower values
When p > 0.5
- Distribution is skewed left
- More probability on higher values
Real-World Applications in Public Administration
| Scenario | n | p | X = |
|---|---|---|---|
| Voter turnout | 100 voters | 0.65 | Number who vote |
| Project approval | 15 proposals | 0.40 | Projects approved |
| Audit findings | 20 files | 0.08 | Files with errors |
| Training completion | 50 employees | 0.85 | Employees who complete |
| Service satisfaction | 30 customers | 0.72 | Satisfied customers |
Common Probability Questions and Formulas
| Question Type | Formula |
|---|---|
| Exactly x successes | $P(X = x)$ |
| At most x successes | $P(X \leq x) = \sum_{i=0}^{x} P(X=i)$ |
| At least x successes | $P(X \geq x) = 1 - P(X < x)$ |
| More than x successes | $P(X > x) = 1 - P(X \leq x)$ |
| Less than x successes | $P(X < x) = P(X \leq x-1)$ |
Practice Problems
Problem 1
A fair die is rolled 6 times. Find the probability of getting: (a) Exactly two 6s (b) No 6s (c) At least one 6
Problem 2
In a district, 25% of households are below poverty line. From a random sample of 12 households, find: (a) P(exactly 3 are below poverty line) (b) P(at most 2 are below poverty line) (c) Expected number below poverty line
Problem 3
A public service office handles complaints with 80% resolution rate. If 10 complaints are filed: (a) What is the probability all are resolved? (b) What is the probability at least 8 are resolved? (c) What is the variance in the number resolved?
Problem 4
True or False: If X ~ B(20, 0.5), then P(X = 10) > P(X = 15). Justify your answer.
Problem 5
An employee passes a test with probability 0.7. Find the minimum number of attempts needed for the probability of at least one pass to exceed 0.95.
Summary
| Concept | Formula |
|---|---|
| Binomial Probability | $P(X=x) = \binom{n}{x}p^x(1-p)^{n-x}$ |
| Mean | $\mu = np$ |
| Variance | $\sigma^2 = np(1-p)$ |
| Standard Deviation | $\sigma = \sqrt{np(1-p)}$ |
| Conditions | Binary, Independent, Fixed n, Same p |
Next Topic
In the next chapter, we will study the Normal Distribution - the most important continuous probability distribution, which forms the foundation for many statistical inference techniques.
