Sec 4 E-Math Probability notes

Probability measures chance on a scale from 0 (impossible) to 1 (certain). For equally likely outcomes, P(event) = favourable outcomes ÷ total outcomes. A probability may be written as a fraction, decimal or percentage — 1/4 = 0.25 = 25%. Every probability satisfies 0 ≤ P(event) ≤ 1, and the probabilities of all possible outcomes together sum to exactly 1.

Words map onto numbers: "impossible" is 0, "certain" is 1, "even chance" is 1/2, "unlikely" sits between 0 and 1/2, "likely" between 1/2 and 1. Exam parts asking you to estimate a probability from a described chance want exactly this mapping — quote a sensible fraction or decimal and say why.

The sample space is the list of all possible outcomes. Listing must be systematic so no outcome is missed or counted twice: fix the first stage and vary the second, then move on. Two coins give HH, HT, TH, TT — four equally likely outcomes, so P(two heads) = 1/4. The classic slip is merging HT and TH into one outcome, which wrongly suggests 1/3.

Examiners reward a *visible* structure: write outcomes in rows or a table, not scattered. If outcomes are not equally likely (say a biased spinner), you cannot count them — probabilities must come from given values or a tree diagram instead.

A possibility diagram is a grid whose rows are the first-stage outcomes and columns the second; each cell holds the combined result. Two ordinary dice give a 6 × 6 grid of 36 equally likely cells, and any "two dice" probability becomes a cell count over 36. The sums along diagonals are constant, which is why 7 is the most likely total (6 of 36 cells) and 2 or 12 the least (1 each).

The same grid works for two spinners, coin-plus-dice mixes or any two independent stages with a countable result written in the cell. If the question asks for "a total of at least 9", shade the matching cells first — shading prevents the classic off-by-one at the boundary.

A tree diagram lays out each stage as a fork of branches labelled with probabilities. Two reading rules do all the work: multiply along a path (the path is a chain of ANDs) and add across distinct paths (the paths are alternatives — OR). The branches from any one fork must sum to 1 — that is the diagram’s built-in check.

With replacement, every stage repeats the same probabilities. Bag with 5 red and 3 blue: P(RR) = 5/8 × 5/8 = 25/64; P(exactly one red) = 2 × (5/8 × 3/8) = 30/64 = 15/32; P(at least one red) = 1 − P(BB) = 1 − 9/64 = 55/64. Without replacement, the second fork shifts: P(RR) = 5/8 × 4/7 = 20/56 = 5/14, because after a red leaves, 4 reds remain among 7.

Three-stage trees work identically — three levels of forks, paths of three multiplied fractions. Keep fractions unrounded until the very end: a path of 4/10 × 6/9 × 5/8 = 1/6 dies if you round each branch to 2 d.p. first.

Mutually exclusive events cannot happen together: on one die, "prime" and "multiple of 3" overlap at 3, but "odd" and "even" do not. For mutually exclusive events, P(A or B) = P(A) + P(B) — no subtraction needed, because the overlap is empty. If events merely overlap, count the overlap once: P(A or B) = P(A) + P(B) − P(A and B).

Independent events do not influence each other: P(A and B) = P(A) × P(B). Dice are independent; draws without replacement are not. To *test* independence numerically, check whether P(A) × P(B) equals the given P(A and B): with P(A) = 1/3, P(B) = 1/4 and P(A and B) = 1/12, the product 1/12 matches — independent. The same product rule powers every tree-diagram path.