14.4 Addition Theorems and Inequalities

Let $E$: Anil qualifies, $F$: Ashima qualifies. Given: $P(E) = 0.05, P(F) = 0.10, P(E \cap F) = 0.02$.

(a) Both not: $P(E' \cap F')$. By De Morgan's Law, $E' \cap F' = (E \cup F)'$. First find $P(E \cup F) = P(E) + P(F) - P(E \cap F) = 0.05 + 0.10 - 0.02 = 0.13$. $P(E' \cap F') = 1 - P(E \cup F) = 1 - 0.13 = 0.87$.

(b) At least one not: $P(E' \cup F')$. By De Morgan's, $E' \cup F' = (E \cap F)'$. $P((E \cap F)') = 1 - P(E \cap F) = 1 - 0.02 = 0.98$.

(c) Only one: $P(E \text{ only}) + P(F \text{ only})$. $= P(E - F) + P(F - E)$ $= [P(E) - P(E \cap F)] + [P(F) - P(E \cap F)]$ $= [0.05 - 0.02] + [0.10 - 0.02]$ $= 0.03 + 0.08 = 0.11$.