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 )+ $ $P ( A \cap B \cap C )$

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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 distribution property of intersection of sets over the 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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The probability that at least one of $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$, then $P(A') + P(B') = $