Learning Objectives
By the end of this chapter, you will be able to:
- Define basic probability terminology
- Construct sample spaces for experiments
- Identify different types of events
- Calculate combinations and permutations
- Apply counting principles to real-world scenarios
Why Study Probability?
Probability is the mathematical foundation for inferential statistics. Understanding probability allows us to:
flowchart LR
A[Sample Data] --> B[Probability<br/>Theory]
B --> C[Inferences about<br/>Population]
D["Examples:<br/>- Election predictions<br/>- Budget forecasting<br/>- Risk assessment"]
Applications in Public Administration
- Election forecasting: What’s the probability a candidate wins?
- Budget planning: How likely are cost overruns?
- Policy evaluation: What’s the chance of program success?
- Risk management: What’s the probability of a disaster?
Basic Terminology
1. Random Experiment
A random experiment is a process whose outcome cannot be predicted with certainty.
Examples:
- Tossing a coin
- Rolling a die
- Selecting a random citizen for survey
- Conducting an election
Characteristics:
- Can be repeated under identical conditions
- All possible outcomes are known
- Actual outcome is unknown beforehand
2. Sample Space (S)
The sample space is the set of all possible outcomes of a random experiment.
Examples:
| Experiment | Sample Space (S) | Size |
|---|---|---|
| Coin toss | {Head, Tail} or {H, T} | 2 |
| Die roll | {1, 2, 3, 4, 5, 6} | 6 |
| Two coin tosses | {HH, HT, TH, TT} | 4 |
| Gender of 2 children | {BB, BG, GB, GG} | 4 |
flowchart TD
A[Two Coin Tosses] --> B["First Coin"]
B --> C["H"]
B --> D["T"]
C --> E["HH"]
C --> F["HT"]
D --> G["TH"]
D --> H["TT"]
I["Sample Space S = {HH, HT, TH, TT}"]
3. Event
An event is any subset of the sample space. It’s an outcome or collection of outcomes we’re interested in.
Examples (Rolling a die, S = {1, 2, 3, 4, 5, 6}):
| Event | Description | Outcomes |
|---|---|---|
| A | Getting an even number | {2, 4, 6} |
| B | Getting a number less than 3 | {1, 2} |
| C | Getting a 5 | {5} |
| D | Getting a number less than 7 | {1, 2, 3, 4, 5, 6} |
Types of Events
flowchart TD
A[Types of Events] --> B[Simple Event]
A --> C[Compound Event]
A --> D[Mutually Exclusive]
A --> E[Exhaustive Events]
A --> F[Complementary Events]
B --> B1["Single outcome<br/>e.g., Getting 3"]
C --> C1["Multiple outcomes<br/>e.g., Getting even number"]
D --> D1["Cannot occur together<br/>e.g., Even and Odd"]
E --> E1["Cover all possibilities"]
F --> F1["Event and its opposite<br/>A and A'"]
1. Simple (Elementary) Event
An event with only one outcome.
Example: Getting exactly 4 when rolling a die: {4}
2. Compound Event
An event with more than one outcome.
Example: Getting an even number when rolling a die: {2, 4, 6}
3. Mutually Exclusive Events
Events that cannot occur simultaneously.
\[P(A \cap B) = 0\]Example:
- Event A: Getting an even number {2, 4, 6}
- Event B: Getting an odd number {1, 3, 5}
- A and B cannot both occur on the same roll
4. Exhaustive Events
Events that together cover the entire sample space.
Example: Even numbers {2, 4, 6} and Odd numbers {1, 3, 5} are exhaustive - they cover all outcomes.
5. Complementary Events
Event A’ (A complement) contains all outcomes NOT in A.
\[P(A) + P(A') = 1\]Example: If A = {getting a 6}, then A’ = {not getting a 6} = {1, 2, 3, 4, 5}
Set Operations in Probability
Union (A ∪ B)
Outcomes in A OR B (or both).
\[A \cup B = \{x : x \in A \text{ or } x \in B\}\]Intersection (A ∩ B)
Outcomes in A AND B (both).
\[A \cap B = \{x : x \in A \text{ and } x \in B\}\]Complement (A’)
Outcomes NOT in A.
\[A' = \{x : x \notin A\}\]Example
Rolling a die:
- S = {1, 2, 3, 4, 5, 6}
- A = Even numbers = {2, 4, 6}
- B = Numbers > 3 = {4, 5, 6}
Find:
- $A \cup B$ = {2, 4, 5, 6}
- $A \cap B$ = {4, 6}
- $A’$ = {1, 3, 5}
Counting Principles
To calculate probabilities, we often need to count outcomes. Two key tools:
Fundamental Counting Principle
If task 1 can be done in $m$ ways and task 2 in $n$ ways, then both tasks together can be done in $m \times n$ ways.
Example: A committee needs to select a president (5 candidates) and secretary (4 candidates).
