Probability of A or B

Adding P(A) and P(B) counts the outcomes in A ∩ B twice. Three fair coins: A = {HHT, HTH, THH} and B = {HTH, THH, HHH} each have P = 3/8, but A ∪ B = {HHT, HTH, THH, HHH} has P = 4/8 = 1/2, not 6/8. The shared HTH and THH were counted twice.

Addition rule for any two events: P(A ∪ B) = P(A) − P(A ∩ B) + P(B). Take A, remove its overlap with B, then add all of B.

Proof through the axioms: A ∪ B = A ∪ (B − A) and B = (A ∩ B) ∪ (B − A), each a union of disjoint pieces. So P(A ∪ B) = P(A) + P(B − A) and P(B) = P(A ∩ B) + P(B − A); subtract to get the rule. A Venn diagram shows the same thing.

For mutually exclusive events A ∩ B = φ, so P(A ∩ B) = 0 and the rule reduces to axiom (iii): P(A ∪ B) = P(A) + P(B).

Two-dice stall (made up): 'doubles' and 'sum at least 10' each have 6 of 36 outcomes and share (5,5) and (6,6). So P(doubles or sum ≥ 10) = 6/36 + 6/36 − 2/36 = 10/36 = 5/18.

'Sum 7' and 'doubles' share nothing, so P(sum 7 or doubles) = 6/36 + 6/36 = 1/3. 'First die 6' and 'sum 7' share only (6,1), giving 6/36 + 6/36 − 1/36 = 11/36.

Three events (NCERT's Example 11): P(A ∪ B ∪ C) = S₁ − S₂ + S₃, where S₁ = P(A) + P(B) + P(C) adds the singles, S₂ = P(A ∩ B) + P(B ∩ C) + P(C ∩ A) removes the pairs and S₃ = P(A ∩ B ∩ C) puts the triple back. It follows by writing B ∪ C = E and using the two-event rule twice.