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Statement of the addition theorem on probability:

If A and B are any two events of a random experiment and P is a probability function then the probability of happening of at least one of the events is defined as $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$.

Now, we have to prove the Addition theorem of probability.

Given: A and B are any two events of a random experiment.

To prove: $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$.

Proof:

From the set theory we know that

$n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right)$.

Suppose $n\left( S \right)$ denote the total number of the possible events of random experiment and then dividing both left hand side and right hand side of the of the above equation we get,

\[\dfrac{{P\left( {A \cup B} \right)}}{{n\left( S \right)}} = \dfrac{{P\left( A \right)}}{{n\left( S \right)}} + \dfrac{{P\left( B \right)}}{{n\left( S \right)}} - \dfrac{{P\left( {A \cap B} \right)}}{{n\left( S \right)}}\]

Now, we know that a formula for probability $P\left( x \right) = \dfrac{{n\left( x \right)}}{{n\left( S \right)}}$. By applying this we can write

\[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\].

Special case: If two events A and B are mutually exclusive, then $A \cap B$ is a null set. That is $n\left( {A \cap B} \right) = 0$ So, the probability of happening of at least one of the events is equal to the probability of happening of event A and the probability of happening of event B. Mathematically it is written as $P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right)$. \[\] Two events A and B are independent events if the equation \[P\left( {A \cap B} \right) = P\left( A \right) \times P\left( B \right)\] holds true.

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