Question

# State and prove addition theorem on probability.

Hint- Since this is probability, so there is occurrence of events, say 2 events and this theorem involves addition of them.

Addition theorem of probability$-$ If $A$ and $B$ are any two events then the probability of happening of at least one of the events is defined as,
$P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$
Proof:-
From set theory, we know that,
$n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$
Dividing the above equation $n(s)$both sides we have
$\begin{array}{*{20}{l}} {\dfrac{{n\left( {A \cup B} \right)}}{{n\left( S \right)}} = \dfrac{{n\left( A \right)}}{{n\left( S \right)}} + \dfrac{{n\left( B \right)}}{{n\left( S \right)}} - \dfrac{{n\left( {A \cap B} \right)}}{{n\left( S \right)}}} \\ {P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)} \\ {\left( {\;\because \;P\left( X \right) = \dfrac{{n\left( X \right)}}{{n\left( S \right)}}} \right)} \end{array}$
Hence proved.

NOTE: If two events $A$ and $B$ are mutually exclusive, then
$A \cap B = \phi$(null set)
$A \cap B = 0$