Probability of not A

A and A′ never occur together and one of them always occurs: A ∩ A′ = φ and A ∪ A′ = S. Axioms (ii) and (iii) give P(A) + P(A′) = 1.

Complement rule: P(not A) = P(A′) = 1 − P(A).

Ten cards numbered 1 to 10, equally likely, with A = {2, 4, 6, 8}: P(A) = 4/10 = 2/5, and A′ = {1, 3, 5, 7, 9, 10} gives P(A′) = 6/10 = 3/5 = 1 − 2/5.

NCERT's Example 5, one card from 52: diamond 13/52 = 1/4; not an ace 1 − 4/52 = 12/13; black 26/52 = 1/2; not a diamond 3/4; not black 1/2.

NCERT's Example 6, a bag of 9 discs (4 red, 3 blue, 2 yellow): red 4/9, yellow 2/9, blue 3/9 = 1/3, not blue 2/3, red or blue 4/9 + 3/9 = 7/9 (red and blue exclude each other).

NCERT's Example 7: P(Anil qualifies) = 0.05, P(Ashima qualifies) = 0.10, both 0.02. P(at least one qualifies) = 0.05 + 0.10 − 0.02 = 0.13, so P(neither qualifies) = 1 − 0.13 = 0.87 (by De Morgan, E′ ∩ F′ = (E ∪ F)′). P(at least one fails) = 1 − 0.02 = 0.98. P(only one qualifies) = (0.05 − 0.02) + (0.10 − 0.02) = 0.03 + 0.08 = 0.11.

Use the complement whenever 'at least one' has many cases: P(at least one 6 on two dice) = 1 − 25/36 = 11/36.