Basic Concepts

B A S I C C O N C E P T S

Chapter 1 of the “Stats 2 Book” establishes the fundamental vocabulary and axiomatic framework necessary for studying probability and statistics. This chapter introduces the structure needed to discuss the likelihood of occurrences.

Here is a detailed explanation of the key concepts, followed by illustrative examples and exercises based on the source material:

1.1 Definitions and Properties

The foundation of probability theory rests on defining what is possible before determining what is likely.

Key Definitions

Concept	Definition	Example
Sample Space ( $S$ )	A set listing all possibilities (outcomes) that might occur.	When rolling a six-sided die, $S = \{1, 2, 3, 4, 5, 6\}$ .
Outcome	An element of the sample space $S$ .	The result of rolling a die, e.g., ‘4’.
Experiment	The process of actually selecting one of the outcomes listed.	Flipping a coin or waiting for the winner of the World Cup.
Event ( $E$ )	Any subset of the sample space $S$ .	Rolling a number > 2: $E = \{3, 4, 5, 6\}$ .

Probability Axioms

A probability ( $P$ ) is a function that assigns a chance (a number between 0 and 1) to each event $E$ . This formally relies on Kolmogorov’s axioms.

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Axiom 1: Certainty

The probability of the entire sample space is 1. $P(S) = 1$ Interpretation: There is a 100% chance that an experiment will result in some outcome included in $S$ .

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Axiom 2: Additivity

For any countable collection of disjoint events ( $E_1, E_2, \dots$ ), their combined probability is the sum of their individual probabilities. $P (E_1 \cup E_2 \cup \dots) = P(E_1) + P(E_2) + \dots$ Interpretation: If events cannot happen simultaneously, their probabilities add up.

Basic Properties

From the two fundamental axioms, several properties can be proven that simplify probability calculations:

Property	Formula	Description
Empty Set	$P(\emptyset) = 0$	The probability of nothing happening is zero.
Finite Additivity	$P(\cup E_i) = \sum P(E_i)$	Sum rule for finite disjoint events.
Monotonicity	If $E \subset F, P(E) \le P(F)$	Subsets cannot be more likely than supersets.
Difference Rule	$P(F \setminus E) = P(F) - P(E)$	Prob. of $F$ occurring but $E$ not (if $E \subset F$ ).
Complement Rule	$P(E^c) = 1 - P(E)$	Prob. that $E$ does NOT occur.
General Addition	$P(E \cup F) = P(E) + P(F) - P(E \cap F)$	Subtract intersection to avoid double counting.

Example and Solution

Coin Flip Axioms

A fair coin flip has a sample space $S = \{\text{heads, tails}\}$ . Use the axioms to show that the probability of observing heads is 0.5.

📝 View Detailed Solution ▼

Let $E = \{\text{heads}\}$ and $F = \{\text{tails}\}$ be two disjoint events.
Since the coin is “fair,” $P(E) = P(F) = p$ for some value $p$ .
The union is the sample space: $S = E \cup F$ .
Using Axiom 1 and Axiom 2: $1 = P(S) = P(E \cup F) = P(E) + P(F)$
Substituting $p$ : $1 = 2p \implies p = 0.5$ .

Fishing Tonnage

A town’s fishing fleet has a 35% chance of catching over 400 tons ( $P(F)=0.35$ ) and a 10% chance of catching over 500 tons ( $P(E)=0.10$ ). How likely is it that they will catch between 400 and 500 tons ?

📝 View Detailed Solution ▼

Variable $E$ (“over 500”) is a subset of $F$ (“over 400”).
“Between 400 and 500” is the difference set $F \setminus E$ .
Using the Difference Rule: $P(F \setminus E) = P(F) - P(E)$

Answer: There is a 25% chance that between 400 and 500 tons of fish will be caught.

1.2 Equally Likely Outcomes

When outcomes are finite and equally likely, we have a Uniform Distribution.

Running probability calculations in this setting becomes a pure counting problem.

Rolling Two Dice

Two dice are rolled. How likely is it that their sum will equal eight?

📝 View Detailed Solution ▼

Total sample space $|S| = 6 \times 6 = 36$ .
Event $E$ (sum is 8): $E = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\}$ .
Count $|E| = 5$ .
Probability: $P(E) = \frac{5}{36}$

Group Selection

A group of 12 people includes Grant and Dilip. Pick 3 randomly. How likely is it to include Grant, but not Dilip?

📝 Combinatorial Solution ▼

Total Outcomes: Choose 3 from 12. $|S| = \binom{12}{3} = 220$ .
Event $E$ : Grant is fixed (1 way). Dilip is out. We need 2 more from the remaining 10.
Count $|E|$ : Choose 2 from 10. $|E| = \binom{10}{2} = 45$ .
Probability: $P(E) = \frac{45}{220} = \frac{9}{44}$

1.3 Conditional Probability

How is likelihood “altered” by knowledge that another event $B$ has occurred?

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Conditional Probability Formula

$P(A|B) = \frac{P(A \cap B)}{P(B)}$ Provided $P(B) > 0$

Bayes’ Theorem

One of the most powerful theorems in statistics allows us to reverse conditional probabilities.

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Bayes' Theorem

$P(B|A) = \frac{P(A|B)P(B)}{\sum P(A|B_j)P(B_j)}$ Updates belief $P(B)$ given new evidence $A$ .

The Swine Flu Test

Detects Flu 95% of the time if infected: $P(Pos|Flu) = 0.95$ .
False Positive rate is 2%: $P(Pos|Healthy) = 0.02$ .
Population rate is 1%: $P(Flu) = 0.01$ .

If a person tests positive, what is the probability they actually have the flu?

📝 Bayesian Check ▼

Let $A = \text{Flu}$ , $B = \text{Positive Test}$ .
We want $P(A|B)$ .
Apply Bayes’ Theorem: $P(A|B) = \frac{P(B|A)P(A)}{P(B|A)P(A) + P(B|A^c)P(A^c)}$ $P(A|B) = \frac{(0.95)(0.01)}{(0.95)(0.01) + (0.02)(0.99)}$ $P(A|B) \approx 0.324$

Result: Only a 32.4% chance they actually have the flu!

1.4 Independence

Events are independent if the occurrence of one has no effect on the other.

$P(A \cap B) = P(A)P(B)$

1.5 Using R

R is essential for complex calculations.

Function	Syntax	Purpose
Vector Creation	`c(1, 2, 3)`	Creates a list of numbers.
Sequence	`1:100`	Creates integers from 1 to 100.
Combinations	`choose(n, k)`	Calculates $\binom{n}{k}$ .
Summation	`sum(x)`	Adds all elements in vector $x$ .