When outcomes are equally likely:

Example: a bag has 50 marbles, 20 of which are blue: $P(\text{blue}) = \tfrac{20}{50} = 0.4$. Probabilities can be given as a fraction, decimal or percentage, and always lie between 0 (impossible) and 1 (certain).

The complement rule, $P(A') = 1 - P(A)$, lets you find "not" probabilities instantly: if $P(\text{blue})=\tfrac{3}{10}$, then $P(\text{not blue}) = 1 - \tfrac{3}{10} = \tfrac{7}{10}$. More generally, if a spinner's outcomes have probabilities that must sum to 1 and two are unknown, set up and solve a simple equation for them — a common Grade 2 skill.

In a two-way table, the grand total is the denominator for an ordinary probability; but when a question says "given that...", the denominator changes to just the row or column total named in the condition — not the grand total.

Example: of 17 people over 25, 12 drive to work. Given a person is over 25, $P(\text{car}) = \tfrac{12}{17}$.

In a Venn diagram, fill in the intersection first when building one from a word problem, then work outward — this prevents double-counting. A probability is then (elements in the region) ÷ (total in the universal set).

Multiply along the branches for a specific sequence: $P(\text{heads and }6) = \tfrac{1}{2}\times\tfrac{1}{6} = \tfrac{1}{12}$. For "at least one" questions, it's almost always faster to find $P(\text{none})$ and subtract from 1, rather than adding up every successful branch separately.

The probability a school bus is late is $0.2$ on any given day, independently from day to day. Find the probability it is late on at least one of Monday and Tuesday.

1. $P(\text{not late}) = 1 - 0.2 = 0.8$.
2. $P(\text{not late both days}) = 0.8 \times 0.8 = 0.64$.
3. $P(\text{at least one late}) = 1 - 0.64 = 0.36$.

Answer: $0.36$. Calculating four separate "late/not late" branches and adding the three that contain at least one "late" would reach the same answer, but the complement method above is faster and less error-prone.

Theoretical probability is calculated by reasoning alone (a fair coin landing heads is $0.5$); relative frequency is an estimate from real trials, $\dfrac{\text{times event occurred}}{\text{total trials}}$. As the number of trials grows, relative frequency tends to settle closer to the true theoretical probability — this is why a question might ask you to compare a small experiment's results with the "expected" value from theory.