Q.A die is thrown twice. Let $A$ be the event, 'first die shows 5' and $B$ be the event, 'second die shows 5'. Find $P(A\cup B)$.

Working directly from the three axioms of probability, four theorems give the tools used to combine probabilities of related events.

Theorem 12.3 (impossible event). $P(\varnothing)=0$ -- since $S=S\cup\varnothing$ with $S,\varnothing$ mutually exclusive, axiom $[P_2]$ gives $P(S)=P(S)+P(\varnothing)$, so $P(\varnothing)=0$.

Theorem 12.4 (complement rule). $P(\bar A)=1-P(A)$ -- since $A\cup\bar A=S$ with $A,\bar A$ mutually exclusive, $P(A)+P(\bar A)=P(S)=1$.

Theorem 12.5 (only-$A$ rule). $P(A\cap\bar B)=P(A)-P(A\cap B)$ -- the part of $A$ that misses $B$ has probability 'all of $A$' minus 'the overlap with $B$', since $(A\cap\bar B)\cup(A\cap B)=A$ with the two pieces mutually exclusive.

Theorem 12.6 -- the Addition Theorem. For any two events $A,B$:

$P(A\cup B)=P(A)+P(B)-P(A\cap B).$ …