Prove that
(i) $P(A) = P(A \cap B) + P(A \cap \bar B)$
(ii) $P(A \cup B) = P(A \cap B) + P(A \cap \bar B) + P(\bar A \cap B)$
Prove that
(i) $P(A) = P(A \cap B) + P(A \cap \bar B)$
(ii) $P(A \cup B) = P(A \cap B) + P(A \cap \bar B) + P(\bar A \cap B)$
Official Solution
(i)
$\therefore$ ${\rm{RHS}} = P(A \cap B) + P(A \cap \bar B)$
$= P(A) \cdot P(B) + P(A) \cdot P(\bar B)$
$= P(A)[P(B) + P(\bar B)]$
$= P(A)[P(B) + 1 - P(B)]$
$= P(A) = LHS$
(ii)
$\therefore$ RHS $= P(A) \cdot P(B) + P(A) \cdot P(\bar B) + P(\bar A) \cdot P(B)$
$= P(A) \cdot P(B) + P(A) \cdot [1 - P(B)] + [1 - P(A)]P(B)$
$= P(A) \cdot P(B) + P(A) - P(A) \cdot P(B) + P(B) - P(A) \cdot P(B)$
$= P(A) + P(B) - P(A) \cdot P(B)$
$= P(A) + P(B) - P(A \cap B)$
$= P(A \cup B) = LHS$
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