When every outcome is equally likely, a probability is an act of counting done twice. $P = \dfrac{f}{n}$ — read aloud P equals f over n. Letter by letter: $f$ is the number of favourable outcomes, the ones that count as a success; $n$ is the number of all equally likely outcomes, winners included; and $P$ is the resulting chance, a naked ratio with no unit attached. Whichever of the three the question leaves blank is the one you solve for — the relation runs in all three directions.

The phrase equally likely is load-bearing, not decoration. It is what lets you count faces instead of weighing dice. A loaded die has six faces and no business in this formula at all.

Its constant companion is the complement rule: $P(A^{c}) = 1 - P(A)$, read P of A-complement equals one minus P of A. The little superscript c means not A — every outcome where A fails to happen. Every trial ends in A or in not-A, never both and never neither, so the two shares fill exactly one whole. That single subtraction is the most profitable move in this chapter, and by the last lesson it will be doing the heavy lifting.