The Addition Rule
Understand the addition rule for probability in Queensland Year 11 Mathematical Methods (QCAA). It gives the probability of the union of two events, subtracting their intersection so the shared outcomes are not double-counted.
You will learn to apply and rearrange the rule, recognise mutually exclusive events where the intersection is empty, use the complement, and read probabilities from Venn diagrams and two-way tables — a reliable technique across QCAA probability.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 1 Topic 5), the addition rule finds the probability of a union: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). This page shows how to apply it, rearrange it for a missing probability, handle mutually exclusive events, and use the complement rule \(P(A')=1-P(A)\) with Venn diagrams and two-way tables.
The addition rule gives the probability that \(A\) or \(B\) (or both) happens. Because \(P(A)\) and \(P(B)\) both include the overlap \(P(A\cap B)\), we subtract it once: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).
Two events are mutually exclusive when they cannot both occur, so \(P(A\cap B)=0\) and the rule simplifies to \(P(A\cup B)=P(A)+P(B)\).
The complement rule \(P(A')=1-P(A)\) and De Morgan's law \((A\cup B)'=A'\cap B'\) let you find the probability of neither event: \(P(\text{neither})=1-P(A\cup B)\).
| \(B\) | \(B'\) | Total | |
|---|---|---|---|
| \(A\) | 0.25 | 0.35 | 0.60 |
| \(A'\) | 0.20 | 0.20 | 0.40 |
| Total | 0.45 | 0.55 | 1 |
The addition rule:
For mutually exclusive events (\(P(A\cap B)=0\)):
Complement and neither:
How to apply the addition rule
- Identify \(P(A)\), \(P(B)\) and \(P(A\cap B)\) from the words, a Venn diagram or a table.
- Check for mutual exclusivity: if \(A\) and \(B\) cannot both occur, set \(P(A\cap B)=0\).
- Substitute into \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\), or rearrange for the unknown.
- Use the complement \(P(\text{neither})=1-P(A\cup B)\) if the question asks for neither.
Substitute into the addition rule:
| \(P(A\cup B)\) | \(=\) | \(P(A)+P(B)-P(A\cap B)\) |
| \(=\) | \(0.5+0.3-0.1\) |
Simplify:
| \(=\) | \(0.7\) |
\(P(A\cup B)=0.7\).
Rearrange the addition rule for the overlap:
| \(P(A\cap B)\) | \(=\) | \(P(A)+P(B)-P(A\cup B)\) |
| \(=\) | \(0.6+0.45-0.8\) |
Simplify:
| \(=\) | \(0.25\) |
\(P(A\cap B)=0.25\).
A card cannot be both, so the events are mutually exclusive: \(P(K\cap Q)=0\).
| \(P(K)\) | \(=\) | \(\dfrac{4}{52}\) |
| \(P(Q)\) | \(=\) | \(\dfrac{4}{52}\) |
Add the probabilities (overlap is \(0\)):
| \(P(K\cup Q)\) | \(=\) | \(\dfrac{4}{52}+\dfrac{4}{52}\) |
| \(=\) | \(\dfrac{8}{52}\) | |
| \(=\) | \(\dfrac{2}{13}\) |
\(P(\text{king or queen})=\dfrac{2}{13}\).
First find the union:
| \(P(A\cup B)\) | \(=\) | \(0.7+0.5-0.35\) |
| \(=\) | \(0.85\) |
Neither \(=(A\cup B)'\) by De Morgan:
| \(P(A'\cap B')\) | \(=\) | \(1-P(A\cup B)\) |
| \(=\) | \(1-0.85\) | |
| \(=\) | \(0.15\) |
\(P(A'\cap B')=0.15\).
Common pitfalls
Frequently asked questions
What is the addition rule in probability?
\(P(A\cup B)=P(A)+P(B)-P(A\cap B)\): the probability of \(A\) or \(B\) is the sum of the two probabilities minus the overlap.
Why do you subtract P(A and B)?
Because the overlap \(P(A\cap B)\) is counted in both \(P(A)\) and \(P(B)\); subtracting it once removes the double count.
What happens for mutually exclusive events?
They cannot both occur, so \(P(A\cap B)=0\) and the rule becomes \(P(A\cup B)=P(A)+P(B)\).
How do you find a missing probability?
Rearrange the rule, for example \(P(A\cap B)=P(A)+P(B)-P(A\cup B)\).
How do you find the probability that neither event occurs?
Use the complement of the union: \(P(A'\cap B')=1-P(A\cup B)\).
What is the difference between ‘neither’ and ‘not both’?
‘Neither’ is \(1-P(A\cup B)\); ‘not both’ is \(1-P(A\cap B)\).