If $A, B, C$ are three events associated with a random experiment,prove that $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)$.
Consider $E = B \cup C$ so that
$P(A \cup B \cup C) = P(A \cup E)$
$= P(A) + P(E) - P(A \cap E)$ ...... $(1)$
Now
$P(E) = P(B \cup C)$
$= P(B) + P(C) - P(B \cap C)$ ......... $(2)$
Also $A \cap E = A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ [using the distributive property of intersection over union].
Thus,$P(A \cap E) = P(A \cap B) + P(A \cap C) - P[(A \cap B) \cap (A \cap C)]$
$= P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)$ ......... $(3)$
Using $(2)$ and $(3)$ in $(1)$,we get
$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(B \cap C) - P(A \cap B) - P(A \cap C) + P(A \cap B \cap C)$.
$P(A \cup B \cup C) = P(A \cup E)$
$= P(A) + P(E) - P(A \cap E)$ ...... $(1)$
Now
$P(E) = P(B \cup C)$
$= P(B) + P(C) - P(B \cap C)$ ......... $(2)$
Also $A \cap E = A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$ [using the distributive property of intersection over union].
Thus,$P(A \cap E) = P(A \cap B) + P(A \cap C) - P[(A \cap B) \cap (A \cap C)]$
$= P(A \cap B) + P(A \cap C) - P(A \cap B \cap C)$ ......... $(3)$
Using $(2)$ and $(3)$ in $(1)$,we get
$P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(B \cap C) - P(A \cap B) - P(A \cap C) + P(A \cap B \cap C)$.
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