Let A, B and C be any three events. Show that

(i) $\displaystyle P(A) = P(B) $if and only if $\displaystyle P(A \cap B^{c}) = P(A^{c} \cap B)$

(ii) Given $\displaystyle P(A) = 0.5$ and $\displaystyle P(A \cup (B^{c} \cap C^{c})^{c})= 0.8$, determine $\displaystyle P(A^{c} \cap (B \cup C))$

Show that for any two events :

(i) If $\displaystyle A \subset B$, then $\displaystyle P(A^{c} \cap B) = P(B) - P(A)$

(ii) If $\displaystyle A \subset B$, then $\displaystyle P(A) \leq P(B)$

(iii) Why is it incorrect to assume that for some events A and B, $\displaystyle P(A) = 0.4, P(B) = 0.3$ and $\displaystyle P (A \cap B) = 0.35$?