Revise Probability, Venn diagrams and tree diagrams for IB Mathematics: Analysis and Approaches (4.3)

An experiment's sample space $U$ is the set of all possible outcomes; an event is any collection of those outcomes. When the outcomes are equally likely, the probability of an event $A$ is the fraction of the sample space it occupies, $P(A) = \dfrac{n(A)}{n(U)}$, and every probability lies between 0 and 1. The complement $A'$ is the event that $A$ does not occur, and because $A$ and $A'$ together fill the whole sample space, $P(A') = 1 - P(A)$. That single relationship is often the quickest route to an answer, especially for 'at least one' questions where computing the opposite ('none') takes just one line.

To find the probability that at least one of two events happens — the union $A \cup B$ — you cannot simply add $P(A)$ and $P(B)$, because any outcome living in both events would be counted twice. The addition rule corrects for this by subtracting the overlap once. When two events cannot happen together they are mutually exclusive, their overlap is empty, and the rule collapses to a plain sum.

$\text{Addition rule: } P(A\cup B) = P(A) + P(B) - P(A\cap B)$ $\text{Mutually exclusive } (P(A\cap B)=0): \; P(A\cup B) = P(A) + P(B)$

Conditional probability measures how one event's probability changes once we know another has occurred. We write $P(A|B)$ for 'the probability of $A$ given $B$', and the formula shrinks the sample space down to $B$. Independence is the special case where that knowledge changes nothing: knowing $B$ leaves $P(A)$ exactly as it was. Do not confuse independence with mutual exclusivity — they are opposite ideas. Mutually exclusive events with non-zero probability are as dependent as events can be, because one occurring forces the other's probability down to zero.

$| \text{Conditional probability: } P(A | B) = \frac{P(A\cap B)}{P(B)} |$ | --- | --- | $| \text{Independence test: } P(A\cap B) = P(A)\,P(B) \;\; (\text{equivalently } P(A | B) = P(A)) |$

When an experiment happens in stages — draw a ball, then another; test a part, then a second — a tree diagram maps every path. Each set of branches is one stage, the probabilities on the branches out of any point sum to 1, and you read the diagram with two moves: multiply along a path for one complete sequence, and add the paths for an event that can happen several ways. The critical distinction is replacement. With replacement, the second-stage branches carry the same fractions as the first. Without replacement, the total falls by one and so does the count of whatever you removed, so the second-stage fractions genuinely change — and that is exactly where marks are won or lost.

In Paper 2 the marks are analytic: each is tied to a specific line of working. A method mark (M) rewards the correct approach; an accuracy mark (A) rewards the correct value and depends on the method being right first. Follow-through (FT) means a value carried correctly from an earlier slip can still earn later marks, 'ignore subsequent working' (ISW) means a correct answer is not un-done by extra scribble, and equivalent forms — $\frac{15}{28}$, $0.536$, $53.6\%$ — are all accepted. But this protection only exists if the method is written down. Study how each mark below is pinned to one line, especially the two mixed paths a tree diagram must add.

These tools recur throughout the statistics unit and beyond. Conditional probability and independence underpin the probability distributions that follow, where 'independent trials' is precisely the condition that lets you multiply. The habit of listing events, choosing the right diagram, and showing every path is the same discipline that earns method marks across all of Paper 2. Master the routine — name the events, pick Venn or tree, adjust for replacement, multiply along and add across, and write each line down — and the probability questions that follow become variations on a method you already own.