# State and prove addition theorem on probability.

Last updated date: 24th Mar 2023

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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)

\[A \cap B = 0\]

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)

\[A \cap B = 0\]

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