Complementary Events

The complement rule

$P(A) = 1 - P(\text{not } A) = 1 - P(A^c)$

This is useful when computing $P(A)$ directly is hard, but computing $P(\text{not } A)$ is easy.

Step-by-step: P(at least one head in 3 flips)

Direct method (hard): List all outcomes with at least one head:

HHH, HHT, HTH, HTT, THH, THT, TTH — that's 7 out of 8. So $P = 7/8$.

Complement method (easy):

$P(\text{at least 1 head}) = 1 - P(\text{no heads}) = 1 - P(\text{all tails})$

$P(\text{all tails}) = (1/2)^3 = 1/8$

$P(\text{at least 1 head}) = 1 - 1/8 = 7/8 = 87.5\%$

When to use the complement

Use it when "at least one" or "at least two" appears in the problem. It's almost always easier to compute the complement (none, zero) and subtract from 1.

More examples

• P(at least one 6 in 4 rolls) = $1 - (5/6)^4 = 1 - 0.482 = 0.518$
• P(at least 2 people share a birthday in a room of 23) ≈ $1 - 0.493 = 0.507$ (the birthday paradox!)

Try it in the visualization

The tree diagram shows all 8 outcomes of 3 flips. The complement (all tails) is highlighted separately, and $1 - P(\text{complement})$ gives the answer more simply.