'What's the chance of at least one six in four rolls?' Listing the ways is a nightmare. Flip the question: what's the chance of NO sixes? Easy. Subtract from 1. Done.
The complement of an event A is 'A does not happen', written A′ or Aᶜ. Always: P(A) + P(A′) = 1, so P(A) = 1 − P(A′).
'At least one' problems, reliability, the birthday paradox, quick sanity checks — the complement trick turns hard counting into easy subtraction.
P(rain) = 0.3. What's P(no rain), as a percentage?
Whenever 'at least one' appears, try computing P(none) instead — it's usually a single product.
Roll a die 3 times. P(at least one 6)?
Flip a coin 10 times. P(at least one head)?
P(no heads) = (1/2)¹⁰ = 1/1024. So P(at least one head) = 1 − 1/1024 = 1023/1024 ≈ 99.9%.
'At least one' is NOT 'exactly one'. Don't compute P(exactly one) when the question says 'at least one' — use the complement of 'none' instead.
Spot the keyword 'at least one' → immediately think 'complement of none'. It collapses a messy sum into one easy product subtracted from 1.
- P(A) = 1 − P(not A) — always.
- For 'at least one' problems, compute P(none) and subtract.
- The complement is the single most useful shortcut in basic probability.