In Grade 7 you listed sample spaces with lists, tables and trees. Probability rules are built on that idea, plus the language of sets: an event is just a set of outcomes, and "or," "and" and "not" are ways of combining sets.
Key idea
The sample space $S$ is the set of all possible outcomes. An event is a subset of $S$. Complement $A^{\prime}$ (not A): outcomes not in A. Union $A \cup B$ (A or B): outcomes in A, in B, or in both. Intersection $A \cap B$ (A and B): outcomes in both. When outcomes are equally likely, $$P(A) = \dfrac{\text{number of outcomes in } A}{\text{number of outcomes in } S}.$$
Why it works
Probability measures how much of the sample space an event covers. If every outcome is equally likely, each outcome carries the same share of probability, so counting outcomes gives the probability. Thinking of events as regions (a Venn diagram) makes "or" and "and" visible: "or" is everything covered by either region, "and" is only the overlap.
Worked example: One die
Roll a fair die. Let A = "even" and B = "greater than 3." Find $P(A \cup B)$, $P(A \cap B)$ and $P(A^{\prime})$.
- $S = \{1, 2, 3, 4, 5, 6\}$, six equally likely outcomes. $A = \{2, 4, 6\}$ and $B = \{4, 5, 6\}$.
- $A \cup B = \{2, 4, 5, 6\}$: 4 outcomes, so $P = \dfrac{4}{6} = \dfrac{2}{3}$. (4 and 6 are in both, but each is listed once.)
- $A \cap B = \{4, 6\}$: $P = \dfrac{2}{6} = \dfrac{1}{3}$.
- $A^{\prime} = \{1, 3, 5\}$: $P = \dfrac{3}{6} = \dfrac{1}{2}$.
Answer: $\frac{2}{3}$, $\frac{1}{3}$ and $\frac{1}{2}$.
Worked example: One card
Draw one card from a standard 52-card deck. What is the probability it is a heart or a face card (jack, queen, king)?
- Hearts: 13 cards. Face cards: 12 cards (3 in each of 4 suits).
- The jack, queen and king of hearts are in both groups, so they were counted twice.
- Cards in "heart or face card": $13 + 12 - 3 = 22$.
- $P = \dfrac{22}{52} = \dfrac{11}{26}$.
Answer: $\frac{11}{26}$
Common mistake
Reading "A or B" as "exactly one of A and B." In probability, "or" is inclusive: outcomes in both events count too. For the die, $A \cup B$ includes 4 and 6. Check with a Venn diagram: shade both circles completely, overlap included.