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AnswersState and prove the Addition theorem on probability.
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Hint: Since this is a question related to probability, so there is occurrence of events say two events and this theorem involves addition on them. Probability of an event is the number of ways events can occur divided by the total number of possible outcomes.
Complete step-by-step answer:
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
\[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\].
Hence, the above given addition theorem of probability is proved.
Note:
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.
Complete step-by-step answer:
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
\[P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)\].
Hence, the above given addition theorem of probability is proved.
Note:
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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