Total ways = $5 \times 4 = 20$ ways
Permutations
A permutation is an arrangement where order matters.
Formula
\[P(n, r) = \frac{n!}{(n-r)!}\]Where:
- $n$ = total items
- $r$ = items being arranged
- $n!$ = n factorial = $n \times (n-1) \times (n-2) \times \cdots \times 1$
Special Cases
- $0! = 1$
- $P(n, n) = n!$ (arranging all n items)
Step-by-Step Example 1
Problem: In how many ways can 3 officers be arranged in a row from 5 candidates?
Solution:
\[P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!}\] \[= \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}\] \[= \frac{120}{2} = 60 \text{ ways}\]Step-by-Step Example 2
Problem: How many different 4-digit passwords can be formed using digits 1, 2, 3, 4, 5 without repetition?
Solution:
\[P(5, 4) = \frac{5!}{(5-4)!} = \frac{5!}{1!}\] \[= 5 \times 4 \times 3 \times 2 = 120 \text{ passwords}\]Combinations
A combination is a selection where order does NOT matter.
Formula
\[C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}\]Also written as: $_nC_r$ or $\binom{n}{r}$
When to Use Permutation vs Combination
| Scenario | Order Matters? | Use |
|---|---|---|
| Arranging books on shelf | Yes | Permutation |
| Selecting committee members | No | Combination |
| Creating passwords | Yes | Permutation |
| Choosing lottery numbers | No | Combination |
| Ranking candidates | Yes | Permutation |
| Forming a team | No | Combination |
flowchart TD
A[Selection Problem] --> B{Does order<br/>matter?}
B -->|Yes| C[Use Permutation<br/>P(n,r)]
B -->|No| D[Use Combination<br/>C(n,r)]
C --> E["Example: Arrange 3 from 5<br/>P(5,3) = 60"]
D --> F["Example: Choose 3 from 5<br/>C(5,3) = 10"]
Step-by-Step Example 3
Problem: A committee of 3 members is to be formed from 7 government officers. How many different committees are possible?
Solution:
Since order doesn’t matter (selecting A, B, C is same as B, C, A), use combination:
\[C(7, 3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3! \times 4!}\] \[= \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!}\] \[= \frac{7 \times 6 \times 5}{6} = 35 \text{ committees}\]Step-by-Step Example 4
Problem: From 10 candidates, how many ways can we select a panel of 4?
Solution:
\[C(10, 4) = \frac{10!}{4! \times 6!}\] \[= \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}\] \[= \frac{5040}{24} = 210 \text{ ways}\]Special Combination Formulas
Property 1: Symmetry
\[C(n, r) = C(n, n-r)\]Example: $C(10, 7) = C(10, 3) = 120$
Property 2: Boundary Values
\[C(n, 0) = C(n, n) = 1\]Property 3: Pascal’s Identity
\[C(n, r) = C(n-1, r-1) + C(n-1, r)\]Exam-Style Example 5
Problem: A public administration department has 8 male and 6 female officers. A committee of 5 is to be formed consisting of 3 males and 2 females. In how many ways can this be done?
Solution:
Step 1: Selecting 3 males from 8
\[C(8, 3) = \frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6}{6} = 56\]Step 2: Selecting 2 females from 6
\[C(6, 2) = \frac{6!}{2! \times 4!} = \frac{6 \times 5}{2} = 15\]Step 3: Total committees (multiply selections)
\[\text{Total} = 56 \times 15 = 840 \text{ ways}\]Factorial Quick Reference
| n | n! |
|---|---|
| 0 | 1 |
| 1 | 1 |
| 2 | 2 |
| 3 | 6 |
| 4 | 24 |
| 5 | 120 |
| 6 | 720 |
| 7 | 5,040 |
| 8 | 40,320 |
| 9 | 362,880 |
| 10 | 3,628,800 |
Practice Problems
Problem 1
In how many ways can 5 students be arranged in a row for a photograph?
Problem 2
A ballot paper has 8 candidates. In how many ways can a voter rank their top 3 choices?
Problem 3
From 12 district offices, 4 are to be selected for an audit. How many different selections are possible?
Problem 4
A committee of 6 is to be formed from 5 economists and 4 administrators, with at least 2 from each group. How many ways can this be done?
Problem 5
Define with examples: (a) Sample space (b) Mutually exclusive events (c) Complementary events
Summary
| Concept | Definition/Formula |
|---|---|
| Sample Space (S) | Set of all possible outcomes |
| Event | Subset of sample space |
| Mutually Exclusive | Events that cannot occur together |
| Complement | $P(A) + P(A’) = 1$ |
| Permutation | $P(n,r) = \frac{n!}{(n-r)!}$ (order matters) |
| Combination | $C(n,r) = \frac{n!}{r!(n-r)!}$ (order doesn’t matter) |
Next Topic
In the next chapter, we will study Approaches to Probability - how to assign numerical probabilities to events using classical, relative frequency, and subjective methods.
