For two events A and B it is given that P(A) = 0.42, P(B) = 0.48 and P(A ∩ B) = 0.16. Find P(A′ ∩ B′), the probability that neither event happens.

By De Morgan's law, A′ ∩ B′ is the complement of A ∪ B. First P(A ∪ B) = 0.42 + 0.48 − 0.16 = 0.74, so P(A′ ∩ B′) = 1 − 0.74 = 0.26. Answering 0.74 stops one step early at P(A ∪ B); 0.10 comes from adding 0.42 and 0.48 to 0.90 and subtracting that from 1 without ever restoring the double-counted overlap; and 0.16 is the overlap itself.