Probability Notes: Simple Events, Notation, and Basic Rules

A die has six sides; the sample space for a single roll is S = {1,2,3,4,5,6}.
A simple event is a single outcome from the sample space (e.g., rolling a 4).
An event is any subset of the sample space (i.e., a collection of simple events).
Probability basics: when all simple events are equally likely, the probability of an event E is the sum of the probabilities of the simple events in E.
- If E contains k outcomes, then for a fair die with 6 sides: $P(E)=\frac{k}{6}$ .
The transcript contrasts an event with a simple event: an event can be a combination of several simple outcomes.
Example concept (qualifying a scenario): the probability of a specific combination when rolling a die multiple times or flipping a coin multiple times.
- If you flip a coin three times, the sample space has 2^3 = 8 equally likely sequences of H/T.
- The event "H on the first flip and T on the third flip" consists of the two sequences {H, H, T} and {H, T, T}, so $P( ext{H on 1st and T on 3rd})=\frac{2}{8}=\frac{1}{4}.$
Summary: simple event is one outcome; event is a set of outcomes; probabilities add over the outcomes in the event.

P(A) denotes the probability of event A.
Intersection: $P(A\cap B)$ is the probability that both A and B occur.
Union: $P(A\cup B)$ is the probability that at least one of A or B occurs.
Complement: $A^c$ (or sometimes \bar{A}) is the event that A does not occur; $P(A^c)=1-P(A)$ .
Conditional probability: $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$ , provided $P(B)>0$.
The vertical bar "|" denotes conditionality; the bar with a superscript c denotes complement; the dot in between events denotes "and".
Key takeaway on notation:
- A occurs with B? use $P(A\mid B)$ .
- Both A and B occur? use $P(A\cap B)$ .
- Either A or B (or both)? use $P(A\cup B)$ .

Axiom of total probability for a finite sample space: the probabilities of all simple events sum to 1.
- If S is the sample space for a finite experiment, then $\sum_{e\in S} P(e)=1.$
For a finite event E (a subset of S): $P(E)=\sum_{e\in E} P(e).$
If all simple outcomes are equally likely, and E contains k outcomes, then $P(E)=\dfrac{k}{|S|}$ .
Complement rule: if A is an event, then $P(A^c)=1-P(A).$
Addition (Union) rule: $P(A\cup B)=P(A)+P(B)-P(A\cap B).$
If A and B are independent, then $P(A\cap B)=P(A)P(B).$
Relationship between independence and conditional probability: if A and B are independent, then $P(A\mid B)=P(A)\;\text{and}\;P(B\mid A)=P(B).$
Example intuition: the probability of having a dog or a cat (A or B) involves considering overlap (A and B) if both conditions can occur simultaneously.

Example setup used in teaching:
- A contingency table with 2 attributes (e.g., gender: Male/Female; smoking status: Smoker/Non-smoker).
- Total individuals: e.g., 250 employees.
- Known margin: number who smoke cigarettes: 130.
- Therefore, non-smokers: 120.
The point: with some margins, you can fill the rest of a 2x2 table and then convert counts to proportions.
Important caution: the transcript indicates a question about the probability that a randomly chosen person is a nonsmoker conditional on gender (e.g., female). To compute this, you typically need the count of female nonsmokers (or total females plus one of the margins).
Probability of a nonsmoker given female would be: $P(\text{Non-smoker}\mid \text{Female})=\dfrac{P(\text{Female} \cap \text{Non-smoker})}{P(\text{Female})}.$ If you do not know the joint count, you cannot compute it without an independence assumption or additional data.
Independence assumption (if justified): if gender and smoking status are independent, then $P(\text{Non-smoker}\mid \text{Female})=P(\text{Non-smoker})=\dfrac{120}{250}=0.48.$
Without independence, you would need the actual joint counts to compute the conditional probability.
The process students should follow:
- Fill margins with known totals.
- Use the relationships a+b=F, c+d=M, a+c=130, b+d=120 (where a=Female & Smoker, b=Female & Non-smoker, c=Male & Smoker, d=Male & Non-smoker).
- Compute the desired probability (e.g., nonsmoker given female) as b/F.

The transcript mentions attrition percentages: e.g., 13.9% leave in the first year and 7% in the second year.
A note on interpretation: summing yearly percentages directly can lead to values that exceed 100% if each year’s percentage is measured as a portion of the initial cohort and treated independently.
Correct cumulative interpretation: if p1 is the probability of leaving in year 1 and p2 in year 2, the probability of leaving by the end of year 2 is:
- $P(\text{leave by year 2}) = p1 + p2 - p1 p2 = 1 - (1-p1)(1-p2).$
In the transcript, the statement that the ten-year total equals 250% appears inconsistent with standard probability; ensure to interpret percentages as probabilities (0 to 1) or as properly scaled rates (e.g., annual attrition rates with a cumulative formula).

The rule for union of two events general form: $P(A\cup B)=P(A)+P(B)-P(A\cap B).$
Example: if you own a dog (A) or a cat (B): the probability of having either a dog or a cat is given by the sum of their individual probabilities minus the overlap where you have both.
If A and B are independent, an alternative form: $P(A\cup B)=P(A)+P(B)-P(A)P(B).$
This is a practical way to connect independence to the union probability.

Distinguish simple events from events:
- Simple event: a single outcome.
- Event: a set of outcomes.
Master the core formulas:
- Complement: $P(A^c)=1-P(A)$
- Intersection: $P(A\cap B)$
- Union: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$
- Conditional: $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$
- Independence: if A and B are independent, $P(A\cap B)=P(A)P(B)$ and hence $P(A\cup B)=P(A)+P(B)-P(A)P(B)$ .
- For a finite, fair die, if E is any event with k outcomes: $P(E)=\dfrac{k}{6}$ ; in general, probabilities of simple events must sum to 1: $\sum_{e\in S} P(e)=1\,.$
Practice applying these ideas to (a) simple experiments like coin flips or die rolls, and (b) contingency tables with margins to compute conditional probabilities.
Always check whether an independence assumption is justified before using P(A|B)=P(A) or P(A|B)=P(B); if not justified, rely on the joint counts or additional data.

This material connects basic probability theory to data interpretation (contingency tables, marginals, and conditional probabilities).
It underpins decision making under uncertainty, risk assessment, and the interpretation of percentages in real-world data (e.g., attrition, demographics).
Ethical and practical implications: assumptions about independence can significantly change conclusions; ensure transparency about assumptions when reporting probabilities.