Probability Diagrams — Venn & Tree DiagramsEdexcel IGCSE Maths: Revision notes

A Venn diagram shows how events overlap. In IGCSE you rarely draw one — you work from a description or table of numbers placed in each region.

For two events $A$ and $B$ in a sample space of size $n(\varepsilon)$:

• The intersection $A \cap B$ is the region where both events happen.
• The union $A \cup B$ is everything in $A$, $B$, or both.
• $A'$ (the complement) is everything not in $A$.

Key formula: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$

This subtracts the overlap once so it isn't double-counted.

When you're given raw frequencies (e.g. "12 students study French, 9 study Spanish, 4 study both, 30 students in total"), always fill in the intersection region first, then work outwards, then find the region outside both circles by subtracting from the total.

Once the diagram (or table) is filled with frequencies, probability is just that region ÷ total.

• $P(A) = \dfrac{n(A)}{n(\varepsilon)}$
• $P(A \cap B) = \dfrac{n(A \cap B)}{n(\varepsilon)}$
• $P(A \mid B)$ (conditional probability) $= \dfrac{n(A \cap B)}{n(B)}$ — restrict the total to only the region $B$, since we are told $B$ has already happened.

Conditional probability questions on Venn diagrams are a common higher-tier trap: students divide by $n(\varepsilon)$ instead of $n(B)$.

A tree diagram shows outcomes of two or more events happening in sequence. Each branch is labelled with a probability, and branches from the same point must sum to 1.

Structure:

1. First event: branches for each outcome, probabilities sum to 1.
2. Second event: a new set of branches drawn from the end of each first-event branch.
3. Multiply along a branch (path) to get the probability of that combined outcome.
4. Add separate paths that lead to the same overall result.

$P(A \text{ then } B) = P(A) \times P(B \text{ given the branch taken})$

Always check: probabilities on branches from the same node add to 1, and probabilities of all end-of-tree outcomes add to 1 — a fast way to self-check your diagram.

Independent events (e.g. with replacement): the probability doesn't change between draws. Second-stage branches are identical regardless of which first branch you followed.

$P(A \text{ and } B) = P(A) \times P(B)$

Dependent events (e.g. without replacement): the probability changes because the sample space shrinks or composition changes after the first event. The denominator (and sometimes numerator) of the second-stage branch depends on what happened first.

Example — a bag has 5 red, 3 blue counters (8 total), no replacement:

• $P(\text{Red then Red}) = \dfrac{5}{8} \times \dfrac{4}{7}$ (one fewer red, one fewer total)
• $P(\text{Red then Blue}) = \dfrac{5}{8} \times \dfrac{3}{7}$ (total reduced, but blue count unchanged)

Always ask: "does the first outcome change what's left?" If yes → dependent → adjust the second-stage fractions.

"At least one" (e.g. at least one head, at least one defective item) is usually fastest via the complement:

$P(\text{at least one}) = 1 - P(\text{none})$

Find $P(\text{none of the event happening})$ by multiplying the "not happening" branches along the tree, then subtract from 1. This avoids adding up many separate paths (e.g. exactly one, exactly two, etc.).

Combined tree + Venn questions often ask you to find a probability from the tree, then use it to complete a two-way table or Venn diagram — read carefully which representation the question wants your final answer in.