Probabilities sum to 1 — and the complement rule
If you list every possible outcome of an experiment without overlap, their probabilities add to 1. From this fall some powerful shortcuts.
The sum-to-1 rule
For mutually exclusive events A₁, A₂, …, Aₙ that partition the sample space (cover every outcome, no overlaps):
P(A₁) + P(A₂) + … + P(Aₙ) = 1.
This lets you find the probability of one outcome from the others.
Worked example: a biased spinner has 4 colours. P(red) = 0.3, P(blue) = 0.25, P(green) = 0.2. Find P(yellow).
- 0.3 + 0.25 + 0.2 + P(yellow) = 1.
- P(yellow) = 0.25.
The complement rule
The complement of A, written A' (or "not A"), is everything that isn't A. Since A and A' are exhaustive and mutually exclusive:
P(A) + P(A') = 1, hence P(A') = 1 − P(A).
The complement rule is invaluable when computing P("at least one") because the complement is "none".
Worked example: a fair coin is tossed 3 times. Find P(at least one head).
- P(no heads) = (1/2)³ = 1/8.
- P(at least one head) = 1 − 1/8 = 7/8.
Recognising mutually exclusive events
Events are mutually exclusive if they cannot both happen at the same trial.
Examples:
- Rolling a 6 and rolling a 2 (on the same single roll).
- Drawing a heart and drawing a club (single card).
Non-examples:
- Rolling an even number and rolling a multiple of 3 (could be 6).
- Drawing a heart and drawing a face card (could be the queen of hearts).
For mutually exclusive events: P(A or B) = P(A) + P(B) (the addition rule). For non-mutually-exclusive events the formula has a correction: P(A ∪ B) = PA + PB − P(A ∩ B). (Higher tier — see P9.)
✦Worked example— Worked example combining both rules
A bag contains red, blue and green balls. P(red) = 2/5 and P(blue) = 1/3. Find P(green).
- P(green) = 1 − P(red) − P(blue) = 1 − 2/5 − 1/3.
- Common denominator 15: 1 − 6/15 − 5/15 = 4/15.
Practical use — surveys
A survey: 80% of people drink tea OR coffee. 35% drink only tea. 30% drink only coffee. Find P(both tea and coffee).
- P(only tea) + P(only coffee) + P(both) = P(tea or coffee) = 0.8.
- 0.35 + 0.30 + P(both) = 0.80 → P(both) = 0.15.
⚠Common mistakes— Common mistakes (examiner traps)
- Adding probabilities of non-mutually-exclusive events without subtracting the overlap.
- Listing outcomes that overlap so the sum exceeds 1.
- Forgetting an outcome so the listed probabilities sum to less than 1 (gives a "missing" probability — sometimes intentional in exam questions).
- Computing P("at least one") the long way when the complement makes it easy.
- Mixing decimals and fractions without a common denominator — convert before adding/subtracting.
➜Try this— Quick check
A bag contains red, blue, green and yellow balls only. P(red) = 0.4, P(blue) = 1/4, P(green) = 0.15. Find P(yellow).
- 0.4 + 0.25 + 0.15 + P(yellow) = 1.
- 0.8 + P(yellow) = 1 → P(yellow) = 0.2.
AI-generated · claude-opus-4-7 · v3-deep-probability