The Addition Rule
Every question with a fully worked solution.
- The Addition Rule - Video - Addition Rule of Probability - Explained Watch
Theory
Two events are mutually exclusive if they cannot happen at the same time. Then \(P(A\text{ or }B)=P(A)+P(B)\). If they can happen together, subtract the overlap: \(P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)\). This Year 10 maths page revises the addition rule and "and" / "or" from Year 9, ready for conditional probability: dice and cards, the rule forwards and backwards, "exactly one", "neither" and counting multiples.
Events \(A\) and \(B\) are mutually exclusive if they have no outcomes in common, so they cannot both happen. Drawing a diamond and drawing a black card are mutually exclusive. On a Venn diagram the circles do not overlap, and \(P(A\text{ and }B)=0\).
Events are non-mutually exclusive if at least one outcome belongs to both. Drawing a king and drawing a club share the king of clubs.
In probability, "\(A\) or \(B\)" normally means the inclusive or: \(A\), or \(B\), or both (\(A\cup B\)). The exclusive or, "exactly one of \(A\) and \(B\)", leaves out the outcomes in both.
Adding \(P(A)\) and \(P(B)\) counts the overlap twice, which is why it is subtracted once. Mutually exclusive is not the same as complementary: complementary events are mutually exclusive and together cover every outcome, so their probabilities add to \(1\).
For mutually exclusive events:
| Wanted | Formula |
|---|---|
| \(A\) or \(B\) (inclusive) | \(P(A)+P(B)-P(A\cap B)\) |
| exactly one (exclusive or) | \(P(A\cup B)-P(A\cap B)\) |
| \(A\) but not \(B\) | \(P(A)-P(A\cap B)\) |
| neither | \(1-P(A\cup B)\) |
| both, from the others | \(P(A)+P(B)-P(A\cup B)\) |
How to find P(A or B)
- List or describe the outcomes in each event.
- Check for overlap: is there any outcome in both? If not, the events are mutually exclusive.
- Add the probabilities, and subtract the probability of the overlap if there is one.
- Check that the answer is between \(0\) and \(1\) and no bigger than \(P(A)+P(B)\).
Other question types
- Explain whether events are mutually exclusive: name a shared outcome to show they are not, or explain why no outcome can be in both.
- Find the overlap from counts: if everyone is in at least one group, both \(=n(A)+n(B)-\) total. If some are in neither, subtract them from the total first.
- Unknown probabilities: substitute expressions such as \(P(A)=x\), \(P(B)=3x\) into the addition rule and solve.
- Multiples up to \(N\): divide and round down, and use the LCM for the overlap (Example 4).
- Three events: check each pair separately. Some pairs may be mutually exclusive while others are not.
(i) \(P(\text{a sum of } 9 \text{ or a double})\)
(ii) \(P(\text{a sum of } 8 \text{ or a double})\)
(i) Sum of \(9\): \((3,6), (4,5), (5,4), (6,3)\), which is \(4\) outcomes. Doubles: \(6\) outcomes. A double has an even sum, so no outcome is in both. The events are mutually exclusive.
| \(P\) | \(=\) | \(\dfrac{4}{36}+\dfrac{6}{36}=\boldsymbol{\dfrac{5}{18}}\) |
(ii) Sum of \(8\): \(5\) outcomes, and \((4,4)\) is also a double. The events are not mutually exclusive.
| \(P\) | \(=\) | \(\dfrac{5}{36}+\dfrac{6}{36}-\dfrac{1}{36}\) |
| \(\) | \(=\) | \(\dfrac{10}{36}=\boldsymbol{\dfrac{5}{18}}\) |
The answers happen to match, but the working is different: (ii) must remove the double-counted \((4,4)\).
(i) \(P(\text{a king or a club})\)
(ii) \(P(\text{a queen or an ace})\)
(i) There are \(4\) kings and \(13\) clubs, and the king of clubs is both.
| \(P(K\cup C)\) | \(=\) | \(\dfrac{4}{52}+\dfrac{13}{52}-\dfrac{1}{52}\) |
| \(\) | \(=\) | \(\dfrac{16}{52}=\boldsymbol{\dfrac{4}{13}}\) |
(ii) No card is both a queen and an ace, so the events are mutually exclusive.
| \(P(Q\cup A)\) | \(=\) | \(\dfrac{4}{52}+\dfrac{4}{52}=\boldsymbol{\dfrac{2}{13}}\) |
(i) Find \(P(A\text{ and }B)\). Are \(A\) and \(B\) mutually exclusive?
(ii) Find the probability that exactly one of \(A\) and \(B\) happens.
(iii) Find \(P(\text{neither})\).
(i) Rearrange the addition rule.
| \(0.82\) | \(=\) | \(0.62+0.3-P(A\text{ and }B)\) |
| \(P(A\text{ and }B)\) | \(=\) | \(0.92-0.82=\boldsymbol{0.1}\) |
\(P(A\text{ and }B)\neq 0\), so \(A\) and \(B\) are not mutually exclusive.
(ii) "Exactly one" is \(A\) or \(B\) without the overlap: \(0.82-0.1=\boldsymbol{0.72}\).
(iii) \(P(\text{neither})=1-0.82=\boldsymbol{0.18}\).
Check with the four regions: \(A\) only \(0.52\), both \(0.1\), \(B\) only \(0.2\), neither \(0.18\). Total \(1\).
Count each set by dividing and rounding down.
| \(\text{of } 6\) | \(:\) | \(500\div 6=83.3\ldots \to 83\) |
| \(\text{of } 15\) | \(:\) | \(500\div 15=33.3\ldots \to 33\) |
Numbers in both are multiples of the LCM of \(6\) and \(15\), which is \(30\) (not \(90\)).
| \(\text{of } 30\) | \(:\) | \(500\div 30=16.6\ldots \to 16\) |
| \(n(\text{win})\) | \(=\) | \(83+33-16=100\) |
| \(P(\text{win})\) | \(=\) | \(\dfrac{100}{500}=\boldsymbol{\dfrac{1}{5}}\) |
Common pitfalls
Frequently asked questions
What does mutually exclusive mean in probability?
Two events are mutually exclusive if they cannot happen at the same time, such as rolling a \(2\) and rolling a \(5\) on one die. The probability that both happen is \(0\).
What is the addition rule for probability?
\(P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)\). For mutually exclusive events the last term is zero, so the probabilities simply add.
Why do you subtract P(A and B)?
Outcomes in both events are included in \(P(A)\) and again in \(P(B)\). Subtracting \(P(A\text{ and }B)\) once means each outcome is counted exactly once.
What is the difference between inclusive or and exclusive or?
Inclusive or means \(A\), \(B\) or both, and is the usual meaning in probability. Exclusive or means exactly one of \(A\) and \(B\), so the outcomes in both are left out.
Are complementary events mutually exclusive?
Yes. An event and its complement cannot both happen, and together they cover every outcome, so their probabilities add to \(1\). But mutually exclusive events need not be complementary.